[RESOLVED] Remove trailing NewLines from string

I cant get my head around this, I have a string that may contain more than one newline at the end, and I want to trim them off, but I cant use String.Trim because the string my contain a space as the last char before the Newlines.
:confused: thanks in advance

Use the string functions to replace the chars you want. .Replace, .IndexOf, .LastIndexOf, and .SubString are what you want to look into using depending on your actual string contents.

mystring = mystring.Substring(0, mystring.lastindexof(" "))

?

Quote:

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Originally Posted by wossname

mystring = mystring.Substring(0, mystring.lastindexof(" "))

?

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no that doesnt work

VB Code:

Data = "abcdefghijkl" & vbNewLine & vbNewLine & vbNewLine
Data = Data.Substring(0, Data.LastIndexOf(" "))
MessageBox.Show(Data)

gives exception

Quote:

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Originally Posted by RobDog888

Use the string functions to replace the chars you want. .Replace, .IndexOf, .LastIndexOf, and .SubString are what you want to look into using depending on your actual string contents.

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yes thats what ive been trying and I cant get it to work. Replace is not an option as there is other newlines within the string

give us an example string as we can not write the code without an example. ;)

Quote:

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Originally Posted by RobDog888

give us an example string as we can not write the code without an example. ;)

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take a look at the post before yours ;)

There is no space in your string?

yes your right, i dont need to use a space??? you mean newline? in my test string because i'm not interested in the replace function as that will remove all newlines, for arguments sake lets say the string looks like this then, but there could be more or less newlines at the end

Code:

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"abc" & vbCrLf & "x zy " & vbcrlf & vbcrlf & "ab c de " & vbcrlf & vbcrlf & vbcrlf

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Why not simply loop through your string

While myString.(mystring.lenth -1) = Vbcrlf
'Remove Vbcrlf with Substring
Next

ive tried this but it does nothing :(

VB Code:

While Data.Substring(Data.Length - 1, 1) = vbCrLf
            Data = Data.Substring(Data.Length - 1)
        End While

So you only want one newline and never have two or more of them together, correct?

Quote:

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Originally Posted by RobDog888

So you only want one newline and never have two or more of them together, correct?

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no sorry I want to remove all new lines at the end of the string but leaving any spaces intact

Newlines are different then spaces. If you want to remove all newlines then Replace is what you want. :confused:

noooo not replace! for it is evil, it will take away all my newlines within the string

Code:

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"abc" & vbCrLf & "x zy " & vbcrlf & vbcrlf & "ab c de " & vbcrlf & vbcrlf & vbcrlf

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:)

You can use Regex.Replace, although I still am not quite sure what you meant by your post just above. Also, Environment.Newline should be used instead of vbcrlf...

VB Code:

Dim MyLines As String = "this " & Environment.NewLine & Environment.NewLine & _
                                "and that" & Environment.NewLine & " and another!" 
MessageBox.Show(System.Text.RegularExpressions.Regex.Replace(MyLines, Environment.NewLine, ""))

Ok, now that will need the use of SubString and LastIndexOf.

VB Code:

Dim str As String = "abc" & vbCrLf & "x zy " & vbCrLf & vbCrLf & "ab c de " & vbCrLf & vbCrLf & vbCrLf
Do While str.LastIndexOf(Environment.NewLine) = str.Length - 2
    str = str.Substring(0, str.LastIndexOf(Environment.NewLine) - 2)
Loop
MessageBox.Show(str)

thx robdogg but that removes the final two chars of my string which are not newlines, it removes "e " from the end, where all other chars including spaces should remain intact

Try this...Should remove the last Newline, regardless of where it is positioned at...

VB Code:

Dim MyLines As String = "this " & Environment.NewLine & Environment.NewLine & _
                                      "and that" & Environment.NewLine & " and another!"
MessageBox.Show(MyLines)
Dim MyBeginning As String = MyLines.Substring(0, MyLines.LastIndexOf(Environment.NewLine))
Dim MyEnd As String = MyLines.Substring(MyLines.LastIndexOf(Environment.NewLine) + 2, _
                              MyLines.Length - MyLines.LastIndexOf(Environment.NewLine) - 2)
MessageBox.Show(MyBeginning & MyEnd)

Of course, you could combine it all into one line, just wanted to make it a little more readable :)

Quote:

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Originally Posted by gigemboy

Try this...Should remove the last Newline, regardless of where it is positioned at...

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thx gigemboy but what I'm actually looking for is some code which will remove all trailing new lines from the string, but leaving any other char (including spaces) intact

This will remove all carriage return and line feed characters from the end of your string:

VB Code:

myString = myString.TrimEnd(Convert.ToChar(Keys.Return), Convert.ToChar(Keys.LineFeed))

this is my example string

Code:

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"abc" & Environment.NewLine & "x zy " & Environment.NewLine & Environment.NewLine & "ab c de " & Environment.NewLine & Environment.NewLine & Environment.NewLine

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I just want to cut off the NewlInes at the end so it will be

Code:

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"abc" & Environment.NewLine & "x zy " & Environment.NewLine & Environment.NewLine & "ab c de "

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although this is only an example string, the string will be generated programatically so there could be only 1 newline at the end, or maybe 5 if you see what I mean

You can actually do it in a couple lines of code using regex if thats the case, since regex has a handly little "$" character that you can add to only match at the end of the string...

