How to split word files by the number of characters - ms-word

Could you anybody help me how to split word file by character!
I can't find any way to split word files by the number of characters on the internet!

For example, to split a document into 500-character blocks:
Sub SplitDocument()
Application.ScreenUpdating = False
Dim Rng As Range, i As Long, j As Long
Const Char As Long = 500
With ActiveDocument
' Process each character block
For i = 1 To Int(.Characters.Count / Char)
j = j + 1
' Get the character block
Set Rng = .Range((i - 1) * Char, i * Char)
' Copy the character block
Rng.Copy
Rng.Collapse wdCollapseEnd
Call NewDoc(ActiveDocument, (i - 1) * Char + 1, j)
Next
If Rng.End < .Range.End Then
i = i + 1: j = j + 1
Rng.End = .Range.End
' Copy the range
Rng.Copy
Rng.Collapse wdCollapseEnd
Call NewDoc(ActiveDocument, (i - 1) * Char + 1, j)
End If
End With
Set Rng = Nothing
Application.ScreenUpdating = True
End Sub
Sub NewDoc(DocSrc As Document, i As Long, j As Long)
Dim DocTgt As Document, HdFt As HeaderFooter
' Create the output document
Set DocTgt = Documents.Add(Visible:=False)
With DocTgt
' Paste contents into the output document, preserving the formatting
.Range.PasteAndFormat (wdFormatOriginalFormatting)
' Replicate the headers & footers
For Each HdFt In DocSrc.Sections(DocSrc.Characters(i).Sections(1).Index).Headers
.Sections(1).Headers(HdFt.Index).Range.FormattedText = HdFt.Range.FormattedText
Next
For Each HdFt In DocSrc.Sections(DocSrc.Characters(i).Sections(1).Index).Footers
.Sections(1).Footers(HdFt.Index).Range.FormattedText = HdFt.Range.FormattedText
Next
' Save & close the output document
.SaveAs FileName:=Split(DocSrc.FullName, ".doc")(0) & "_" & j & ".docx", _
FileFormat:=wdFormatXMLDocument, AddToRecentFiles:=False
.Close SaveChanges:=False
End With
Set DocTgt = Nothing: Set DocSrc = Nothing
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

Is there a MS Word wildcard for frequency?

