VBA UserForm in Excel--mismatched rows on spreadsheet data entry - forms

I have a userform created via VBA that is supposed to populate rows on an Excel sheet. It does, but they're one off:
Name | Race | Agency
Black
Joe Asian B
White
Joanne C
Joe's races are black and Asian, at agency B; Joanne's is white, and she's at agency C. Somehow, the entries are staggered.
Name is a textbox, race and agency are listboxes, with race as a multiselect and agency as a single select.
Here's my code:
Private Sub CommandButton1_Click()
Dim j As Long
Dim i As Integer
With ListBox2
ReDim arr(.ListCount - 1)
For i = 0 To .ListCount - 1
If .Selected(i) = True Then
.Selected(i) = False
arr(j) = .List(i)
j = j + i
End If
Next i
End With
ReDim Preserve arr(j)
With ActiveSheet
.Range("B" & .Rows.Count).End(xlUp). _
Offset(1, 0).Resize(j + 1, 1).Value = Application.Transpose(arr)
End With
i = 1
While ThisWorkbook.Worksheets("Sheet1").Range("B" & i).Value <> ""
i = i + 1
Wend
ThisWorkbook.Worksheets("Sheet1").Range("A" & i).Value = TextBox1.Value
ThisWorkbook.Worksheets("Sheet1").Range("C" & i).Value = ListBox1.Value
End Sub
Private Sub CommandButton2_Click()
Dim ctl As MSForms.Control
For Each ctl In Me.Controls
Select Case TypeName(ctl)
Case "TextBox"
ctl.Text = ""
Case "CheckBox", "OptionButton", "ToggleButton"
ctl.Value = False
Case "ComboBox", "ListBox"
ctl.ListIndex = -1
End Select
Next ctl
End Sub
Sub UserForm_Initialize()
ListBox1.List = Array("A", "B", "C")
With ListBox2
.Clear
.AddItem "White"
.AddItem "Black"
.AddItem "Asian"
.AddItem "Am Indian/Al Native"
.AddItem "Native Hawaiian/Pac Islander"
.AddItem "Other"
End With
End Sub
I would love any ideas ya'll have on how to fix that! Ideally, it would come out in one of the following ways:
Name | Race | Agency
Joe Black B
Asian B
Joanne White C
or
Name | Race | Agency
Joe Black, Asian B
Joanne White C
or
Name | Race | Agency
Joe Black B
Joe Asian B
Joanne White C
(I prefer the second, but any would work.)

If I understand the code appropriately, the refactored CommandButton1_Click procedure below should produce the preferred result for you.
Private Sub CommandButton1_Click()
Dim j As Long
Dim i As Integer
'load races into array
With ListBox2
ReDim arr(.ListCount - 1)
For i = 0 To .ListCount - 1
If .Selected(i) = True Then
.Selected(i) = False
arr(j) = .List(i)
j = j + i
End If
Next i
End With
ReDim Preserve arr(j)
'build "," separated string of races
For i = LBound(arr) To UBound(arr)
Dim sRace As String
sRace = sRace & "," & arr(i)
Next
sRace = Mid(sRace, 2) 'to remove first comma
'place info on next available line in sheet.
With ThisWorkbook.Worksheets("Sheet1")
Dim lRow As Long
lRow = .Range("A" & .Rows.Count).End(xlUp).Offset(1).Row
.Range("A" & lRow).Value = TextBox1.Value
.Range("B" & lRow).Value = sRace
.Range("C" & lRow).Value = ListBox1.Value
End With
End Sub

Related

Can you Print the wavy lines generated by Spell check in writer?

As per google group, this macro can be used to print mis-spelled words in MS office.
https://groups.google.com/g/microsoft.public.word.spelling.grammar/c/OiFYPkLAbeU
Is there similar option in libre-office writer?
The following Subroutine replicates what the code in the Google group does. It is more verbose than the MS version but that is to be expected with LibreOffice / OpenOffice. It only does the spellchecker lines and not the green grammar checker ones, which is also the case with the MS version in the Google group.
Sub UnderlineMisspelledWords
' From OOME Listing 315 Page 336
GlobalScope.BasicLibraries.loadLibrary( "Tools" )
Dim sLocale As String
sLocale = GetRegistryKeyContent("org.openoffice.Setup/L10N", FALSE).getByName("ooLocale")
' ooLocale appears to return a string that consists of the language and country
' seperated by a dash, e.g. en-GB
Dim nDash As Integer
nDash = InStr(sLocale, "-")
Dim aLocale As New com.sun.star.lang.Locale
aLocale.Language = Left(sLocale, nDash - 1)
aLocale.Country = Right(sLocale, Len(sLocale) -nDash )
Dim oSpeller As Variant
oSpeller = createUnoService("com.sun.star.linguistic2.SpellChecker")
Dim emptyArgs() as new com.sun.star.beans.PropertyValue
Dim oCursor As Object
oCursor = ThisComponent.getText.createTextCursor()
oCursor.gotoStart(False)
oCursor.collapseToStart()
Dim s as String, bTest As Boolean
Do
oCursor.gotoEndOfWord(True)
s = oCursor.getString()
bTest = oSpeller.isValid(s, aLocale, emptyArgs())
If Not bTest Then
With oCursor
.CharUnderlineHasColor = True
.CharUnderlineColor = RGB(255, 0,0)
.CharUnderline = com.sun.star.awt.FontUnderline.WAVE
' Possible alternatives include SMALLWAVE, DOUBLEWAVE and BOLDWAVE
End With
End If
Loop While oCursor.gotoNextWord(False)
End Sub
This will change the actual formatting of the font to have a red wavy underline, which will print out like any other formatting. If any of the misspelled words in the document already have some sort of underlining then that will be lost.
You will probably want to remove the underlining after you have printed it. The following Sub removes underlining only where its style exactly matches that of the line added by the first routine.
