My memo line that looks something like this:
Return: #999 100.00\NSF|Balance=$242.00. Available Balance=$50.00`
or
Return: #888 45.90\WD REST.
I need it to return everything before the | if there is one, or the whole memo line if there isn't a |. Right now I have two formulas that give everything before the | or nothing if there isn't a |
memo1: left({table.memo},instr(table.memo},"|"))
memo2: replace({#Memo1},"|"," ")
How can I configure this such that if a | is present, return #Memo2, else return {table.memo}?
If I understand correctly, just make one more formula like this:
IF instr({table.memo},"|")>0
THEN {#Memo2}
ELSE {table.memo}
This displays one formula or the other based on the presence of that vertical line character.
I would use an if statement with instr:
memo1: left({table.memo},instr(table.memo},"|"))
memo2: replace({#Memo1},"|"," ")
if instr(table.memo} > 0
left({table.memo},instr(table.memo},"|"))
else
replace({#Memo1},"|"," ")
Please try something like that. (I did not check the syntax, so please check my work.)
Example for if
Related
I'm using AutoHotkey for this as the code is the most understandable to me. So I have a document with numbers and text, for example like this
120344 text text text
234000 text text
and the desired output is
12:03:44 text text text
23:40:00 text text
I'm sure StrReplace can be used to insert the colons in, but I'm not sure how to specify the position of the colons or ask AHK to 'find' specific strings of 6 digit numbers. Before, I would have highlighted the text I want to apply StrReplace to and then press a hotkey, but I was wondering if there is a more efficient way to do this that doesn't need my interaction. Even just pointing to the relevant functions I would need to look into to do this would be helpful! Thanks so much, I'm still very new to programming.
hfontanez's answer was very helpful in figuring out that for this problem, I had to use a loop and substring function. I'm sure there are much less messy ways to write this code, but this is the final version of what worked for my purposes:
Loop, read, C:\[location of input file]
{
{ If A_LoopReadLine = ;
Continue ; this part is to ignore the blank lines in the file
}
{
one := A_LoopReadLine
x := SubStr(one, 1, 2)
y := SubStr(one, 3, 2)
z := SubStr(one, 5)
two := x . ":" . y . ":" . z
FileAppend, %two%`r`n, C:\[location of output file]
}
}
return
Assuming that the "timestamp" component is always 6 characters long and always at the beginning of the string, this solution should work just fine.
String test = "012345 test test test";
test = test.substring(0, 2) + ":" + test.substring(2, 4) + ":" + test.substring(4, test.length());
This outputs 01:23:45 test test test
Why? Because you are temporarily creating a String object that it's two characters long and then you insert the colon before taking the next pair. Lastly, you append the rest of the String and assign it to whichever String variable you want. Remember, the substring method doesn't modify the String object you are calling the method on. This method returns a "new" String object. Therefore, the variable test is unmodified until the assignment operation kicks in at the end.
Alternatively, you can use a StringBuilder and append each component like this:
StringBuilder sbuff = new StringBuilder();
sbuff.append(test.substring(0,2));
sbuff.append(":");
sbuff.append(test.substring(2,4));
sbuff.append(":");
sbuff.append(test.substring(4,test.length()));
test = sbuff.toString();
You could also use a "fancy" loop to do this, but I think for something this simple, looping is just overkill. Oh, I almost forgot, this should work with both of your test strings because after the last colon insert, the code takes the substring from index position 4 all the way to the end of the string indiscriminately.
I'm trying to parse a file that (apparently) ends its lines with carriage returns, but they aren't being matched as such in Swift, despite having the same UTF8 value. I can see possible fixes for the problem, but I'm curious as to what these characters actually are.
Here's some sample code, with the output below. (CR is set using Character("\r"), although I've tried it using "\r" as well.
try f.forEach() { c in
print(c, terminator:" ") // DBG
if (c == "\r") {
print("Carriage return found!")
}
print(String(c).utf8.first!, terminator:" ")//DBG
print(String(describing:pstate)) // DBG
...
case .field:
switch c {
case CR,LF :
self.endline()
pstate = .eol
When it reaches the end of line (which shows up as such in my text editors), I get this:
. 46 field
0 48 field
13 field
I 73 field
It doesn't seem to be matching using == or in the switch statement. Is there another approach I should be using for this character?
(I'll note that the parsing works fine with files that terminate in newlines.)
I determined what the problem was. By looking at c.unicodeScalars I discovered that the end of line character was in fact "\r\n", not just "\r". As seen in my code I was only taking the first when printing it out as UTF-8. I don't know if that's something from String.forEach or in the file itself.
I know that there are tests to determine if something is a newline. Swift 5 has them directly (c.isNewline), and there is also the CharacterSet approach as noted by Bill Nattaner.
I'm happier with something that will work in my switch statement (and thus I'll define each one explicitly), but that might change if I expect to deal with a wider variety of files.
