I have this text:
2|#Favo|Name||26.0000|50.10000|_GRE|||||City|Road||||
I want to capture anything between those special chars: ||
For example, I want to capture "Name" only or I want to capture "City"
I've spent many hours and all I came up with is this regex:
([^|].*[$|])\w+
Here are the required values:
How can I capture one of them?
Thank you.
You may split the string with | removing empty entries and also all those that are blank or consisting only of digits:
Dim strng As String = "2|#Favo|Name||26.0000|50.10000|_GRE|||||City|Road||||"
Dim reslt As List(Of String) = strng.Split(New String() {"|"}, StringSplitOptions.RemoveEmptyEntries).Where(
Function(m) m.All(AddressOf Char.IsDigit) = False And String.Equals(m.Trim(), String.Empty) = False).ToList()
Console.Write(String.Join(", ", reslt))
Output:
Related
Need some help writing a regular expression to count the number of words in a string (Please note the data is a html string, which needs to be placed into a spreadsheet) when separated either by any special characters like . , - , +, /, Tab etc. Count should exclude special characters.
**Original String** **End Result**
Ex : One -> 1
One. -> 1
One Two -> 2
One.Two -> 2
One Two. -> 2
One.Two. -> 2
One.Tw.o -> 3
Updated
I think you asked a valuable question and this downvoting is not fair!
Function WCount(ByVal strWrd As String) As Long
'Variable declaration
Dim Delimiters() As Variant
Dim Delimiter As Variant
'Initialization
Delimiters = Array("+", "-", ".", "/", Chr(13), Chr(9)) 'Define your delimiter characters here.
'Core
For Each Delimiter In Delimiters
strWrd = Replace(strWrd, Delimiter, " ")
Next Delimiter
strWrd = Trim(strWrd)
Do While InStr(1, strWrd, " ") > 0
strWrd = Replace(strWrd, " ", " ")
Loop
WCount = UBound(Split(strWrd, " ")) + 1
End Function
________________
You can use this function as a UDF in excel formulas or can use in another VBA codes.
Using in formula
=WCOUNT("One.Two.Three.") or =WCOUNT($A$1") assuming your string is in A1 cell.
Using in VBA
(With assume passing your string with Str argument.)
Sub test()
Debug.Print WCount(Str)
End Sub
Regards.
Update
I have test your text as shown below.
copy your text in a Cell of Excel as shown.
The code updated for Line break and Tab characters and count your string words correctly now.
Try this code, all necessary comments are in code:
Sub SpecialSplit()
Dim i As Long
Dim str As String
Dim arr() As String
Dim delimeters() As String
'here you define all special delimeters you want to use
delimetres = Array(".", "+", "-", "/")
For i = 1 To 9
str = Cells(i, 1).Value
'this will protect us from situation where last character is delimeter and we have additional empty string
str = Left(str, Len(str) - 1)
'here we replace all special delimeters with space to simplify
For Each delimeter In delimetres
str = Replace(str, delimeter, " ")
Next
arr = Split(str)
Cells(i, 2).Value = UBound(arr) - LBound(arr) + 1
Next
End Sub
With your posted data following RegExp is working correctly. Put this in General Module in Visual Basic Editor.
Public Function CountWords(strInput As String) As Long
Dim objMatches
With CreateObject("VBScript.RegExp")
.Global = True
.MultiLine = True
.Pattern = "\w+"
Set objMatches = .Execute(strInput)
CountWords = objMatches.Count
End With
End Function
You have to use it like a normal formula. e.g. assuming data is in cell A1 function would be:
=CountWords(A1)
For your information, it can be also achieved through formula if number of characters are specific like so:
=LEN(TRIM(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(TRIM(A1),"."," "),","," "),"-"," "),"+"," "),"/"," "),"\"," ")))-LEN(SUBSTITUTE(TRIM(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(TRIM(A1),"."," "),","," "),"-"," "),"+"," "),"/"," "),"\"," "))," ",""))+1
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 $, you are trying to match the whole string - not a part as your are stating in the question.
So you either need to remove them or rephrase your question.
"\:(.*)$"
Hi all i am using above expression to remove all the string before : (colon), but it is giving me all the string before this. how can i do this. Thanks a lot.
My string is:
This is text: Hi here we go
I am getting: This is text
I want : Hi here we go
Updated code
Sub Main()
Dim input As String = "This is text with : far too much "
Dim pattern As String = "\:(.*)$"
Dim replacement As String = " "
Dim rgx As New Regex(pattern)
Dim result As String = rgx.Replace(input, replacement)
Console.WriteLine("Original String: {0}", input)
' MsgBox("Original String: {0}")
Console.WriteLine("Replacement String: {0}", result)
MsgBox("Original String: {0}")
End Sub
Try this pattern. This will help you to match string after colon
/?:(.)/
or
/: (.+)/
It should be:
Dim pattern As String = "(.*)\:"
' in vb if above one doesn't work, then try this one
' Dim pattern As String = "^(.*)\:"
' also i don't think we need to use any brackets here as well.
This regex means, anything before the colon(:), Where you were using anything after the colon(:) in your example.
If you are not dead set on RegEx then you can also use
Dim result As String
result = Strings.Split(Input, ":", 2)(1)
This splits the input into an array with two elements. First element is the text before the first ":", the second element is the text after.
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