Sub RemoveUnderlining
Dim oCursor As Object
oCursor = ThisComponent.getText.createTextCursor()
oCursor.gotoStart(False)
oCursor.collapseToStart()
Dim s as String, bTest As Boolean
Do
oCursor.gotoEndOfWord(True)
Dim bTest1 As Boolean
bTest1 = False
If oCursor.CharUnderlineHasColor = True Then
bTest1 = True
End If
Dim bTest2 As Boolean
bTest2 = False
If oCursor.CharUnderlineColor = RGB(255, 0,0) Then
bTest2 = True
End If
Dim bTest3 As Boolean
bTest3 = False
If oCursor.CharUnderline = com.sun.star.awt.FontUnderline.WAVE Then
bTest3 = True
End If
If bTest1 And bTest2 And bTest3 Then
With oCursor
.CharUnderlineHasColor = False
.CharUnderline = com.sun.star.awt.FontUnderline.NONE
End With
End If
Loop While oCursor.gotoNextWord(False)
End Sub
This will not restore any original underlining that was replaced by red wavy ones. Other ways of removing the wavy lines that would restore these are:
Pressing undo (Ctrl Z) but you will need to do that once for every word in your document, which could be a bit of a pain.
Running the subroutine UnderlineMisspelledWords on a temporary copy of the document and then discarding it after printing.
I hope this is what you were looking for.
In response to your above comment, it is straightforward to modify the above subroutine to do that instead of drawing wavy lines. The code below opens a new Writer document and writes into it a list of the misspelled words together with the alternatives that the spellchecker suggests:
Sub ListMisSpelledWords
' From OOME Listing 315 Page 336
GlobalScope.BasicLibraries.loadLibrary( "Tools" )
Dim sLocale As String
sLocale = GetRegistryKeyContent("org.openoffice.Setup/L10N", FALSE).getByName("ooLocale")
' ooLocale appears to return a string that consists of the language and country
' seperated by a dash, e.g. en-GB
Dim nDash As Integer
nDash = InStr(sLocale, "-")
Dim aLocale As New com.sun.star.lang.Locale
aLocale.Language = Left(sLocale, nDash - 1)
aLocale.Country = Right(sLocale, Len(sLocale) -nDash )
Dim oSource As Object
oSource = ThisComponent
Dim oSourceCursor As Object
oSourceCursor = oSource.getText.createTextCursor()
oSourceCursor.gotoStart(False)
oSourceCursor.collapseToStart()
Dim oDestination As Object
oDestination = StarDesktop.loadComponentFromURL( "private:factory/swriter", "_blank", 0, Array() )
Dim oDestinationText as Object
oDestinationText = oDestination.getText()
Dim oDestinationCursor As Object
oDestinationCursor = oDestinationText.createTextCursor()
Dim oSpeller As Object
oSpeller = createUnoService("com.sun.star.linguistic2.SpellChecker")
Dim oSpellAlternatives As Object, emptyArgs() as new com.sun.star.beans.PropertyValue
Dim sMistake as String, oSpell As Object, sAlternatives() as String, bTest As Boolean, s As String, i as Integer
Do
oSourceCursor.gotoEndOfWord(True)
sMistake = oSourceCursor.getString()
bTest = oSpeller.isValid(sMistake, aLocale, emptyArgs())
If Not bTest Then
oSpell = oSpeller.spell(sMistake, aLocale, emptyArgs())
sAlternatives = oSpell.getAlternatives()
s = ""
for i = LBound(sAlternatives) To Ubound(sAlternatives) - 1
s = s & sAlternatives(i) & ", "
Next i
s = s & sAlternatives(Ubound(sAlternatives))
oDestinationText.insertString(oDestinationCursor, sMistake & ": " & s & Chr(13), False)
End If
Loop While oSourceCursor.gotoNextWord(False)
End Sub
I don't know about the dictionaries but, in answer to your previous comment, if you paste the following code below Loop While and above End Sub it will result in the text in the newly opened Writer document being sorted without duplicates. It's not very elegant but it works on the text I've tried it on.
oDestinationCursor.gotoStart(False)
oDestinationCursor.gotoEnd(True)
Dim oSortDescriptor As Object
oSortDescriptor = oDestinationCursor.createSortDescriptor()
oDestinationCursor.sort(oSortDescriptor)
Dim sParagraphToBeChecked As String
Dim sThisWord As String
sThisWord = ""
Dim sPreviousWord As String
sPreviousWord = ""
oDestinationCursor.gotoStart(False)
oDestinationCursor.collapseToStart()
Dim k As Integer
Do
oDestinationCursor.gotoEndOfParagraph(True)
sParagraphToBeChecked = oDestinationCursor.getString()
k = InStr(sParagraphToBeChecked, ":")
If k <> 0 Then
sThisWord = Left(sParagraphToBeChecked, k-1)
End If
If StrComp(sThisWord, sPreviousWord, 0) = 0 Then
oDestinationCursor.setString("")
End If
sPreviousWord = sThisWord
Loop While oDestinationCursor.gotoNextParagraph(False)
Dim oReplaceDescriptor As Object
oReplaceDescriptor = oDestination.createReplaceDescriptor()
oReplaceDescriptor.setPropertyValue("SearchRegularExpression", TRUE)
oReplaceDescriptor.setSearchString("^$")
oReplaceDescriptor.setReplaceString("")
oDestination.replaceAll(oReplaceDescriptor)
It seems I didn't spot that because the text I tested it on contained only words that were either correct or had more than zero alternatives. I managed to replicate the error by putting in a word consisting of random characters for which the spellchecker was unable to suggest any alternatives. If no alternatives are found the function .getAlternatives() returns an array of size -1 so the error can be avoided by testing for this condition before the array is used. Below is a modified version of the first Do loop in the subroutine with such a condition added. If you replace the existing loop with that it should eliminate the error.
