How to Remove Line Break in String

Question

I want to Remove the Line Break from the string if my string Ends with Line Break.

Sub linebreak(myString)
    If Len(myString) <> 0 Then
        If Rig         


        

Answer 1

I know this post is very old but I could not find anything that would do this. So I finally put this together from all the forums.

This will remove All line breaks from the left and right, and removes any blank lines. Then Trim it all to look good. Alter as you see fit. I also made one that removes All Line Breaks, if interested. TY

    Sub RemoveBlankLines()
    Application.ScreenUpdating = False
    Dim rngCel As Range
    Dim strOldVal As String
    Dim strNewVal As String

    For Each rngCel In Selection
        If rngCel.HasFormula = False Then
            strOldVal = rngCel.Value
            strNewVal = strOldVal
            Debug.Print rngCel.Address

            Do
            If Left(strNewVal, 1) = vbLf Then strNewVal = Right(strNewVal, Len(strNewVal) - 1)
            If strNewVal = strOldVal Then Exit Do
                strOldVal = strNewVal
            Loop

            Do
            If Right(strNewVal, 1) = vbLf Then strNewVal = Left(strNewVal, Len(strNewVal) - 1)
            If strNewVal = strOldVal Then Exit Do
                strOldVal = strNewVal
            Loop

            Do
            strNewVal = Replace(strNewVal, vbLf & vbLf, "^")
            strNewVal = Replace(strNewVal, Replace(String(Len(strNewVal) - _
                        Len(Replace(strNewVal, "^", "")), "^"), "^", "^"), "^")
            strNewVal = Replace(strNewVal, "^", vbLf)

            If strNewVal = strOldVal Then Exit Do
                strOldVal = strNewVal
            Loop

            If rngCel.Value <> strNewVal Then
                rngCel = strNewVal
            End If

        rngCel.Value = Application.Trim(rngCel.Value)
        End If
    Next rngCel
    Application.ScreenUpdating = True
End Sub

Answer 2

I had the exact same issue. I made a separate function I can call easily when needed:

Function removeLineBreakIfAtEnd(s As String) As String
    If Right(s, 1) = vbLf Then s = Left(s, Len(s) - 2)
    removeLineBreakIfAtEnd = s
End Function

I found that I needed to check the last character only and do -2 to remove the line break. I also found that checking for vbLf was the ONLY way to detect the line break. The function can be called like this:

Sub MainSub()
    Dim myString As String
    myString = "Hello" & vbCrLf
    myString = removeLineBreakIfAtEnd(myString)
    MsgBox ("Here is the resulting string: '" & myString & "'")
End Sub

Answer 3

Clean function can be called from VBA this way:

Range("A1").Value = Application.WorksheetFunction.Clean(Range("A1"))

However as written here, the CLEAN function was designed to remove the first 32 non-printing characters in the 7 bit ASCII code (values 0 through 31) from text. In the Unicode character set, there are additional nonprinting characters (values 127, 129, 141, 143, 144, and 157). By itself, the CLEAN function does not remove these additional nonprinting characters.

Rick Rothstein have written code to handle even this situation here this way:

Function CleanTrim(ByVal S As String, Optional ConvertNonBreakingSpace As Boolean = True) As String
  Dim X As Long, CodesToClean As Variant
  CodesToClean = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, _
                       21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 127, 129, 141, 143, 144, 157)
  If ConvertNonBreakingSpace Then S = Replace(S, Chr(160), " ")
  For X = LBound(CodesToClean) To UBound(CodesToClean)
    If InStr(S, Chr(CodesToClean(X))) Then S = Replace(S, Chr(CodesToClean(X)), "")
  Next
  CleanTrim = WorksheetFunction.Trim(S)
End Function

Answer 4

mystring = Replace(mystring, Chr(10), "")

Answer 5

Private Sub Form_Load()
    Dim s As String

    s = "test" & vbCrLf & "this" & vbCrLf & vbCrLf
    Debug.Print (s & Len(s))

    s = Left(s, Len(s) - Len(vbCrLf))
    Debug.Print (s & Len(s))
End Sub

Answer 6

just this:

str = Replace(str, vbCrLf, "")