How do I grab the position of the first matched result in a regular expression? See below.
Function MYMATCH(strValue As String, strPattern As String, Optional blnCase As Boolean = True, Optional blnBoolean = True) As String
Dim objRegEx As Object
Dim strPosition As Integer
' Create regular expression.
Set objRegEx = CreateObject("VBScript.RegExp")
objRegEx.Pattern = strPattern
objRegEx.IgnoreCase = blnCase
' Do the search match.
strPosition = objRegEx.Match(strValue)
MYMATCH = strPosition
End Function
For one, I'm not entirely certain what .Match is returning (string, integer, etc.). The one solution I found said I should create a Match object to and then grab the position from there, but unlike vb, vba does not recognize the Match object. I've also seen some code like the following, but I'm not necessarily looking for the value, just the first string placement:
If allMatches.count <> 0 Then
result = allMatches.Item(0).submatches.Item(0)
End If
Somewhat ignoring any of the possible syntax errors above (mostly due to me changing variable types right and left), how do I easily/simply accomplish this?
Thanks!
You can use FirstIndex to return the position of matches using the Execute method, ie
Function MYMATCH(strValue As String, strPattern As String, Optional blnCase As Boolean = True, Optional blnBoolean = True) As String
Dim objRegEx As Object
Dim strPosition As Integer
Dim RegMC
' Create regular expression.
Set objRegEx = CreateObject("VBScript.RegExp")
With objRegEx
.Pattern = strPattern
.IgnoreCase = blnCase
If .test(strValue) Then
Set RegMC = .Execute(strValue)
MYMATCH = RegMC(0).firstindex + 1
Else
MYMATCH = "no match"
End If
End With
End Function
Sub TestMe()
MsgBox MYMATCH("test 1", "\d+")
End Sub
For the benefit of others who may be having this problem, I finally figured it out.
Option Explicit
Function CHAMATCH(strValue As String, strPattern As String, Optional blnCase As Boolean = True, Optional blnBoolean = True) As String
Dim objRegEx As Object
Dim objPosition As Object
Dim strPosition As String
' Create regular expression.
Set objRegEx = CreateObject("VBScript.RegExp")
objRegEx.Pattern = strPattern
objRegEx.IgnoreCase = blnCase
' Do the search match.
Set objPosition = objRegEx.Execute(strValue)
strPosition = objPosition(0).FirstIndex
CHAMATCH = strPosition
End Function
Instead of a Match type, just a regular Object type will do (considering all it's returning is a class). Then, if you want to grab the index location, just use .FirstIndex on the match [of your choice], or if you want the value, us .Value
Related
I'm using the following VBA code from a related question in my Excel spreadsheet, and when I use it in a cell, it always fails (returns nothing). Even if I call it on a string literal in the function call (i.e. =RegexExtract("ABC1_DEF","ABC[0-9]")), it still fails. I've enabled the "Microsoft Visual Basic Regular Expressions 5.0" feature in the MSVBA application, so I'm not sure why these results are always empty. How can I resolve this?
Function RegexExtract(ByVal text As String, _
ByVal extract_what As String, _
Optional separator As String = ", ") As String
Dim allMatches As Object
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
Dim i As Long, j As Long
Dim result As String
RE.pattern = extract_what
RE.Global = True
Set allMatches = RE.Execute(text)
For i = 0 To allMatches.count - 1
For j = 0 To allMatches.Item(i).submatches.count - 1
result = result & (separator & allMatches.Item(i).submatches.Item(j))
Next
Next
If Len(result) <> 0 Then
result = Right$(result, Len(result) - Len(separator))
End If
RegexExtract = result
End Function
Edit
I tried yet another function from a separate question, and it just returns #VALUE!:
Function RegexExtract(ByVal text As String, _
ByVal extract_what As String) As String
Dim allMatches As Object
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
RE.Pattern = extract_what
RE.Global = True
Set allMatches = RE.Execute(text)
RegexExtract = allMatches.Item(0).submatches.Item(0)
End Function
Note you are trying to access .Submatches that stores capturing group values, but you have not defined any capturing groups in the pattern.
If you use (ABC[0-9]) you will get your match with the current function. Else, access the allMatches.Item(i) for full match values and discard the code to get the captured groups.
My VBA function should take a string referencing a range of units (i.e. "WWW1-5") and then return another string.
I want to take the argument, and put it in a comma separated string,
So "WWW1-5" should become "WWW1, WWW2, WWW3, WWW4, WWW5".
It's not always going to be a single digit. For example, I might need to separate "XXX11-18" or something similar.
I have never used regular expressions, but keep trying different things to make this work and it seems to only be finding 1 match instead of 3.