Do
oSourceCursor.gotoEndOfWord(True)
sMistake = oSourceCursor.getString()
bTest = oSpeller.isValid(sMistake, aLocale, emptyArgs())
If Not bTest Then
oSpell = oSpeller.spell(sMistake, aLocale, emptyArgs())
sAlternatives = oSpell.getAlternatives()
s = ""
If Ubound(sAlternatives) >= 0 Then
for i = LBound(sAlternatives) To Ubound(sAlternatives) - 1
s = s & sAlternatives(i) & ", "
Next i
s = s & sAlternatives(Ubound(sAlternatives))
End If
oDestinationText.insertString(oDestinationCursor, sMistake & ": " & s & Chr(13), False)
End If
Loop While oSourceCursor.gotoNextWord(False)
On re-reading the whole subroutine I think it would improve its readability if the variable sMistake were renamed to something like sWordToBeChecked, as the string this variable contains isn't always misspelled. This would of course need to be changed everywhere in the routine and not just in the above snippet.
Below is a modified version that uses the dispatcher as suggested by Jim K in his answer go to end of word is not always followed. I have written it out in its entirety because the changes are more extensive than just adding or replacing a block. In particular, it is necessary to get the view cursor before creating the empty destination document, otherwise the routine will spell check that.
Sub ListMisSpelledWords2
' From OOME Listing 315 Page 336
GlobalScope.BasicLibraries.loadLibrary( "Tools" )
Dim sLocale As String
sLocale = GetRegistryKeyContent("org.openoffice.Setup/L10N", FALSE).getByName("ooLocale")
' ooLocale appears to return a string that consists of the language and country
' seperated by a dash, e.g. en-GB
Dim nDash As Integer
nDash = InStr(sLocale, "-")
Dim aLocale As New com.sun.star.lang.Locale
aLocale.Language = Left(sLocale, nDash - 1)
aLocale.Country = Right(sLocale, Len(sLocale) -nDash )
Dim oSourceDocument As Object
oSourceDocument = ThisComponent
Dim nWordCount as Integer
nWordCount = oSourceDocument.WordCount
Dim oFrame As Object, oViewCursor As Object
With oSourceDocument.getCurrentController
oFrame = .getFrame()
oViewCursor = .getViewCursor()
End With
Dim oDispatcher as Object
oDispatcher = createUnoService("com.sun.star.frame.DispatchHelper")
oDispatcher.executeDispatch(oFrame, ".uno:GoToStartOfDoc", "", 0, Array())
Dim oDestinationDocument As Object
oDestinationDocument = StarDesktop.loadComponentFromURL( "private:factory/swriter", "_blank", 0, Array() )
Dim oDestinationText as Object
oDestinationText = oDestinationDocument.getText()
Dim oDestinationCursor As Object
oDestinationCursor = oDestinationText.createTextCursor()
Dim oSpeller As Object
oSpeller = createUnoService("com.sun.star.linguistic2.SpellChecker")
Dim oSpellAlternatives As Object, emptyArgs() as new com.sun.star.beans.PropertyValue
Dim sMistake as String, oSpell As Object, sAlternatives() as String, bTest As Boolean, s As String, i as Integer
For i = 0 To nWordCount - 1
oDispatcher.executeDispatch(oFrame, ".uno:WordRightSel", "", 0, Array())
sWordToBeChecked = RTrim( oViewCursor.String )
bTest = oSpeller.isValid(sWordToBeChecked, aLocale, emptyArgs())
If Not bTest Then
oSpell = oSpeller.spell(sWordToBeChecked, aLocale, emptyArgs())
sAlternatives = oSpell.getAlternatives()
s = ""
If Ubound(sAlternatives) >= 0 Then
for i = LBound(sAlternatives) To Ubound(sAlternatives) - 1
s = s & sAlternatives(i) & ", "
Next i
s = s & sAlternatives(Ubound(sAlternatives))
End If
oDestinationText.insertString(oDestinationCursor, sWordToBeChecked & ": " & s & Chr(13), False)
End If
oDispatcher.executeDispatch(oFrame, ".uno:GoToPrevWord", "", 0, Array())
oDispatcher.executeDispatch(oFrame, ".uno:GoToNextWord", "", 0, Array())
Next i
oDestinationCursor.gotoStart(False)
oDestinationCursor.gotoEnd(True)
' Sort the paragraphs
Dim oSortDescriptor As Object
oSortDescriptor = oDestinationCursor.createSortDescriptor()
oDestinationCursor.sort(oSortDescriptor)
' Remove duplicates
Dim sParagraphToBeChecked As String, sThisWord As String, sPreviousWord As String
sThisWord = ""
sPreviousWord = ""
oDestinationCursor.gotoStart(False)
oDestinationCursor.collapseToStart()
Dim k As Integer
Do
oDestinationCursor.gotoEndOfParagraph(True)
sParagraphToBeChecked = oDestinationCursor.getString()
k = InStr(sParagraphToBeChecked, ":")
If k <> 0 Then
sThisWord = Left(sParagraphToBeChecked, k-1)
End If
If StrComp(sThisWord, sPreviousWord, 0) = 0 Then
oDestinationCursor.setString("")
End If
sPreviousWord = sThisWord
Loop While oDestinationCursor.gotoNextParagraph(False)
' Remove empty paragraphs
Dim oReplaceDescriptor As Object
oReplaceDescriptor = oDestinationDocument.createReplaceDescriptor()
oReplaceDescriptor.setPropertyValue("SearchRegularExpression", TRUE)
oReplaceDescriptor.setSearchString("^$")
oReplaceDescriptor.setReplaceString("")
oDestinationDocument.replaceAll(oReplaceDescriptor)
End Sub
It looks like the problem is caused by one of the while loops in the function TrimWord, which removes punctuation from before and after a word before it is fed to the spell-check service. If the word is only one character long and if it is a valid punctuation character then the condition at the beginning of the loop is true, so the loop is entered and the counter n is decremented to zero. Then at the beginning of the next traversal of the loop, even though the condition is false anyway, it still asks the Mid function to return the 0th character of the word, which it can't do because the characters are numbered from 1, so it throws an error. Some languages would ignore the error if the truth value of the condition could be unambiguously determined from the other parts of the expression. It looks like BASIC doesn't do that.