I learning how to use Microsoft Word wildcards and codes to help me in my position as a medical editor. A big part of my job is submitting manuscripts to medical journals for review, and each journal has very specific requirements.
Most of the journals we submit manuscripts to require that medical terms/phrases be abbreviated only if they are used three or more times. For example, the term “Overall Survival” can be abbreviated to OS if the term is referenced at least three times in the text. If the text only mentions “Overall Survival” once or twice, it is preferred that the term remain expanded, and it should not be abbreviated to OS.
We have been using the PerfectIt system, by Intelligent Editing. This Word widget scans for abbreviations that are only used once and will flag them for our review, but does not pick up if an abbreviation is only used twice in the selected text. We are hoping to find some solution (my thought would be some sort of wildcard search or macro) that will be able to detect if an abbreviation is used only one or two times.
I saw this similar post on stackoverflow, but it seemed to do with code. I will need this to be on a company computer that I do not have administrative access to, and furthermore, I know nothing about code. I appreciate any help, guidance, or directions for further research!
Thank you!
Edit: I could use a wildcard search to make all of the two+ capitalized letters highlighted by using <[A-Z]{2,}>, then formatting them as highlighted, if this would help with any macros.
For any given abbreviation, you could use a macro like:
Sub Demo()
Application.ScreenUpdating = False
Dim i As Long
With ActiveDocument.Range
With .Find
.ClearFormatting
.Replacement.ClearFormatting
.Text = InputBox("What is the Text to Find")
.Replacement.Text = ""
.Forward = True
.Wrap = wdFindStop
.Format = False
.MatchCase = True
.MatchWholeWord = True
.MatchWildcards = False
.MatchSoundsLike = False
.MatchAllWordForms = False
.Execute
End With
Do While .Find.Found
i = i + 1
.Collapse wdCollapseEnd
.Find.Execute
Loop
End With
Application.ScreenUpdating = True
MsgBox i & " instances found."
End Sub
For PC macro installation & usage instructions, see: http://www.gmayor.com/installing_macro.htm
For Mac macro installation & usage instructions, see: https://wordmvp.com/Mac/InstallMacro.html
Provided there's at least one occurrence of the abbreviation in parens you could use a macro like the following. The macro checks the contents of a document for upper-case/numeric parenthetic abbreviations it then looks backwards to try to determine what term they abbreviate. For example:
World Wide Web (WWW)
Naturally, given the range of acronyms in use, it’s not foolproof and, if a match isn’t made, the preceding sentence (in VBA terms) is captured so the user can edit the output. A table is then built at the end of the document, which is then searched for all references to the acronym (other than for the definition) and the counts and page numbers added to the table.
Note that the macro won't tell you how many times 'World Wide Web' appears in the document, though. After all, given your criteria, it's impossible to know what terms should have been reduced to an acronym but weren't.
Sub AcronymLister()
Application.ScreenUpdating = False
Dim StrTmp As String, StrAcronyms As String, i As Long, j As Long, k As Long, Rng As Range, Tbl As Table
StrAcronyms = "Acronym" & vbTab & "Term" & vbTab & "Page" & vbTab & "Cross-Reference Count" & vbTab & "Cross-Reference Pages" & vbCr
With ActiveDocument
With .Range
With .Find
.ClearFormatting
.Replacement.ClearFormatting
.MatchWildcards = True
.Wrap = wdFindStop
.Text = "\([A-Z0-9]{2,}\)"
.Replacement.Text = ""
.Execute
End With
Do While .Find.Found = True
StrTmp = Replace(Replace(.Text, "(", ""), ")", "")
If (InStr(1, StrAcronyms, .Text, vbBinaryCompare) = 0) And (Not IsNumeric(StrTmp)) Then
If .Words.First.Previous.Previous.Words(1).Characters.First = Right(StrTmp, 1) Then
For i = Len(StrTmp) To 1 Step -1
.MoveStartUntil Mid(StrTmp, i, 1), wdBackward
.Start = .Start - 1
If InStr(.Text, vbCr) > 0 Then
.MoveStartUntil vbCr, wdForward
.Start = .Start + 1
End If
If .Sentences.Count > 1 Then .Start = .Sentences.Last.Start
If .Characters.Last.Information(wdWithInTable) = False Then
If .Characters.First.Information(wdWithInTable) = True Then
.Start = .Cells(.Cells.Count).Range.End + 1
End If
ElseIf .Cells.Count > 1 Then
.Start = .Cells(.Cells.Count).Range.Start
End If
Next
End If
StrTmp = Replace(Replace(Replace(.Text, " (", "("), "(", "|"), ")", "")
StrAcronyms = StrAcronyms & Split(StrTmp, "|")(1) & vbTab & Split(StrTmp, "|")(0) & vbTab & .Information(wdActiveEndAdjustedPageNumber) & vbTab & vbTab & vbCr
End If
.Collapse wdCollapseEnd
.Find.Execute
Loop
StrAcronyms = Replace(Replace(Replace(StrAcronyms, " (", "("), "(", vbTab), ")", "")
Set Rng = .Characters.Last
With Rng
If .Characters.First.Previous <> vbCr Then .InsertAfter vbCr
.InsertAfter Chr(12)
.Collapse wdCollapseEnd
.Style = "Normal"
.Text = StrAcronyms
Set Tbl = .ConvertToTable(Separator:=vbTab, NumRows:=.Paragraphs.Count, NumColumns:=5)
With Tbl
.Columns.AutoFit
.Rows(1).HeadingFormat = True
.Rows(1).Range.Style = "Strong"
.Rows.Alignment = wdAlignRowCenter
End With
.Collapse wdCollapseStart
End With
End With
Rng.Start = ActiveDocument.Range.Start
For i = 2 To Tbl.Rows.Count
StrTmp = "": j = 0: k = 0
With .Range
With .Find
.ClearFormatting
.Replacement.ClearFormatting
.Format = False
.Forward = True
.Text = "[!\(]" & Split(Tbl.Cell(i, 1).Range.Text, vbCr)(0) & "[!\)]"
.MatchWildcards = True
.Execute
End With
Do While .Find.Found
If Not .InRange(Rng) Then Exit Do
j = j + 1
If k <> .Duplicate.Information(wdActiveEndAdjustedPageNumber) Then
k = .Duplicate.Information(wdActiveEndAdjustedPageNumber)
StrTmp = StrTmp & k & " "
End If
.Collapse wdCollapseEnd
.Find.Execute
Loop
End With
Tbl.Cell(i, 4).Range.Text = j
StrTmp = Replace(Trim(StrTmp), " ", ",")
If StrTmp <> "" Then
'Add the current record to the output list (StrOut)
StrTmp = Replace(Replace(ParseNumSeq(StrTmp, "&"), ",", ", "), " ", " ")
End If
Tbl.Cell(i, 5).Range.Text = StrTmp
Next
End With
Set Rng = Nothing: Set Tbl = Nothing
Application.ScreenUpdating = True
End Sub
Function ParseNumSeq(StrNums As String, Optional StrEnd As String)
'This function converts multiple sequences of 3 or more consecutive numbers in a
' list to a string consisting of the first & last numbers separated by a hyphen.
' The separator for the last sequence can be set via the StrEnd variable.
Dim ArrTmp(), i As Long, j As Long, k As Long
ReDim ArrTmp(UBound(Split(StrNums, ",")))
For i = 0 To UBound(Split(StrNums, ","))
ArrTmp(i) = Split(StrNums, ",")(i)
Next
For i = 0 To UBound(ArrTmp) - 1
If IsNumeric(ArrTmp(i)) Then
k = 2
For j = i + 2 To UBound(ArrTmp)
If CInt(ArrTmp(i) + k) <> CInt(ArrTmp(j)) Then Exit For
ArrTmp(j - 1) = ""
k = k + 1
Next
i = j - 2
End If
Next
StrNums = Join(ArrTmp, ",")
StrNums = Replace(Replace(Replace(StrNums, ",,", " "), ", ", " "), " ,", " ")
While InStr(StrNums, " ")
StrNums = Replace(StrNums, " ", " ")
Wend
StrNums = Replace(Replace(StrNums, " ", "-"), ",", ", ")
If StrEnd <> "" Then
i = InStrRev(StrNums, ",")
If i > 0 Then
StrNums = Left(StrNums, i - 1) & Replace(StrNums, ",", " " & Trim(StrEnd), i)
End If
End If
ParseNumSeq = StrNums
End Function