I'm a little hazy as to what the f.forEach represents, but if your variable c is of type Character then you could replace your if statement with:
if "\(c)".rangeOfCharacter( from: CharacterSet.newlines ) != nil
{
print("Carriage return found!")
}
That way you won't have to invent a list of all-possible new line characters.
I am trying to write a perl script to search a config file for the following line:
remote_phonebook.data.1.url =
and do 1 of 2 things:
if the right of side of the = is blank add someString
if there is something there, replace anything there with someString
This will insert just fine:
s/remote_phonebook\.data\.1\.url = /remote_phonebook.data.1.url = someString/;
however if someString already exists, it will append it to look like this:
remote_phonebook.data.1.url = someString someString
This will replace just fine if someString already exists, but wont insert if its blank.
s/remote_phonebook\.data\.1\.url = someString/remote_phonebook.data.1.url = someString/;
.* is your friend, here. It means "match 0 or more (*) of any character (.)":
s/remote_phonebook\.data\.1\.url =.*/remote_phonebook.data.1.url = someString/;
So whether or not there is anything after the =, you'll end up with the contents you want. To make sure that you're matching from the start of the line (so "xxxremote_phonebook..." won't match), and to allow for more (or less) space before the "=", I'd use:
s/^remote_phonebook\.data\.1\.url\s*=.*/remote_phonebook.data.1.url = someString/;
s/^\s*remote_phonebook\.data\.1\.url\s*=\K.*/someString/;
The .* will match anything up to a newline.
The \K makes it so you don't have to repeat everything.
I have a list of pdf files in this format "123 - Test - English.pdf". I want to be able to set "111", "Test" and "English.pdf" in their own individual variables. I tried running the code below but I don't think it accounts for multiple dashes "-". How can I do this? Please help Thanks in advance.
Loop,C:\My Documents\Notes\*.pdf, 0, 0
{
NewVariable = Trim(Substr(A_LoopFileName,1, Instr(A_LoopFileName, "-")-1))
I would recommend using a parse loop to get your variables. The following loops through values between the dashes and removes the whitespace.
FileName = Test - file - name.pdf
Loop, parse, FileName, `-
MyVar%A_Index% := RegExReplace(A_LoopField, A_Space, "")
msgbox % Myvar1 "`n" Myvar2 "`n" MyVar3
First, I don't know if it was a typo, but if you use a { under your loop statement, you also need to close it. If your next statement is just one line, you don't need any brackets at all.
Second, if you just use = then your code will output as just that very code text. You need to use a :=
Third, your present code, if coded correctly would result in this:
somepdffile.pd
if it found any pdf files without a dash. Instr() will return the position of a dash. If there is no dash, it returns 0 - in which case, your substr() statement will add 0 and your -1 which adds up to -1 and if you use a negative number with substr(), it will search from the end of the string instead of the beginning - which is why your string would get cut off.
Loop, C:\My Documents\Notes\*.pdf, 0, 0
{
;look at the docs (http://www.autohotkey.com/docs/) for `substr`
}
So there is an explanation of why your code doesn't work. To get it to do what you want to do, can you explain a bit more as to how you want NewVariable to look like?
; here is another way (via RegExMatch)
src:="123 - Test - English.pdf", pat:="[^\s|-]+"
While, mPos:=RegExMatch(src, pat, match, mPos ? mPos+StrLen(match):1)
match%A_Index%:=match
MsgBox, 262144, % "result", % match1 ", "match2 ", "match3
I have a string that looks like this:
17/07/2013 TEXTT TEXR 1 Text 1234567 456.78 987654
I need to separate this so I only end up with 2 values (in this example it's 1234567 and 456.78). The rest is unneeded.
I tried using string split with %A_Space% but as the whole middle area between values is filled with spaces, it doesn't really work.
Anyone got an idea?
src:="17/07/2013 TEXTT TEXR 1 Text "
. " 1234567 456.78 987654", pattern:="([\d\.]+)\s+([\d\.]+)"
RegexMatch(src, pattern, match)
MsgBox, 262144, % "result", % match1 "`n"match2
You should look at RegExMatch() and RegexReplace().
So, you will need to build a regex needle (I'm not an expert regexer, but this will work)
First, remove all of the string up to the end of "1 Text" since "1 Text" as you say, is constant. That will leave you with the three number values.
Something like this should find just the numbers you want:
needle:= "iO)1\s+Text"
partialstring := RegexMatch(completestring, needle, results)
lenOfFrontToRemove := results.pos() + results.len()
lastthreenumbers := substr(completestring, lenOfFrontToRemove, strlen(completestring) )
lastthreenumbers := trim(lastthreenumbers)
msgbox % lastthreenumbers
To explain the regex needle:
- the i means case insensitive
- the O stands for options - it lets us use results.pos and results.len
- the \s means to look for whitespace; the + means to look for more than one if present.
Now you have just the last three numbers.
1234567 456.78 987654
But you get the idea, right? You should able to parse it from here.
Some hints: in a regex needle, use \d to find any digit, and the + to make it look for more than one in a row. If you want to find the period, use \.