The following modified version of the function gets round the problem in a rather inelegant way, but it seems to work:
Function TrimWord(sWord As String) As String
Dim n as Long
n = Len(sWord)
If n > 0 Then
Dim m as Long : m = 1
Dim bTest As Boolean
bTest = m <= n
Do While IsPermissiblePrefix( ASC(Mid(sWord, m, 1) ) ) And bTest
if (m < n) Then
m = m + 1
Else
bTest = False
End If
Loop
bTest = n > 0
Do While IsPermissibleSuffix( ASC(Mid(sWord, n, 1) ) ) And bTest
if (n > 1) Then
n = n - 1
Else
bTest = False
End If
Loop
If n > m Then
TrimWord = Mid(sWord, m, (n + 1) - m)
Else
TrimWord = sWord
End If
Else
TrimWord = ""
End If
End Function
This works for me.
Firstly, in response to your question about the bug, I'm not a maintainer so I can't fix that. However, as the bug concerns moving a text cursor to the start and end of a word it should be possible to get round it by searching for the white-space between words instead. Since the white-space characters are (I think) the same in all languages, any problems recognising certain characters from certain alphabets shouldn't matter. The easiest way to do it would be to first read the entire text of the document into a string but LibreOffice strings have a maximum length of 2^16 = 65536 characters and while this seems like a lot it could easily be too small for a reasonable sized document. The limit can be avoided by navigating through the text one paragraph at a time. According to Andrew Pitonyak (OOME Page 388): "I found gotoNextSentence() and gotoNextWord() to be unreliable, but the paragraph cursor worked well."
The code below is yet another modification of the subroutines in previous answers. This time it gets a string from a paragraph and splits it up into words by finding the white-space between the words. It then spell checks the words as before. The subroutine depends on some other functions that are listed below it. These allow you to specify which characters to designate as word separators (i.e. white-space) and which characters to ignore if they are found at the beginning or end of a word. This is necessary so that, for example, the quotes surrounding a quoted word are not counted as part of the word, which would lead to it being recognised as a spelling mistake even if the word inside the quotes is correctly spelled.
I am not familiar with non-latin alphabets and I don't have an appropriate dictionary installed, but I pasted the words from your question go to end of word is not always followed, namely testी, भारत and इंडिया and they all appeared unmodified in the output document.
On the question of looking up synonyms, as each misspelled word has multiple suggestions, and each of those will have multiple synonyms, the output could rapidly become very large and confusing. It may be better for your user to look them up individually if they want to use a different word.
Sub ListMisSpelledWords3
' From OOME Listing 315 Page 336
GlobalScope.BasicLibraries.loadLibrary( "Tools" )
Dim sLocale As String
sLocale = GetRegistryKeyContent("org.openoffice.Setup/L10N", FALSE).getByName("ooLocale")
' ooLocale appears to return a string that consists of the language and country
' seperated by a dash, e.g. en-GB
Dim nDash As Integer
nDash = InStr(sLocale, "-")
Dim aLocale As New com.sun.star.lang.Locale
aLocale.Language = Left( sLocale, nDash - 1)
aLocale.Country = Right( sLocale, Len(sLocale) - nDash )
Dim oSource As Object
oSource = ThisComponent
Dim oSourceCursor As Object
oSourceCursor = oSource.getText.createTextCursor()
oSourceCursor.gotoStart(False)
oSourceCursor.collapseToStart()
Dim oDestination As Object
oDestination = StarDesktop.loadComponentFromURL( "private:factory/swriter", "_blank", 0, Array() )
Dim oDestinationText as Object
oDestinationText = oDestination.getText()
Dim oDestinationCursor As Object
oDestinationCursor = oDestinationText.createTextCursor()
Dim oSpeller As Object
oSpeller = createUnoService("com.sun.star.linguistic2.SpellChecker")
Dim oSpellAlternatives As Object, emptyArgs() as new com.sun.star.beans.PropertyValue
Dim sWordToCheck as String, oSpell As Object, sAlternatives() as String, bTest As Boolean
Dim s As String, i as Integer, j As Integer, sParagraph As String, nWordStart As Integer, nWordEnd As Integer
Dim nChar As Integer
Do
oSourceCursor.gotoEndOfParagraph(True)
sParagraph = oSourceCursor.getString() & " " 'It is necessary to add a space to the end of
'the string otherwise the last word of the paragraph is not recognised.
nWordStart = 1
nWordEnd = 1
For i = 1 to Len(sParagraph)
nChar = ASC(Mid(sParagraph, i, 1))
If IsWordSeparator(nChar) Then '1
If nWordEnd > nWordStart Then '2
sWordToCheck = TrimWord( Mid(sParagraph, nWordStart, nWordEnd - nWordStart) )
bTest = oSpeller.isValid(sWordToCheck, aLocale, emptyArgs())
If Not bTest Then '3
oSpell = oSpeller.spell(sWordToCheck, aLocale, emptyArgs())
sAlternatives = oSpell.getAlternatives()
s = ""
If Ubound(sAlternatives) >= 0 Then '4
for j = LBound(sAlternatives) To Ubound(sAlternatives) - 1
s = s & sAlternatives(j) & ", "
Next j
s = s & sAlternatives(Ubound(sAlternatives))
End If '4
oDestinationText.insertString(oDestinationCursor, sWordToCheck & " : " & s & Chr(13), False)
End If '3
End If '2
nWordEnd = nWordEnd + 1
nWordStart = nWordEnd
Else
nWordEnd = nWordEnd + 1
End If '1
Next i
Loop While oSourceCursor.gotoNextParagraph(False)
oDestinationCursor.gotoStart(False)
oDestinationCursor.gotoEnd(True)
Dim oSortDescriptor As Object
oSortDescriptor = oDestinationCursor.createSortDescriptor()
oDestinationCursor.sort(oSortDescriptor)
Dim sParagraphToBeChecked As String
Dim sThisWord As String
sThisWord = ""
Dim sPreviousWord As String
sPreviousWord = ""
oDestinationCursor.gotoStart(False)
oDestinationCursor.collapseToStart()
Dim k As Integer
Do
oDestinationCursor.gotoEndOfParagraph(True)
sParagraphToBeChecked = oDestinationCursor.getString()
k = InStr(sParagraphToBeChecked, ":")
If k <> 0 Then
sThisWord = Left(sParagraphToBeChecked, k-1)
End If
If StrComp(sThisWord, sPreviousWord, 0) = 0 Then
oDestinationCursor.setString("")
End If
sPreviousWord = sThisWord
Loop While oDestinationCursor.gotoNextParagraph(False)
Dim oReplaceDescriptor As Object
oReplaceDescriptor = oDestination.createReplaceDescriptor()
oReplaceDescriptor.setPropertyValue("SearchRegularExpression", TRUE)
oReplaceDescriptor.setSearchString("^$")
oReplaceDescriptor.setReplaceString("")
oDestination.replaceAll(oReplaceDescriptor)
End Sub
'----------------------------------------------------------------------------
' From OOME Listing 360.