Excel Range object throws error 0x800A03EC because of a string longer than 255 characters

Using an ActiveX server from MATLAB, I am trying to highlight many cells in an Excel sheet at once. These are not in specific columns or rows so I use Range('A1,B2,...') to access them. However the string accepted by the Range object has to be less than 255 characters or an error:
Error: Object returned error code: 0x800A03EC
is thrown. The following code reproduces this error with an empty Excel file.
hActX = actxserver('Excel.Application');
hWB = hActX.Workbooks.Open('C:\Book1.xlsx');
hSheet = hWB.Worksheets.Item('Sheet1');
col = repmat('A', 100, 1);
row = num2str((1:100)'); %'
cellInd = strcat(col, strtrim(cellstr(row)));
str1 = strjoin(cellInd(1:66), ','); %// 254 characters
str2 = strjoin(cellInd(1:67), ','); %// 258 characters
hSheet.Range(str1).Interior.Color = 255; %// Works
hSheet.Range(str2).Interior.Color = 255; %// Error 0x800A03EC
hWB.Save;
hWB.Close(false);
hActX.Quit;
How can I get around this? I found no other relevant method of calling Range, or of otherwise getting the cells I want to modify.
If you start with a String, you can test its length to determine if Range() can handle it. Here is an example of building a diagonal range:
Sub DiagonalRange()
Dim BigString As String, BigRange As Range
Dim i As Long, HowMany As Long, Ln As String
HowMany = 100
For i = 1 To HowMany
BigString = BigString & "," & Cells(i, i).Address(0, 0)
Next i
BigString = Mid(BigString, 2)
Ln = Len(BigString)
MsgBox Ln
If Ln < 250 Then
Set BigRange = Range(BigString)
Else
Set BigRange = Nothing
arr = Split(BigString, ",")
For Each a In arr
If BigRange Is Nothing Then
Set BigRange = Range(a)
Else
Set BigRange = Union(BigRange, Range(a))
End If
Next a
End If
BigRange.Select
End Sub
For i = 10, the code will the the direct method, but if the code were i=100, the array method would be used.
The solution, as Rory pointed out, is to use the Union method. To minimize the number of calls from MATLAB to the ActiveX server, this is what I did:
str = strjoin(cellInd, ',');
isep = find(str == ',');
isplit = diff(mod(isep, 250)) < 0;
isplit = [isep(isplit) (length(str) + 1)];
hRange = hSheet.Range(str(1:(isplit(1) - 1)));
for ii = 2:numel(isplit)
hRange = hActX.Union(hRange, ...
hSheet.Range(str((isplit(ii-1) + 1):(isplit(ii) - 1))));
end
I used 250 in the mod to account for the cell names being up to 6 characters long, which is sufficient for me.

