I just want to get the part of string that matches the regular expression but trying with match.Value or with groups it always returns "". It's driving me crazy.
EDIT:
This worked:
Private Function NormalizeValue(ByVal fieldValue As String) As String
Dim result As String = ""
Dim pattern As String = "[a-zA-Zñ'-]*"
Dim matches As Match
matches = Regex.Match(fieldValue, pattern)
While (matches.Success = True)
result = result & matches.Value
matches = matches.NextMatch()
End While
Return result
End Function
If your regex starts with ^ and ends with $, you are trying to match the whole string - not a part as your are stating in the question.
So you either need to remove them or rephrase your question.
Related
I've got a string such as :
Dim initialString As String = "Some text here is f(42,foo,bar) and maybe some other here."
And want to replace the "f(42,foo,bar)" part to the evaluation of a function with following prototype :
Function myLittleFunction(ByVal number As Integer, ByVal string0 As String = "NA0", ByVal string1 As String = "NA1")
Witch I did with this regex :
finalString = System.Text.RegularExpressions.Regex.Replace(initialString, "f\((\d+),([a-zA-Z0-9_ ]+),([a-zA-Z0-9_ ]+)\)", myLittleFunction(Convert.ToUInt32("${1}"), "$2", "$3"))
But that's not working because Convert.ToUInt32("${1}") fails. If a replace it by any integer by hand and run the code, I've got the correct evaluation and replacement in my string.
How can I correctly cast "$1" to appropriate integer ?
String replacement pattern cannot be interpolated for use as variables to a method.
You may use a match evaluator:
Dim rx = New Regex("f\((\d+),([a-zA-Z0-9_\s]+),([a-zA-Z0-9_\s]+)\)")
Dim result = rx.Replace(s, New MatchEvaluator(Function(m As Match)
Return myLittleFunction(Convert.ToUInt32(m.Groups(1).Value), m.Groups(2).Value, m.Groups(3).Value)
End Function))
The m is a Match object, the one that is found by the Regex.Replace method. You may access all the groups captured with the regex using m.Groups(N).Value.
How do I access the .index method in regex, so in this case it should output the location of the first instance of a number
Dim sourceString As String = "abcdefg12345"
Dim textboxregex As Regex = New Regex("^(d)$")
If textboxregex.IsMatch(sourceString) Then
Console.WriteLine(Match.Index) 'this should display the location of the first occurrence of the pattern within the sourcestring
End If
In this case you dont need regex:
Dim digits = From chr In sourceString Where Char.IsDigit(chr)
Dim index = -1
If digits.Any() Then index = sourceString.IndexOf(digits.First())
or in one statement with the ugly method syntax:
Dim index As Int32 = "abcdefg12345".
Select(Function(chr, ix) New With {chr, ix}).
Where(Function(x) Char.IsDigit(x.chr)).
Select(Function(x) x.ix).
DefaultIfEmpty(-1).
First()
Try a Lookbehind:
Dim textboxregex As Regex = New Regex("(?<=\D)\d")
If textboxregex.IsMatch(sourceString) Then
Console.WriteLine(textboxregex.Match(sourceString).Index)
End If
This will match the first occurence of a digit after all non digit characters.
Your Regex expression is wrong (you need the '\' before the d) and you haven't defined Match
Dim sourceString As String = "abcdefg12345"
Dim textboxregex As Regex = New Regex("\d")
Dim rxMatch as Match = textboxregex.Match(sourceString)
If rxMatch.success Then
Console.WriteLine(rxMatch.Index) 'this should display the location of the first occurrence of the pattern within the sourcestring
End If
This is probably a simple problem, but unfortunately I wasn't able to get the results I wanted...
Say, I have the following line:
"Wouldn't It Be Nice" (B. Wilson/Asher/Love)
I would have to look for this pattern:
" (<any string>)
In order to retrieve:
B. Wilson/Asher/Love
I tried something like "" (([^))]*)) but it doesn't seem to work. Also, I'd like to use Match.Submatches(0) so that might complicate things a bit because it relies on brackets...
Edit: After examining your document, the problem is that there are non-breaking spaces before the parentheses, not regular spaces. So this regex should work: ""[ \xA0]*\(([^)]+)\)
"" 'quote (twice to escape)
[ \xA0]* 'zero or more non-breaking (\xA0) or a regular spaces
\( 'left parenthesis
( 'open capturing group
[^)]+ 'anything not a right parenthesis
) 'close capturing group
\) 'right parenthesis
In a function:
Public Function GetStringInParens(search_str As String)
Dim regEx As New VBScript_RegExp_55.RegExp
Dim matches
GetStringInParens = ""
regEx.Pattern = """[ \xA0]*\(([^)]+)\)"
regEx.Global = True
If regEx.test(search_str) Then
Set matches = regEx.Execute(search_str)
GetStringInParens = matches(0).SubMatches(0)
End If
End Function
Not strictly an answer to your question, but sometimes, for things this simple, good ol' string functions are less confusing and more concise than Regex.