Function IsWordSeparator(iChar As Integer) As Boolean
' Horizontal tab \t 9
' New line \n 10
' Carriage return \r 13
' Space 32
' Non-breaking space 160
Select Case iChar
Case 9, 10, 13, 32, 160
IsWordSeparator = True
Case Else
IsWordSeparator = False
End Select
End Function
'-------------------------------------
' Characters to be trimmed off beginning of word before spell checking
Function IsPermissiblePrefix(iChar As Integer) As Boolean
' Symmetric double quote " 34
' Left parenthesis ( 40
' Left square bracket [ 91
' Back-tick ` 96
' Left curly bracket { 123
' Left double angle quotation marks « 171
' Left single quotation mark ‘ 8216
' Left single reversed 9 quotation mark ‛ 8219
' Left double quotation mark “ 8220
' Left double reversed 9 quotation mark ‟ 8223
Select Case iChar
Case 34, 40, 91, 96, 123, 171, 8216, 8219, 8220, 8223
IsPermissiblePrefix = True
Case Else
IsPermissiblePrefix = False
End Select
End Function
'-------------------------------------
' Characters to be trimmed off end of word before spell checking
Function IsPermissibleSuffix(iChar As Integer) As Boolean
' Exclamation mark ! 33
' Symmetric double quote " 34
' Apostrophe ' 39
' Right parenthesis ) 41
' Comma , 44
' Full stop . 46
' Colon : 58
' Semicolon ; 59
' Question mark ? 63
' Right square bracket ] 93
' Right curly bracket } 125
' Right double angle quotation marks » 187
' Right single quotation mark ‘ 8217
' Right double quotation mark “ 8221
Select Case iChar
Case 33, 34, 39, 41, 44, 46, 58, 59, 63, 93, 125, 187, 8217, 8221
IsPermissibleSuffix = True
Case Else
IsPermissibleSuffix = False
End Select
End Function
'-------------------------------------
Function TrimWord( sWord As String) As String
Dim n as Integer
n = Len(sWord)
If n > 0 Then
Dim m as Integer : m = 1
Do While IsPermissiblePrefix( ASC(Mid(sWord, m, 1) ) ) And m <= n
m = m + 1
Loop
Do While IsPermissibleSuffix( ASC(Mid(sWord, n, 1) ) ) And n >= 1
n = n - 1
Loop
If n > m Then
TrimWord = Mid(sWord, m, (n + 1) - m)
Else
TrimWord = sWord
End If
Else
TrimWord = ""
End If
End Function

custom data validation in a VBA form

I have this form to enter new data to a table.
I would like to warn the user when he is entering an invoice number that already exist. Here is the code I have but its not working:
Private Sub CommandButton1_Click()
Dim L As Long
Dim Code As String
Dim TextBox2 As Long
Dim valFormula As String
valFormula = "=COUNTIFS($F12:$F1702,F1702,$D12:$D1702,D1702)=1"
If MsgBox("Confirm?", vbYesNo, "Confirming new invoice") = vbYes Then
With Worksheets("FACTURE")
L = Sheets("FACTURE").Range("D65535").End(xlUp).Row + 1 'Pour placer le nouvel enregistrement _ la premi_re ligne de tableau non vide
End With
With Me
Range("D" & L).Validation
.Add Type:=xlValidateCustom, _
AlertStyle:=xlValidAlertWarning, _
Formula1:="=COUNTIFS($F12:$F1702,F1702,$D12:$D1702,D1702)=1"
.InputTitle = ""
.ErrorTitle = "Duplicate alert"
.InputMessage = ""
.ErrorMessage = "This invoice number already exist. Continue?"
Range("B" & L).Value = .ComboBox2 & .ComboBox3
Range("C" & L).Value = (Now)
Range("D" & L).Value = .TextBox2
Range("E" & L).Value = .TextBox3
Range("F" & L).Value = .TextBox4
Range("G" & L).Value = .TextBox5
Range("K" & L).Value = .ComboBox1
Range("L" & L).Value = .ComboBox2
Range("M" & L).Value = .ComboBox3
Range("N" & L).Value = .TextBox9
Range("O" & L).Value = .TextBox10
Range("R" & L).Value = .TextBox39
Range("P" & L).Value = .TextBox40
Range("C" & L).Interior.ColorIndex = 0
If .OptionButton1 Then
FormatCell Range("B" & L), xlThemeColorAccent3
ElseIf .OptionButton2 Then
FormatCell Range("B" & L), xlThemeColorAccent1
ElseIf .OptionButton3 Then
FormatCell Range("B" & L), xlThemeColorAccent4
Else
FormatCell Range("B" & L), xlThemeColorAccent2
End If
End With
End If
End Sub
Any advice?