Align series by value in DevExpress XtraCharts

I have a DevExpress XtraChart object with several series, all of type: line.
I have a requirement from a client to align the series by the max value of each series. This is without regard to the attached axis, has anyone done this before?
Although I was interested in displaying this on the chart element, the 'work' was done on the underlying DataSet.
I was able to achieve this by looping through the series collection, deriving the max value of the first series, through a SQL query, and then each subsequent series, and noting the difference, and then adding or subtracting the difference at the DataSet level.
Here's the code:
Private Sub cbAlignPeaks_CheckedChanged(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles cbAlignPeaks.CheckedChanged
Dim dt As DataTable = chart.DataSource
Dim Row() As Data.DataRow
Dim s As Series = Nothing
Dim dtr As DataTableReader = Nothing
Dim maxTimeGet, maxTimeSet, diff As Decimal
Me.Cursor = Cursors.WaitCursor
If cbAlignPeaksPre.Checked Then
For i As Integer = 0 To chartPreTim.Series.Count - 1
s = chartPreTim.Series(i)
If _offsets.Count = chartPreTim.Series.Count - 1 Then
If i > 0 Then
diff = _offsets(s.DataFilters(0).Value)
Row = dt.Select("BORING_NAME = '" & s.Name & "'")
For k As Integer = 0 To Row.Count - 1
Row(k)("TIME_SEC") = Row(k)("TIME_SEC") + diff
Next
End If
Else
If i = 0 Then 'get adjustment info
dtr = getDetectorMax(s.DataFilters(0).Value, cbDetectors.Text, timType.PRE) ' <-- getDetectorMax runs a SQL query returning the max value
If dtr.Read Then
maxTimeGet = dtr("TIME_SEC")
End If
Else 'set adjustment info
dtr = Nothing
dtr = getDetectorMax(s.DataFilters(0).Value, cbDetectors.Text, timType.PRE)
If dtr.Read Then
maxTimeSet = dtr("TIME_SEC")
End If
If maxTimeGet > maxTimeSet Then
diff = maxTimeGet - maxTimeSet
_offsets.Add(s.DataFilters(0).Value, diff)
Row = dt.Select("BORING_NAME = '" & s.DataFilters(0).Value & "'")
For k As Integer = 0 To Row.Count - 1
Row(k)("TIME_SEC") = Row(k)("TIME_SEC") + diff
Next
Else
diff = maxTimeSet - maxTimeGet
_offsets.Add(s.DataFilters(0).Value, diff * -1)
Row = dt.Select("BORING_NAME = '" & s.DataFilters(0).Value & "'")
For k As Integer = 0 To Row.Count - 1
Row(k)("TIME_SEC") = Row(k)("TIME_SEC") - diff
Next
End If
End If
End If
Next
Else
For i As Integer = 1 To chartPreTim.Series.Count - 1 ' We skip item 0 as that's the baseline
s = chartPreTim.Series(i)
diff = _offsets(s.DataFilters(0).Value)
Row = dt.Select("BORING_NAME = '" & s.DataFilters(0).Value & "'")
For k As Integer = 0 To Row.Count - 1
Row(k)("TIME_SEC") = Row(k)("TIME_SEC") - diff
Next
Next
End If
chartPreTim.RefreshData()
chartPreTim.Refresh()
Me.Cursor = Cursors.Default
End Sub

