I have a string variable.
Dim str As String = "ABBCD"
I want to replace only the second 'B' character of str (I mean the second occurrence)
my code
Dim regex As New Regex("B")
Dim result As String = regex.Replace(str, "x", 2)
'result: AxxCD
'but I want: ABxCD
What's the easiest way to do this with Regular Expressions.
thanks
Dim str As String = "ABBCD"
Dim matches As MatchCollection = Regex.Matches(str, "B")
If matches.Count >= 2 Then
str = str.Remove(matches(1).Index, matches(1).Length)
str = str.Insert(matches(1).Index, "x")
End If
First we declare the string 'str', then find the matches of "B". If we found two results or more, replace the second result with "x".
How about:
resultString = Regex.Replace(subjectString, #"(B)\1", "$+x");
Use a positive lookbehind:
Dim regex As New Regex("(?<=B)B")
Live demo
If ABCABCABC should produce ABCAxCABC, then the following regex will work:
(?<=^[^B]*B[^B]*)B
Usage:
Dim result As String = Regex.Replace(str, "(?<=^[^B]*B[^B]*)B", "x")
I assume BB was just an example, it can be CC, DD, EE, etc..
Based on that, the regex below will replace any repeated character in the string.
resultString = Regex.Replace(subjectString, #"(\w)\1", "$1x");
'Alternative way to replace the second occurrence
'only of B in the string with X
Dim str As String = "ABBCD"
Dim pattern As String = "B"
Dim reg As Regex = New Regex(pattern)
Dim replacement As String = "X"
'find position of second B
Dim secondBpos As Integer = Regex.Matches(str, pattern)(1).Index
'replace that B with X
Dim result As String = reg.Replace(str, replacement, 1, secondBpos)
MessageBox.Show(result)
Related
Sorry for my simple question but I don't know how to do.
I have this string:
Dim SourceString = "{capital?} has a bridge for {people?}"
Now I want ResultString like this:
ResultString = "capital_Den has a bridge for people_Den"
I used
Dim str As String = "{capital?} has a bridge for {people?}"
Dim str1 As String str1 = Regex.Replace(str, "\{?\?\}", "_DEN}")
Result: {capital_DEN} has a bridge for {people_DEN}
But I want this result: capital_DEN has a bridge for people_DEN
The \{?\?\} pattern matches an optional {, a ? and then a } char.
You may use
str1 = Regex.Replace(str, "\{(\w+)\?\}", "$1_DEN")
Or, if there can be more than just word chars inside:
str1 = Regex.Replace(str, "\{([^{}]+)\?\}", "$1_DEN")
See the VB.NET demo online and the regex demo. The pattern matches:
\{ - a { char
(\w+) - Group 1: one or more word chars
[^{}]+ - 1+ chars other than { and }
\?\} - a ?} substring.
Full VB.NET code snippet:
Dim str As String = "{capital?} has a bridge for {people?}"
Dim str1 As String
str1 = Regex.Replace(str, "\{(\w+)\?\}", "$1_DEN")
Console.WriteLine(str1)
' -> capital_DEN has a bridge for people_DEN
First Make a ConsoleApplication
Module Module1
Sub Main()
Console.Title = "Combine"
Dim a As String = "capital_Den"
Dim b As String = "people_Mar"
Dim ResultString As String = a & " has a bridge for " & b
Console.ForegroundColor = ConsoleColor.Green
Console.WriteLine(ResultString)
Console.ReadKey()
End Sub
End Module
My application stores the path of the userdata.dll in a String.
I need to convert this string: C:\Applications\User\userdata.dll
into this: C:\\Applications\\User\\userdata.dll
All \ will need to be duplicated, independent on how many \ the path have.
Something like:
Dim defaultPath As String = "C:\Applications\User\userdata.dll"
' Regex
Dim r As Regex = New Regex( ... )
' This is the replacement string
Dim Replacement As String = " ... $1 ... "
' Replace the matched text in the InputText using the replacement pattern
Dim modifiedPath As String = r.Replace(defaultPath,Replacement)
Any help on this? I am trying to follow this question:
How to replace some part of this string with vb.net?
But cant find out how to make this Regex...
You can use
Dim pattern As String = "\\"
Dim rgx As New Regex(pattern)
Dim input As String = "C:\Applications\User\userdata.dll"
Dim result As String = rgx.Replace(input, "\\")
Console.WriteLine(result)
Ideone Demo
If you mean to say that replace any number of \ to \\, then you can use
Dim pattern As String = "\\+"
Dim rgx As New Regex(pattern)
Dim input As String = "C:\\\\Applications\User\userdata.dll"
Dim result As String = rgx.Replace(input, "\\")
Ideone Demo
How do I access the .index method in regex, so in this case it should output the location of the first instance of a number
Dim sourceString As String = "abcdefg12345"
Dim textboxregex As Regex = New Regex("^(d)$")
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
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 am trying to separate numbers from a string which includes %,/,etc for eg (%2459348?:, or :2434545/%). How can I separate it, in VB.net
you want only the numbers right?
then you could do it like this
Dim theString As String = "/79465*44498%464"
Dim ret = Regex.Replace(theString, "[^0-9]", String.Empty)
hth
edit:
or do you want to split by all non number chars?
then it would go like this
Dim ret = Regex.Split(theString, "[^0-9]")
You could loop through each character of the string and check the .IsNumber() on it.
This should do:
Dim test As String = "%2459348?:"
Dim match As Match = Regex.Match(test, "\d+")
If match.Success Then
Dim result As String = match.Value
' Do something with result
End If
Result = 2459348
Here's a function which will extract all of the numbers out of a string.
Public Function GetNumbers(ByVal str as String) As String
Dim builder As New StringBuilder()
For Each c in str
If Char.IsNumber(c) Then
builder.Append(c)
End If
Next
return builder.ToString()
End Function