As Comintern suggested, use Find() method of Range object, with code like:
Set f = rngToSerachIn.Find(what:=factureNo, LookIn:=xlValues, lookat:=xlWhole)
where
f is a range variable where to store the range with the searched value
rngToSerachIn is the range where to search the value
factureNo is the value to search for
furthermore it seems to me your invoices will be stored in rows from 12 downwards, so it could be useful to write a generic function to get first empty cell in a given column of a given worksheet ranging from a certain row
Since it'd be a good practice to demand specific tasks to Sub/Function to improve both code readability and maintenance, you could do that for:
getting first empty row after last non empty one starting from a given row in a given column of a given worksheet
validating invoice number
filling worksheet ranges
formatting invoice cell
as follows:
Option Explicit
Private Sub CommandButton1_Click()
Dim L As Long
Dim factureWs As Worksheet
If MsgBox("Confirm?", vbYesNo, "Confirming new invoice") = vbNo Then Exit Sub
Set factureWs = Worksheets("FACTURE") '<--| set the worksheet you want to work with
L = GetLastNonEmptyRow(factureWs, "D", 12) + 1 '<--| get passed worksheet first empty row after last non empty one in column "D" from row 12 (included)
If L > 12 Then If Not CheckDuplicate(Me.TextBox2, factureWs.Range("D12:D" & L - 1)) Then Exit Sub '<--| exit if duplicated non accepted by the user
FillRanges factureWs, L '<--| fill worksheet ranges with userfom controls values
FormatInvoice factureWs.Range("B" & L) '<--| color invoice cell depending on option buttons values
End Sub
Function GetLastNonEmptyRow(ws As Worksheet, colIndex As String, firstRow As Long) As Long
Dim lastRow As Long
With ws
lastRow = .Cells(.Rows.Count, colIndex).End(xlUp).row ' <--| get last non empty row in given column
If lastRow = 1 Then If IsEmpty(.Range(colIndex & 1)) Then lastRow = 0 '<--| handle the case of an empty column
If lastRow < firstRow Then lastRow = firstRow - 1 '<--| handle the case the last non empty row is above the first passed one
End With
GetLastNonEmptyRow = lastRow
End Function
Function CheckDuplicate(factureNo As String, rng As Range) As Boolean
Dim f As Range
Set f = rng.Find(what:=factureNo, LookIn:=xlValues, lookat:=xlWhole)
If Not f Is Nothing Then
CheckDuplicate = MsgBox("This invoice number already exist!" & vbCrLf & vbCrLf & "Continue?", vbExclamation + vbYesNo, "Duplicate alert") = vbYes
Else
CheckDuplicate = True
End If
End Function
Sub FormatInvoice(rng As Range)
Dim thColor As XlThemeColor
With Me
Select Case True
Case .OptionButton1
thColor = xlThemeColorAccent3
Case .OptionButton2
thColor = xlThemeColorAccent1
Case .OptionButton3
thColor = xlThemeColorAccent4
Case Else
thColor = xlThemeColorAccent2
End Select
End With
FormatCell rng, thColor
End Sub
Sub FillRanges(ws As Worksheet, L As Long)
With ws
.Range("C" & L).Value = (Now)
.Range("D" & L).Value = Me.TextBox2
.Range("E" & L).Value = Me.TextBox3
.Range("F" & L).Value = Me.TextBox4
.Range("G" & L).Value = Me.TextBox5
.Range("K" & L).Value = Me.ComboBox1
.Range("L" & L).Value = Me.ComboBox2
.Range("M" & L).Value = Me.ComboBox3
.Range("N" & L).Value = Me.TextBox9
.Range("O" & L).Value = Me.TextBox10
.Range("R" & L).Value = Me.TextBox39
.Range("P" & L).Value = Me.TextBox40
End With
End Sub
you may find it useful and follow this pattern in your subsequent coding

Loop subroutine for every used row using multiple dynamic cell references

Basically what I am trying to do is, sending an email for every used row on the target worksheet, each row has the details of the addresses, subject line, table with values etc.
So I can't seem to get it working, as it only dispatches one email from the first target row (2nd row).
I have tried using a combination of For Each and For i = 1 to LR which aren't working. I suspect it is to do with the cell references.
Here is the code:
Sub TestEmail1()
Application.ScreenUpdating = False
Dim aOutlook As Object
Dim aEmail As Object
Dim rngeAddresses As Range, rngeCell As Range, strRecipients As String
Dim ccAddresses As Range, ccCell As Range, ccRecipients As String
Dim rngeSubject As Range, SubjectCell As Range, SubjectContent As Variant
Dim rngeBody As Range, bodyCell As Range, bodyContent As Variant
Dim Table1 As Range
Dim i As Integer
For Each c In ActiveSheet.UsedRange.Columns("A").Cells
Set rng = ActiveSheet.UsedRange
LRow = rng.Rows.Count
For i = 2 To LRow
Set Table1 = Worksheets(1).Range("K1:R1")
Set Table2 = Worksheets(2).Range("K" & i & ":" & "R" & i)
Set aOutlook = CreateObject("Outlook.Application")
Set aEmail = aOutlook.CreateItem(0)
'set sheet to find address for e-mails as I have several people to
'mail to
Set rngeAddresses = ActiveSheet.Range("B" & i)
For Each rngeCell In rngeAddresses.Cells
strRecipients = strRecipients & ";" & rngeCell.Value
Next
Set ccAddresses = ActiveSheet.Range("C" & i)
For Each ccCell In ccAddresses.Cells
ccRecipients = ccRecipients & ";" & ccCell.Value
Next
Set rngeSubject = ActiveSheet.Range("D" & i)
For Each SubjectCell In rngeSubject.Cells
SubjectContent = SubjectContent & SubjectCell.Value
Next
Set rngeBody = ActiveSheet.Range("E" & i)
For Each bodyCell In rngeBody.Cells
bodyContent = bodyContent & bodyCell.Value
Next
'set Importance
'aEmail.Importance = 2
'Set Subject
aEmail.Subject = rngeSubject
'Set Body for mail
'aEmail.Body = bodyContent
aEmail.HTMLBody = bodyContent & "<br><br><br>" & RangetoHTML_ (Table1)
aEmail.To = strRecipients
aEmail.CC = ccRecipients
aEmail.Send
Exit Sub
Next i
Next c
End Sub
There is an Exit Sub at the end of your inner loop that makes the code exit from the procedure after the first iteration:
Sub TestEmail1()
...