Renaming a Word document and saving its filename with its first 10 letters

I have recovered some Word documents from a corrupted hard drive using a piece of software called photorec. The problem is that the documents' names can't be recovered; they are all renamed by a sequence of numbers. There are over 2000 documents to sort through and I was wondering if I could rename them using some automated process.
Is there a script I could use to find the first 10 letters in the document and rename it with that? It would have to be able to cope with multiple documents having the same first 10 letters and so not write over documents with the same name. Also, it would have to avoid renaming the document with illegal characters (such as '?', '*', '/', etc.)
I only have a little bit of experience with Python, C, and even less with bash programming in Linux, so bear with me if I don't know exactly what I'm doing if I have to write a new script.
How about VBScript? Here is a sketch:
FolderName = "C:\Docs\"
Set fs = CreateObject("Scripting.FileSystemObject")
Set fldr = fs.GetFolder(Foldername)
Set ws = CreateObject("Word.Application")
For Each f In fldr.Files
If Left(f.name,2)<>"~$" Then
If InStr(f.Type, "Microsoft Word") Then
MsgBox f.Name
Set doc = ws.Documents.Open(Foldername & f.Name)
s = vbNullString
i = 1
Do While Trim(s) = vbNullString And i <= doc.Paragraphs.Count
s = doc.Paragraphs(i)
s = CleanString(Left(s, 10))
i = i + 1
Loop
doc.Close False
If s = "" Then s = "NoParas"
s1 = s
i = 1
Do While fs.FileExists(s1)
s1 = s & i
i = i + 1
Loop
MsgBox "Name " & Foldername & f.Name & " As " & Foldername & s1 _
& Right(f.Name, InStrRev(f.Name, "."))
'' This uses copy, because it seems safer
f.Copy Foldername & s1 & Right(f.Name, InStrRev(f.Name, ".")), False
'' MoveFile will copy the file:
'' fs.MoveFile Foldername & f.Name, Foldername & s1 _
'' & Right(f.Name, InStrRev(f.Name, "."))
End If
End If
Next
msgbox "Done"
ws.Quit
Set ws = Nothing
Set fs = Nothing
Function CleanString(StringToClean)
''http://msdn.microsoft.com/en-us/library/ms974570.aspx
Dim objRegEx
Set objRegEx = CreateObject("VBScript.RegExp")
objRegEx.IgnoreCase = True
objRegEx.Global = True
''Find anything not a-z, 0-9
objRegEx.Pattern = "[^a-z0-9]"
CleanString = objRegEx.Replace(StringToClean, "")
End Function
Word documents are stored in a custom format which places a load of binary cruft on the beginning of the file.
The simplest thing would be to knock something up in Python that searched for the first line beginning with ASCII chars. Here you go:
#!/usr/bin/python
import glob
import os
for file in glob.glob("*.doc"):
f = open(file, "rb")
new_name = ""
chars = 0
char = f.read(1)
while char != "":
if 0 < ord(char) < 128:
if ord("a") <= ord(char) <= ord("z") or ord("A") <= ord(char) <= ord("Z") or ord("0") <= ord(char) <= ord("9"):
new_name += char
else:
new_name += "_"
chars += 1
if chars == 100:
new_name = new_name[:20] + ".doc"
print "renaming " + file + " to " + new_name
f.close()
break;
else:
new_name = ""
chars = 0
char = f.read(1)
if new_name != "":
os.rename(file, new_name)
NOTE: if you want to glob multiple directories you'll need to change the glob line accordingly. Also this takes no account of whether the file you're trying to rename to already exists, so if you have multiple docs with the same first few chars then you'll need to handle that.
I found the first chunk of 100 ASCII chars in a row (if you look for less than that you end up picking up doc keywords and such) and then used the first 20 of these to make the new name, replacing anything that's not a-z A-Z or 0-9 with underscores to avoid file name issues.