VB Code:

Dim MyLines As String = "this" & Environment.NewLine & _
                                      "and that" & Environment.NewLine & "and another!" & _
                                      Environment.NewLine & Environment.NewLine
        'just shows you the string so you can compare with the result string
        'to see if it looks the same
        MessageBox.Show(MyLines)
        Dim MyMatches As System.Text.RegularExpressions.MatchCollection = _
                System.Text.RegularExpressions.Regex.Matches(MyLines, Environment.NewLine)
        'just displays original count of newlines to see if it works
        MessageBox.Show(MyMatches.Count.ToString)
        MyLines = System.Text.RegularExpressions.Regex.Replace(MyLines, "(" & Environment.NewLine & ")*$", "")
        'just results, shows count of newline character sequence in result
        'as well as result string
        Dim MyMatches2 As System.Text.RegularExpressions.MatchCollection = _
                System.Text.RegularExpressions.Regex.Matches(MyLines, Environment.NewLine)
        MessageBox.Show(MyMatches2.Count.ToString)
        MessageBox.Show(MyLines & " ")

Quote:

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Originally Posted by jmcilhinney

This will remove all carriage return and line feed characters from the end of your string:

VB Code:

myString = myString.TrimEnd(Convert.ToChar(Keys.Return), Convert.ToChar(Keys.LineFeed))

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this works perfectly, thanks a lot mate :thumb:

ps I cant +++++rep you again yet so I will when I can :)

well a summary of my above post can be done in one line doing this:

VB Code:

Dim MyLines As String = "this" & Environment.NewLine & _
                                      "and that" & Environment.NewLine & "and another!" & _
                                      Environment.NewLine & Environment.NewLine
'removes trailing newlines
MyLines = System.Text.RegularExpressions.Regex.Replace(MyLines, "(" & Environment.NewLine & ")*$", "")

The rest was just results to make sure you could see to test.. But JM's post was fairly simple :)

thx to gigemboy and robdog +rep when I can :)

I guess everyone want to do it in one line of vba... i like loops.

bSTR = Trim(bSTR)
While Right(bSTR, Len(bSTR)) = Asc(10) Or Right(bSTR, Len(bSTR)) = Asc(13)
bSTR = Left(bSTR, Len(bSTR) - 1)
Wend

Crud... rushing is never good. But this works:
While Asc(Right(bSTR, 1)) = 10 Or Asc(Right(bSTR, 1)) = 13
bSTR = Left(bSTR, Len(bSTR) - 1)
Debug.Print Len(bSTR)
Wend

Quote:

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Originally Posted by vb1der

Crud... rushing is never good. But this works:
While Asc(Right(bSTR, 1)) = 10 Or Asc(Right(bSTR, 1)) = 13
bSTR = Left(bSTR, Len(bSTR) - 1)
Debug.Print Len(bSTR)
Wend

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This thread is 11.5 years old and it's in the VB.NET forum. Was posting some VBA code really all that useful?

Hello from 2021 !

jmcihinney, I have just created an account for the sole purpose of answering your question.
I am currently at work, half an hour after the end of my shift and I wanted a quick answer
on how to quickly trim any and all vbcrlf s at the end or beginning of my string

I have search google for "VBA trim trailing vbcrlf" and this forum thread is the second result.

The elegant answer you have given involving myString.TrimEnd, does not work because, yes, this is a vb.net answer.
Apparently no one asked this question before for VBA or maybe it is in the catacombs on the second page of google's search results.

So, I'm pretty happy that vb1der has posted a VBA answer.

And I will now test this answer

From VB immediate console

Code:

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bSTR = "mystringblabla" & vbcrlf & vbcrlf & vbclf
print bSTR
mystringblabla


While Asc(Right(bSTR, 1)) = 10 Or Asc(Right(bSTR, 1)) = 13 : bSTR = Left(bSTR, Len(bSTR) - 1) : Wend
print bSTR
mystringblabla
marker------------------marker

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And I made a function out of it too

Code:

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Function TrimEnd(ByVal myString As String, myTrim As String) As String
    If myString = "" Or myTrim = "" Then TrimEnd = myString: Exit Function
    While Right(myString, Len(myTrim)) = myTrim: myString = Left(myString, Len(myString) - Len(myTrim)): Wend: TrimEnd = myString
End Function

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Quote:

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Originally Posted by shodanx

Hello from 2021 !

jmcihinney, I have just created an account for the sole purpose of answering your question.
I am currently at work, half an hour after the end of my shift and I wanted a quick answer
on how to quickly trim any and all vbcrlf s at the end or beginning of my string

I have search google for "VBA trim trailing vbcrlf" and this forum thread is the second result.

The elegant answer you have given involving myString.TrimEnd, does not work because, yes, this is a vb.net answer.
Apparently no one asked this question before for VBA or maybe it is in the catacombs on the second page of google's search results.