For Each c In ActiveSheet.UsedRange.Columns("A").Cells
...
For i = 2 To LRow
...
Exit Sub
Next i
Next c
End Sub
Remove it and processing should continue as desired.

Email excel data range when target cell changes

This macro works on line 5 ,so i need this macro to work on all lines in one sheet instead of one macro for each line. Row X and email range A:L are copy paste in all lines i.e.( X1 A1:L1 | X2 ,A2:L2 ...)
Dim X5 As Variant
Private Sub Worksheet_Change(ByVal Target As Range)
If Range("X5").Value = 1 And X5 <> 1 Then
ActiveSheet.Range("A5:L5").Select
ActiveWorkbook.EnvelopeVisible = True
With ActiveSheet.MailEnvelope
.Introduction = " send thru macro "
.Item.To = "email#gmail.com"
.Item.Subject = "ALERT"
.Item.Send
End With
End If
X5 = Range("X5").Value
End Sub
Not sure if you got your answer or not so I am attempting to answer this question.
To make it flexible for any row, you can store the row of the current cell in a variable using Target.Row and then simply use that to construct your range.
Also to understand how Worksheet_Change works, you may want to see THIS
Is this what you are trying?
Dim X5 As Variant
Private Sub Worksheet_Change(ByVal Target As Range)
On Error GoTo Whoa
'~~> Check if the chnage happened to multiple cells
If Target.cell.CountLarge > 1 Then Exit Sub
Dim Rw As Long
'~~> Get the row number of the cell that was changed
Rw = Target.Row
If Range("X" & Rw).Value = 1 And X5 <> 1 Then
Application.EnableEvents = False
Range("A" & Rw & ":L" & Rw).Select
ActiveWorkbook.EnvelopeVisible = True
With ActiveSheet.MailEnvelope
.Introduction = " send thru macro "
.Item.To = "email#gmail.com"
.Item.Subject = "ALERT"
.Item.Send
End With
End If
X5 = Range("X" & Rw).Value
Letscontinue:
Application.EnableEvents = True
Exit Sub
Whoa:
MsgBox Err.Description
Resume Letscontinue
End Sub

Restrict what someone types in a textbox

Here's what I want to do and I have a problem.
I want to restrict what an user types in certain textboxes. I want to leave him type only numbers but after 3 numbers to add a ";". (eg. 007;123;003;005;).
The problem is that my textbox Controls are generated through a bunch of code. So I can't or I don't know how to set an action to these controls.
The code I'm using to generate the controls is:
Set cControl = form.Controls("io" & masina).Add(
"Forms.Label.1", "lreper" & l & pagina, True)
With cControl
.Caption = "Reper"
.Width = 35
.Height = 9
.Top = 25 + k
.Left = 5
End With
Any ideas?
Thanks a lot!
You can use the key press event to restrict only numbers and the ";". Along with check conditions.
Private Sub TextBox1_KeyPress(ByVal KeyAscii As MSForms.ReturnInteger)
Select Case KeyAscii
'// Numbers 0-9
Case 48 To 57
If Len(TextBox1.Text) = 3 And Right(TextBox1.Text, 3) Like "###" Then
KeyAscii = 0
GoTo DisplayFormatError
End If
'// Key ;
Case 59
If Len(TextBox1.Text) < 3 Or Not Right(TextBox1.Text, 3) Like "###" Then
KeyAscii = 0
GoTo DisplayFormatError
End If
Case Else
KeyAscii = 0
GoTo DisplayFormatError
End Select
Exit Sub
DisplayFormatError:
MsgBox "Please enter serial number in the format '000;000;000'", vbInformation, "Alert!"
End Sub
A better way would be to use a regular expression instead of the like method.
If you need help adding the events for your controls at runtime have a look at:
Add controls and events to form at runtime
EDIT (REQUEST BY TIAGO)
Dynamic creation of Userform and Textbox with keypress event. Uses modified example of above link. Credit to original author.
Add reference - Under Available References, click "Microsoft Visual Basic for Applications Extensibility" and click OK.
Option Explicit
Sub MakeForm()
Dim TempForm As Object ' VBComponent
Dim FormName As String
Dim NewTextBox As MSForms.TextBox
Dim TextLocation As Integer
Dim TextBoxName As String
'** Additional variable
Dim X As Integer
'Locks Excel spreadsheet and speeds up form processing
Application.VBE.MainWindow.Visible = False
Application.ScreenUpdating = False
'Create the UserForm
Set TempForm = ThisWorkbook.VBProject.VBComponents.Add(vbext_ct_MSForm)
'Set Properties for TempForm
With TempForm
.Properties("Caption") = "Temporary Form"
.Properties("Width") = 200
.Properties("Height") = 100
End With
FormName = TempForm.Name
TextBoxName = "MyTextBox"
'Add a CommandButton
Set NewTextBox = TempForm.Designer.Controls _
.Add("Forms.TextBox.1")
With NewTextBox
.Name = TextBoxName
.Left = 60
.Top = 40
End With
'Add an event-hander sub for the CommandButton
With TempForm.CodeModule
'** Add/change next 5 lines
'This code adds the commands/event handlers to the form
X = .CountOfLines
.InsertLines X + 1, "Private Sub " & TextBoxName & "_KeyPress(ByVal KeyAscii As MSForms.ReturnInteger)"
.InsertLines X + 2, "KeyAscii = KeyPress(" & TextBoxName & ".Text, KeyAscii)"
.InsertLines X + 3, "End Sub"
End With
'Show the form
VBA.UserForms.Add(FormName).Show
'Delete the form
ThisWorkbook.VBProject.VBComponents.Remove VBComponent:=TempForm
End Sub
Public Function KeyPress(ByVal strText As String, ByVal KeyAscii As Integer) As Integer
Select Case KeyAscii
'// Numbers 0-9
Case 48 To 57
If Len(strText) = 3 And Right(strText, 3) Like "###" Then
GoTo DisplayFormatError
End If
'// Key ;
Case 59
If Len(strText) < 3 Or Not Right(strText, 3) Like "###" Then
GoTo DisplayFormatError
End If
Case Else
GoTo DisplayFormatError
End Select
KeyPress = KeyAscii
Exit Function
DisplayFormatError:
KeyPress = 0
MsgBox "Please enter serial number in the format '000;000;000'", vbInformation, "Alert!"