Function BetweenParentheses(s As String) As String
BetweenParentheses = Mid(s, InStr(s, "(") + 1, _
InStr(s, ")") - InStr(s, "(") - 1)
End Function
Usage:
Debug.Print BetweenParentheses("""Wouldn't It Be Nice"" (B. Wilson/Asher/Love)")
'B. Wilson/Asher/Love
EDIT #alan points our that this will falsely match the contents of parentheses in the song title. This is easily circumvented with a little modification:
Function BetweenParentheses(s As String) As String
Dim iEndQuote As Long
Dim iLeftParenthesis As Long
Dim iRightParenthesis As Long
iEndQuote = InStrRev(s, """")
iLeftParenthesis = InStr(iEndQuote, s, "(")
iRightParenthesis = InStr(iEndQuote, s, ")")
If iLeftParenthesis <> 0 And iRightParenthesis <> 0 Then
BetweenParentheses = Mid(s, iLeftParenthesis + 1, _
iRightParenthesis - iLeftParenthesis - 1)
End If
End Function
Usage:
Debug.Print BetweenParentheses("""Wouldn't It Be Nice"" (B. Wilson/Asher/Love)")
'B. Wilson/Asher/Love
Debug.Print BetweenParentheses("""Don't talk (yell)""")
' returns empty string
Of course this is less concise than before!
This a nice regex
".*\(([^)]*)
In VBA/VBScript:
Dim myRegExp, ResultString, myMatches, myMatch As Match
Dim myRegExp As RegExp
Set myRegExp = New RegExp
myRegExp.Pattern = """.*\(([^)]*)"
Set myMatches = myRegExp.Execute(SubjectString)
If myMatches.Count >= 1 Then
Set myMatch = myMatches(0)
If myMatch.SubMatches.Count >= 3 Then
ResultString = myMatch.SubMatches(3-1)
Else
ResultString = ""
End If
Else
ResultString = ""
End If
This matches
Put Your Head on My Shoulder
in
"Don't Talk (Put Your Head on My Shoulder)"
Update 1
I let the regex loose on your doc file and it matches as requested. Quite sure the regex is fine. I'm not fluent in VBA/VBScript but my guess is that's where it goes wrong
If you want to discuss the regex some further that's fine with me. I'm not eager to start digging into this VBscript API which looks arcane.
Given the new input the regex is tweaked to
".*".*\(([^)]*)
So that it doesn't falsely match (Put Your Head on My Shoulder) which appears inside the quotes.
This function worked on your example string:
Function GetArtist(songMeta As String) As String
Dim artist As String
' split string by ")" and take last portion
artist = Split(songMeta, "(")(UBound(Split(songMeta, "(")))
' remove closing parenthesis
artist = Replace(artist, ")", "")
End Function
Ex:
Sub Test()
Dim songMeta As String
songMeta = """Wouldn't It Be Nice"" (B. Wilson/Asher/Love)"
Debug.Print GetArtist(songMeta)
End Sub
prints "B. Wilson/Asher/Love" to the Immediate Window.
It also solves the problem alan mentioned. Ex:
Sub Test()
Dim songMeta As String
songMeta = """Wouldn't (It Be) Nice"" (B. Wilson/Asher/Love)"
Debug.Print GetArtist(songMeta)
End Sub
also prints "B. Wilson/Asher/Love" to the Immediate Window. Unless of course, the artist names also include parentheses.
This another Regex tested with a vbscript (?:\()(.*)(?:\)) Demo Here
Data = """Wouldn't It Be Nice"" (B. Wilson/Asher/Love)"
wscript.echo Extract(Data)
'---------------------------------------------------------------
Function Extract(Data)
Dim strPattern,oRegExp,Matches
strPattern = "(?:\()(.*)(?:\))"
Set oRegExp = New RegExp
oRegExp.IgnoreCase = True
oRegExp.Pattern = strPattern
set Matches = oRegExp.Execute(Data)
If Matches.Count > 0 Then Extract = Matches(0).SubMatches(0)
End Function
'---------------------------------------------------------------
I think you need a better data file ;) You might want to consider pre-processing the file to a temp file for modification, so that outliers that don't fit your pattern are modified to where they'll meet your pattern. It's a bit time consuming to do, but it is always difficult when a data file lacks consistency.
I need help extracting the value of a wildcard from a Regular Expressions match. For example:
Regex: "I like *"
Input: "I like chocolate"
I would like to be able to extract the string "chocolate" from the Regex match (or whatever else is there). If possible, I also want to be able to retrieve several wildcard values from a single wildcard match. For example:
Regex: "I play the * and the *"
Input: "I play the guitar and the bass"
I want to be able to extract both "guitar" and "bass". Is there a way to do it?
In general regex utilize the concepts of groups. Groups are indicated by parenthesis.