So, I'm pretty happy that vb1der has posted a VBA answer.

And I will now test this answer

From VB immediate console

Code:

---

bSTR = "mystringblabla" & vbcrlf & vbcrlf & vbclf
print bSTR
mystringblabla


While Asc(Right(bSTR, 1)) = 10 Or Asc(Right(bSTR, 1)) = 13 : bSTR = Left(bSTR, Len(bSTR) - 1) : Wend
print bSTR
mystringblabla
marker------------------marker

---

And I made a function out of it too

Code:

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Function TrimEnd(ByVal myString As String, myTrim As String) As String
    If myString = "" Or myTrim = "" Then TrimEnd = myString: Exit Function
    While Right(myString, Len(myTrim)) = myTrim: myString = Left(myString, Len(myString) - Len(myTrim)): Wend: TrimEnd = myString
End Function

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Does the fact that someone not interested in VB.NET justify polluting the VB.NET forum with code that is of no use to anyone interested in VB.NET? No, it does not. We could pollute this forum with all sorts of other languages and even non-programming information and it might be useful to all sorts of people, but this is a VB.NET forum, so it's purpose is not to help those people. Maybe vb1der should have posted that code to a VBA site somewhere and then it would have been just as useful to you without crapping on this thread and this forum. I'm happy that you found an answer to your question but I don't want to have to wade through all sorts of garbage to get to the VB.NET stuff in this forum and I doubt anyone else does either.

I understand that however, this is the most one of the enticing google result for this query.

And it contains the correct answer for this question.

The problem is that there is no reliable way to change the google results to point toward the right answer in the right forum section.

The fact that this thread in particular has 17 thousand views is an indicator that a LOT of people are searching for this answer and hopefully are finding it here.

here is what the search result page looks from my end. I immediately knew I was on the right track when I saw it !

https://imgur.com/LxCD4Aj

Quote:

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Originally Posted by shodanx

I understand that however

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Let me finish that for you:

Quote:

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I don't care because it helped me so screw what this forum is actually for.

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Have you made the effort to put this information in a logical place for other VBA developers? No, I didn't think so. I guess I'll join a knitting circle and post VBA answers on their message board because who cares what that board is actually for, right? This thread is a question about a VB.NET problem. That is its purpose. As such, an answer using VBA code is, by definition, not useful. The fact that it might be useful for some other purpose isn't relevant to that. Should we start posting Béarnaise sauce recipes here because someone might find that useful? I get that that post was useful for you and probably other VBA developers but that is completely irrelevant to my comment. This thread and this forum exist to help people with VB.NET problems so VBA code is NOT useful and should be posted somewhere that exists to help people with VBA problems. Maybe do that instead of wasting your time telling me that I should be OK with this VB.NET forum being polluted with irrelevant garbage by people who couldn't be bothered posting it in a sensible location because that might take a bit of extra effort on their part.

Quote:

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Originally Posted by shodanx

And I made a function out of it too

Code:

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Function TrimEnd(ByVal myString As String, myTrim As String) As String
    If myString = "" Or myTrim = "" Then TrimEnd = myString: Exit Function
    While Right(myString, Len(myTrim)) = myTrim: myString = Left(myString, Len(myString) - Len(myTrim)): Wend: TrimEnd = myString
End Function

---

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the Function might work, but ever heard of replace

Quote:

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Originally Posted by jmcilhinney

This will remove all carriage return and line feed characters from the end of your string:

VB Code:

myString = myString.TrimEnd(Convert.ToChar(Keys.Return), Convert.ToChar(Keys.LineFeed))

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Quote:

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Originally Posted by the182guy

this works perfectly, thanks a lot mate :thumb:

ps I cant +++++rep you again yet so I will when I can :)

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@jm… I know this is old but…

myString = myString.TrimEnd(vbCr, vbLf)

:D:D:D

Quote:

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Originally Posted by .paul.

@jm… I know this is old but…

myString = myString.TrimEnd(vbCr, vbLf)

:D:D:D

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Yeah, I wouldn't do it that way myself any more. I would tend to use this:

vb.net Code:

myString = myString.TrimEnd(ControlChars.Cr, ControlChars.Lf)

but it amounts to the same thing. It's the one thing I've learned in the last decade and a half. ;)

Just thought I'd throw in my two pennyworth... I'd've tried:

Code:

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txt = txt.Replace(vbNewline & vbNewline, vbNewline)

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Poppa

Quote:

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Originally Posted by Poppa Mintin

Just thought I'd throw in my two pennyworth... I'd've tried:

Code:

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txt = txt.Replace(vbNewline & vbNewline, vbNewline)

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Poppa

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You wasted two pence there. That doesn't remove trailing newline characters

Quote:

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Originally Posted by Poppa Mintin

Just thought I'd throw in my two pennyworth... I'd've tried:

Code:

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txt = txt.Replace(vbNewline & vbNewline, vbNewline)

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Poppa

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this is what I used in some applications:

Code:

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txt = txt.Replace(vbCr, "").Replace(vbLf, "")

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but with the trim.end version, my future code will look more elegant ;)