Any ideas? Here is my code:
Private Function split_group(ByVal group As String) As String
Dim re As Object
Dim matches As Object
Dim result As String
Dim prefix As String
Dim startVar As Integer
Dim endVar As Integer
Dim i As Integer
Set re = CreateObject("vbscript.regexp")
re.Pattern = "([A-Z]+)(\d+)[-](\d+)"
re.IgnoreCase = False
Set matches = re.Execute(group)
Debug.Print matches.Count
If matches.Count <> 0 Then
prefix = matches.Item(0)
startVar = CInt(matches.Item(1)) 'error occurs here
endVar = CInt(matches.Item(2))
result = ""
For i = startVar To endVar - 1
result = result & prefix & i & ","
Next i
split_group = result & prefix & endVar
Else
MsgBox "There is an error with splitting a group."
split_group = "ERROR"
End If
End Function
I tried setting global = true but I realized that wasn't the problem. The error actually occurs on the line with the comment but I assume it's because there was only 1 match.
I tried googling it but everyone's situation seemed to be a little different than mine and since this is my first time using RE I don't think I understand the patterns enough to see if maybe that was the problem.
Thanks!
Try the modified Function below:
Private Function split_metergroup(ByVal group As String) As String
Dim re As Object
Dim matches As Variant
Dim result As String
Dim prefix As String
Dim startVar As Integer
Dim endVar As Integer
Dim i As Integer
Set re = CreateObject("VBScript.RegExp")
With re
.Global = True
.IgnoreCase = True
.Pattern = "[0-9]{1,20}" '<-- Modified the Pattern
End With
Set matches = re.Execute(group)
If matches.Count > 0 Then
startVar = CInt(matches.Item(0)) ' <-- modified
endVar = CInt(matches.Item(1)) ' <-- modified
prefix = Left(group, InStr(group, startVar) - 1) ' <-- modified
result = ""
For i = startVar To endVar - 1
result = result & prefix & i & ","
Next i
split_metergroup = result & prefix & endVar
Else
MsgBox "There is an error with splitting a meter group."
split_metergroup = "ERROR"
End If
End Function
The Sub I've tested it with:
Option Explicit
Sub TestRegEx()
Dim Res As String
Res = split_metergroup("DEV11-18")
Debug.Print Res
End Sub
Result I got in the immediate window:
DEV11,DEV12,DEV13,DEV14,DEV15,DEV16,DEV17,DEV18
Another RegExp option, this one uses SubMatches:
Test
Sub TestRegEx()
Dim StrTst As String
MsgBox WallIndside("WAL7-21")
End Sub
Code
Function WallIndside(StrIn As String) As String
Dim objRegex As Object
Dim objRegMC As Object
Dim lngCnt As Long
Set objRegex = CreateObject("VBScript.RegExp")
With objRegex
.Global = True
.IgnoreCase = True
.Pattern = "([a-z]+)(\d+)-(\d+)"
If .test(StrIn) Then
Set objRegMC = .Execute(StrIn)
For lngCnt = objRegMC(0).submatches(1) To objRegMC(0).submatches(2)
WallIndside = WallIndside & (objRegMC(0).submatches(0) & lngCnt & ", ")
Next
WallIndside = Left$(WallIndside, Len(WallIndside) - 2)
Else
WallIndside = "no match"
End If
End With
End Function
#Shai Rado 's answer worked. But I figured out on my own WHY my original code was not working, and was able to lightly modify it.
The pattern was finding only 1 match because it was finding 1 FULL MATCH. The full match was the entire string. The submatches were really what I was trying to get.
And this is what I modified to make the original code work (asking for each submatch of the 1 full match):
I've created a function that will return the Nth reference which includes a sheetname (if it's there), however it's not working for all instances. The regex string I'm using is
'[\w ]+[']!([$]{0,1})([A-Z]{1,2})([$]{0,1})(\d{1,5})
I'm finding though it won't find the first reference in either of the below examples:
='Biscuits Raw Data'!G783/'Biscuits Raw Data'!E783
=IF('Biscuits Raw Data'!G705="","",'Biscuits Raw Data'!G723/'Biscuits Raw Data'!G7005*100)
Below is my Function code:
Function GrabNthreference(Rng As range, NthRef As Integer) As String
Dim patrn As String
Dim RegX
Dim Matchs
Dim RegEx
Dim FinalMatch
Dim Subm
Dim i As Integer
Dim StrRef As String
patrn = "'[\w ]+[']!([$]{0,1})([A-Z]{1,2})([$]{0,1})(\d{1,5})"
StrRef = Rng.Formula
Set RegEx = CreateObject("vbscript.regexp") ' Create regular expression.
RegEx.Global = True
RegEx.Pattern = patrn ' Set pattern.
RegEx.IgnoreCase = True ' Make case insensitive.
Set RegX = RegEx.Execute(StrRef)
If RegX.Count < NthRef Then
GrabNthreference = StrRef
Exit Function
End If
i= -1
For Each Matchs In RegX ' Iterate Matches collection.