End Function
ANOTHER METHOD (Using an event handler class)
Code in Userform:
Private colEventHandlers As Collection
Private Sub UserForm_Initialize()
'// New collection of events
Set colEventHandlers = New Collection
'// Add dynamic textbox
Set tbxNewTextbox = Me.Controls.Add("Forms.TextBox.1", "MyNewTextbox", True)
With tbxNewTextbox
.Top = 25
.Left = 5
End With
'// Add the event handler
Dim objEventHandler As TextboxEventHandler
Set objEventHandler = New TextboxEventHandler
Set objEventHandler.TextBox = tbxNewTextbox
colEventHandlers.Add objEventHandler
End Sub
And add a class module and rename it too "TextBoxEventHandler", then add the following code:
Private WithEvents tbxWithEvent As MSForms.TextBox
Public Property Set TextBox(ByVal oTextBox As MSForms.TextBox)
Set tbxWithEvent = oTextBox
End Property
Private Sub tbxWithEvent_Change()
End Sub
Private Sub tbxWithEvent_KeyPress(ByVal KeyAscii As MSForms.ReturnInteger)
Select Case KeyAscii
'// Numbers 0-9
Case 48 To 57
If Len(tbxWithEvent.Text) = 3 And Right(tbxWithEvent.Text, 3) Like "###" Then
GoTo DisplayFormatError
End If
'// Key ;
Case 59
If Len(tbxWithEvent.Text) < 3 Or Not Right(tbxWithEvent.Text, 3) Like "###" Then
GoTo DisplayFormatError
End If
Case Else
GoTo DisplayFormatError
End Select
Exit Sub
DisplayFormatError:
KeyAscii = 0
MsgBox "Please enter serial number in the format '000;000;000'", vbInformation, "Alert!"
End Sub
Try Dataannotations / metadata
More here: http://msdn.microsoft.com/en-us/library/ee256141.aspx
AFAIK and if i understood well, there is no way to handle this before user input.
Yet, you can use the TextBox_Exit event to format it afterwards. You can adapt this sample of code.
Although I'd never use dynamic controls unless strictly required, I got puzzled by this question... so I'm thinking of it as a challenge. :-)
Googled around and most answers falls into the same solution, however most of them comes with a 'I couldn't make it work' comment as well, including this one here in SO Assign on-click VBA function to a dynamically created button on Excel Userform.
Here's the code I built... which obviously does not work, otherwise I'd say it could be a solution. The problem on it is that the keypress method it creates dynamically is not called when should be. To test it, just paste the code into a VBA form named 'myForm'.
I kept the TextBox1_KeyPress only for testing purposes, to prove the usability of the text field validator (I'm sorry #Readfidy, your code didn't work for me as expected. I was able to add more than 3 numbers in a row).
In case anyone else is interested in making this code works... I'd be happy to thank ;-)
Option Explicit
Private Sub UserForm_Activate()
Dim sTextBoxName As String
Dim cControl As MSForms.TextBox
Dim sMetaFunction As String
Dim CodeModule
sTextBoxName = "lreper"
Set cControl = myForm.Controls.Add("Forms.TextBox.1", sTextBoxName, True)
With cControl
.Top = 25
.Left = 5
End With
sMetaFunction = "Private Sub " & sTextBoxName & "_KeyPress(ByVal KeyAscii As MSForms.ReturnInteger)" & vbCrLf & _
vbCrLf & _
vbTab & "Set KeyAscii = EvaluateText(myForm.Controls(" & sTextBoxName & "), KeyAscii)" & vbCrLf & _
vbCrLf & _
"End Sub"
Set CodeModule = ActiveWorkbook.VBProject.VBComponents.VBE.ActiveCodePane.CodeModule
CodeModule.InsertLines 60, sMetaFunction
End Sub
Private Sub TextBox1_KeyPress(ByVal KeyAscii As MSForms.ReturnInteger)
Set KeyAscii = EvaluateText(myForm.Controls("TextBox1"), KeyAscii)
End Sub
Private Function EvaluateText(ByRef oTextBox As MSForms.TextBox, ByVal KeyAscii As MSForms.ReturnInteger) As MSForms.ReturnInteger
If ((Len(oTextBox.Text) + 1) / 4 = CInt((Len(oTextBox.Text) + 1) / 4)) Then
If KeyAscii <> 59 Then KeyAscii = 0
Else
If KeyAscii < 48 Or KeyAscii > 57 Then KeyAscii = 0
End If
If KeyAscii = 0 Then
MsgBox "Please enter serial number in the format '000;000;000'", vbInformation, "Alert!"
End If
End Function
USE THIS CODE : NO CLASS NEEDED !
USERFORM CODE
Private Function QNumber(ByRef oTextBox As MSForms.TextBox, ByVal KeyAscii As MSForms.ReturnInteger) As MSForms.ReturnInteger
On Error Resume Next
Select Case KeyAscii
Case 45 '"-"
If InStr(1, oTextBox.Text, "-") > 0 Or oTextBox.SelStart > 0 Then
KeyAscii = 0
End If
Case 46 '"."
If InStr(1, oTextBox.Text, ".") > 0 Then
KeyAscii = 0
End If
Case 48 To 57 '0-9"
Case Else
KeyAscii = 0
End Select
End Function
TEXTBOX CODE
Private Sub TextBox1_KeyPress(ByVal KeyAscii As MSForms.ReturnInteger)
Set KeyAscii = QNumber(Me.Controls("TextBox1"), KeyAscii)
End Sub