So I like
Would be I like (.) . = All character * meaning as many or none of the preceding character
Sub Main()
Dim s As String = "I Like hats"
Dim rxstr As String = "I Like(.*)"
Dim m As Match = Regex.Match(s, rxstr)
Console.WriteLine(m.Groups(1))
End Sub
The above code will work for and string that has I Like and will print out all characters after including the ' ' as . matches even white space.
Your second case is more interesting because the first rx will match the entire end of the string you need something more restrictive.
I Like (\w+) and (\w+) : this will match I Like then a space and one or more word characters and then an and a space and one or more word characters
Sub Main()
Dim s2 As String = "I Like hats and dogs"
Dim rxstr2 As String = "I Like (\w+) and (\w+)"
Dim m As Match = Regex.Match(s2, rxstr2)
Console.WriteLine("{0} : {1}", m.Groups(1), m.Groups(2))
End Sub
For a more complete treatment of regex take a look at this site which has a great tutorial.
Here is my RegexExtract Function in VBA. It will return just the sub match you specify (only the stuff in parenthesis). So in your case, you'd write:
=RegexExtract(A1, "I like (.*)")
Here is the code.
Function RegexExtract(ByVal text As String, _
ByVal extract_what As String) As String
Application.ScreenUpdating = False
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)
Application.ScreenUpdating = True
End Function
Here is a version that will allow you to use multiple groups to extract multiple parts at once:
Function RegexExtract(ByVal text As String, _
ByVal extract_what As String) As String
Application.ScreenUpdating = False
Dim allMatches As Object
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
Dim i As Long
Dim result As String
RE.Pattern = extract_what
RE.Global = True
Set allMatches = RE.Execute(text)
For i = 0 To allMatches.Item(0).submatches.count - 1
result = result & allMatches.Item(0).submatches.Item(i)
Next
RegexExtract = result
Application.ScreenUpdating = True
End Function
I would like to split a string into an array according to a regular expression similar to what can be done with preg_split in PHP or VBScript Split function but with a regex in place of delimiter.
Using VBScript Regexp object, I can execute a regex but it returns the matches (so I get a collection of my splitters... that's not what I want)
Is there a way to do so ?
Thank you
If you can reserve a special delimiter string, i.e. a string that you can choose that will never be a part of the real input string (perhaps something like "###"), then you can use regex replacement to replace all matches of your pattern to "###", and then split on "###".
Another possibility is to use a capturing group. If your delimiter regex is, say, \d+, then you search for (.*?)\d+, and then extract what the group captured in each match (see before and after on rubular.com).
You can alway use the returned array of matches as input to the split function. You split the original string using the first match - the first part of the string is the first split, then split the remainder of the string (minus the first part and the first match)... continue until done.
I wrote this for my use. Might be what you're looking for.
Function RegSplit(szPattern, szStr)
Dim oAl, oRe, oMatches
Set oRe = New RegExp
oRe.Pattern = "^(.*)(" & szPattern & ")(.*)$"
oRe.IgnoreCase = True
oRe.Global = True
Set oAl = CreateObject("System.Collections.ArrayList")
Do
Set oMatches = oRe.Execute(szStr)
If oMatches.Count > 0 Then
oAl.Add oMatches(0).SubMatches(2)
szStr = oMatches(0).SubMatches(0)
Else
oAl.Add szStr
Exit Do
End If
Loop
oAl.Reverse
RegSplit = oAl.ToArray
End Function
'**************************************************************
Dim A
A = RegSplit("[,|;|#]", "bob,;joe;tony#bill")
WScript.Echo Join(A, vbCrLf)
Returns:
bob
joe
tony
bill
I think you can achieve this by using Execute to match on the required splitter string, but capturing all the preceding characters (after the previous match) as a group. Here is some code that could do what you want.
'// Function splits a string on matches
'// against a given string
Function SplitText(strInput,sFind)
Dim ArrOut()
'// Don't do anything if no string to be found
If len(sFind) = 0 then
redim ArrOut(0)
ArrOut(0) = strInput
SplitText = ArrOut
Exit Function
end If
'// Define regexp
Dim re
Set re = New RegExp
'// Pattern to be found - i.e. the given
'// match or the end of the string, preceded
'// by any number of characters
re.Pattern="(.*?)(?:" & sFind & "|$)"
re.IgnoreCase = True
re.Global = True
'// find all the matches >> match collection
Dim oMatches: Set oMatches = re.Execute( strInput )
'// Prepare to process
Dim oMatch
Dim ix
Dim iMax
'// Initialize the output array
iMax = oMatches.Count - 1
redim arrOut( iMax)
'// Process each match
For ix = 0 to iMax
'// get the match
Set oMatch = oMatches(ix)
'// Get the captured string that precedes the match
arrOut( ix ) = oMatch.SubMatches(0)
Next
Set re = nothing
'// Check if the last entry was empty - this
'// removes one entry if the string ended on a match
if arrOut(iMax) = "" then Redim Preserve ArrOut(iMax-1)
'// Return the processed output
SplitText = arrOut
End Function