Set Subm = RegX(i).submatches
i = i + 1
If i = NthRef -1 Then
GrabNthreference = RegX(i)
Exit Function
End If
'Debug.Print RegX(i)
Next
End Function
Here's my final code
Function GrabNthreference(R As range, NthRef As Integer) As String 'based on http://stackoverflow.com/questions/13835466/find-all-used-references-in-excel-formula
Dim result As Object
Dim testExpression As String
Dim objRegEx As Object
Dim i As Integer
i = 0
Set objRegEx = CreateObject("VBScript.RegExp")
objRegEx.IgnoreCase = True
objRegEx.Global = True
objRegEx.Pattern = """.*?""" ' remove expressions
testExpression = CStr(R.Formula)
testExpression = objRegEx.Replace(testExpression, "")
'objRegEx.Pattern = "(([A-Z])+(\d)+)" 'grab the address think this is an old attempt so remming out
objRegEx.Pattern = "(['].*?['!])?([[A-Z0-9_]+[!])?(\$?[A-Z]+\$?(\d)+(:\$?[A-Z]+\$?(\d)+)?|\$?[A-Z]+:\$?[A-Z]+|(\$?[A-Z]+\$?(\d)+))"
If objRegEx.Test(testExpression) Then
Set result = objRegEx.Execute(testExpression)
If result.Count > 0 Then
For Each Match In result
Debug.Print Match.Value
If i = NthRef - 1 Then
GrabNthreference = result(i)
Exit Function
End If
i = i + 1
Next Match
Else
GrabNthreference = "No precedencies found"
End If
End If
End Function
This code did lead me onto thinking about using the simple activecell.precedences method but I think the problem is that it won't report offsheet and won't indicate if the formula is relative or absolute.
Any comments welcome but I think I've answered my own question :)
This is my sample record in a Text format with comma delimited
901,BLL,,,BQ,ARCTICA,,,,
i need to replace ,,, to ,,
The Regular expression that i tried
With regex
.MultiLine = False
.Global = True
.IgnoreCase = False
.Pattern="^(?=[A-Z]{3})\\,{3,}",",,"))$ -- error
Now i want to pass Line from file to Regex to correct the record, can some body guide me to fix this i am very new to VBA
I want to read the file line by line pass it to Regex
Looking at your original pattern I tried using .Pattern = "^\d{3},\D{3},,," which works on the sample record as with the 3 number characters , 3 letters,,,
In the answer I have used a more generalised pattern .Pattern = "^\w*,\w*,\w*,," This also works on the sample and mathces 3 commas each preceded with 0 or more alphanumeric characters followed directly by a fourth comma. Both patterns require a match to be from the begining of the string.
Pattern .Pattern = "^\d+,[a-zA-Z]+,\w*,," also works on the sample record. It would specify that before the first comma there should be 1 or greater numeric characters (and only numeric characters) and before the second comma ther should be 1 or more letters (and only letters). Before the 3rd comma there could be 0 or more alphanumeric characters.
The left function removes the rightmost character in the match ie. the last comma to generate the string used by the Regex.Replace.
Sub Test()
Dim str As String
str = "901,BLL,,,BQ,ARCTICA,,,,"
Debug.Print
Debug.Print str
str = strConv(str)
Debug.Print str
End Sub
Function strConv(ByVal str As String) As String
Dim objRegEx As Object
Dim oMatches As Object
Dim oMatch As Object
Set objRegEx = CreateObject("VBScript.RegExp")
With objRegEx
.MultiLine = False
.IgnoreCase = False
.Global = True
.Pattern = "^\w*,\w*,\w*,,"
End With
Set oMatches = objRegEx.Execute(str)
If oMatches.Count > 0 Then
For Each oMatch In oMatches
str = objRegEx.Replace(str, Left(oMatch.Value, oMatch.Length - 1))
Next oMatch
End If
strConv = str
End Function
Try this
Sub test()
Dim str As String
str = "901,BLL,,,BQ,ARCTICA,,,,"
str = strConv(str)
MsgBox str
End Sub
Function strConv(ByVal str As String) As String
Dim objRegEx As Object, allMatches As Object
Set objRegEx = CreateObject("VBScript.RegExp")
With objRegEx
.MultiLine = False
.IgnoreCase = False
.Global = True
.Pattern = ",,,"
End With
strConv = objRegEx.Replace(str, ",,")
End Function
I am trying to extract the 1st group of digits from this expression "CSTAR ADJ REF # A-3080101078AZ Keying error Deposit $ 938,710.33 on 05/20/2011 item keying". That is, i want to get only "3080101078" as result in an adjacent cell.
I have tried using this:
Function CleanString(strIn As String) As String
Dim objRegex
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "[^0-9\s]"
CleanString= Trim(.Replace(strIn, vbNullString))
End With
End Function
This is giving "3080101078 93871033 05202011". How can i get the first group of number only (3080101078) ?
Try this
Function CleanString(strIn As String) As String
Dim objRegex As Object
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "\d+"
CleanString = .Execute(strIn)(0)
End With
End Function
\d+ = any digit, one or more
.Execute(strIn) executes the regex against the input, returnining a MatchCollection
.Execute(strIn)(0) returns the first item in the match collection