by jjo (5 Submissions)
Category: String Manipulation
Compatability: Visual Basic 3.0
Difficulty: Unknown Difficulty
Originally Published: Wed 26th June 2002
Date Added: Mon 8th February 2021
Rating: 
(1 Votes)
Various functions related to trim(). The difference is that you can trim a string for newline characters, whitespace or any character you wish.
API Declarations
' Trim functions
' jjo 2002
VBcoders
ByVal fjern As String, _
Optional OptComp As Integer = 0) As String
If ((Len(s) = 0) Or (Len(fjern) = 0)) Then
LTrimChr = s: Exit Function
End If
Dim idx As Long
fjern = Left(fjern, 1)
For idx = 1 To Len(s)
If (StrComp(Mid(s, idx, 1), fjern, OptComp) <> 0) Then Exit For
Next
LTrimChr = Mid(s, idx)
End Function
Public Function RTrimChr(ByVal s As String, _
ByVal fjern As String, _
Optional OptComp As Integer = 0) As String
If ((Len(s) = 0) Or (Len(fjern) = 0)) Then
RTrimChr = s: Exit Function
End If
Dim idx As Long
fjern = Left(fjern, 1)
For idx = Len(s) To 1 Step -1
If (StrComp(Mid(s, idx, 1), fjern, OptComp) <> 0) Then Exit For
Next
RTrimChr = Mid(s, 1, idx)
End Function
Public Function TrimChr(s As String, _
tegn As String, _
Optional OptComp As Integer = 0) As String
TrimChr = LTrimChr(RTrimChr(s, tegn, OptComp), tegn, OptComp)
End Function
Public Function LTrimChr2(ByVal s As String, _
ByVal chr1 As String, _
ByVal chr2 As String, _
Optional OptComp As Integer = 0) As String
If ((Len(s) = 0) Or ((Len(chr1) = 0) Or (Len(chr2) = 0))) Then
LTrimChr2 = s: Exit Function
End If
Dim c As String, idx As Long
chr1 = Left(chr1, 1)
chr2 = Left(chr2, 1)
For idx = 1 To Len(s)
c = Mid(s, idx, 1)
If (StrComp(c, chr1, OptComp) <> 0) And _
(StrComp(c, chr2, OptComp) <> 0) Then
Exit For
End If
Next
LTrimChr2 = Mid(s, idx)
End Function
Public Function RTrimChr2(ByVal s As String, _
ByVal chr1 As String, _
ByVal chr2 As String, _
Optional OptComp As Integer = 0) As String
If ((Len(s) = 0) Or ((Len(chr1) = 0) Or (Len(chr2) = 0))) Then
RTrimChr2 = s: Exit Function
End If
Dim c As String, idx As Long
chr1 = Left(chr1, 1)
chr2 = Left(chr2, 1)
For idx = Len(s) To 1 Step -1
c = Mid(s, idx, 1)
If (StrComp(c, chr1, OptComp) <> 0) And _
(StrComp(c, chr2, OptComp) <> 0) Then
Exit For
End If
Next
RTrimChr2 = Mid(s, 1, idx)
End Function
Public Function TrimChr2(s As String, _
chr1 As String, _
chr2 As String, _
Optional OptComp As Integer = 0) As String
TrimChr2 = LTrimChr2(RTrimChr2(s, chr1, chr2, OptComp), chr1, chr2, OptComp)
End Function
Public Function LTrimLF(s As String) As String
LTrimLF = LTrimChr(s, Chr$(10), 0)
End Function
Public Function RTrimLF(s As String) As String
RTrimLF = RTrimChr(s, Chr$(10), 0)
End Function
'Line feed
'vbLf: chr(10)
Public Function TrimLF(s As String) As String
TrimLF = LTrimLF(RTrimLF(s))
End Function
Public Function LTrimCR(s As String) As String
LTrimCR = LTrimChr(s, Chr$(13), 0)
End Function
Public Function RTrimCR(s As String) As String
RTrimCR = RTrimChr(s, Chr$(13), 0)
End Function
'Carriage return
'vbCr: chr(13)
Public Function TrimCR(s As String) As String
TrimCR = LTrimCR(RTrimCR(s))
End Function
Public Function LTrimNL(s As String) As String
LTrimNL = LTrimChr2(s, Chr(10), Chr(13), 0)
End Function
Public Function RTrimNL(s As String) As String
RTrimNL = RTrimChr2(s, Chr(10), Chr(13), 0)
End Function
'New line
'vbLf : chr(10)
'vbCr : chr(13) and vbCrLf
Public Function TrimNL(s As String) As String
TrimNL = LTrimNL(RTrimNL(s))
End Function
Public Function LTrimWS(s As String) As String
LTrimWS = LTrimChr2(s, Chr$(9), Chr$(32), 0)
End Function
Public Function RTrimWS(s As String) As String
RTrimWS = RTrimChr2(s, Chr$(9), Chr$(32), 0)
End Function
'White space
'Tab : chr(9)
'Space : chr(32)
Public Function TrimWS(s As String) As String
TrimWS = LTrimWS(RTrimWS(s))
End Function
Public Function LTrimPATH(s As String) As String
LTrimPATH = LTrimChr2(s, "/", "\", 0)
End Function
Public Function RTrimPATH(s As String) As String
RTrimPATH = RTrimChr2(s, "/", "\", 0)
End Function
'Path separators
Public Function TrimPATH(s As String) As String
TrimPATH = LTrimPATH(RTrimPATH(s))
End Function
Sub TrimWS_test()
Dim a As String, b As String, c As Integer
b = vbTab & Space(1)
For c = 1 To 4: b = b & b: Next
a = b & "_" & b
MsgBox "'" & TrimWS(a) & "'", 0, "TrimWS()"
'MsgBox a, 0, ""
End Sub
Sub TrimPATH_test()
Dim a As String, b As String, c As Integer
b = "\/"
For c = 1 To 4: b = b & b: Next
a = b & "_" & b
MsgBox "'" & TrimPATH(a) & "'", 0, "TrimPATH()"
'MsgBox a, 0, ""
End Sub
Sub TrimNL_test()
Dim a As String, b As String, c As Integer
b = vbLf & vbCr & vbCrLf
For c = 1 To 2: b = b & b: Next
a = b & "_" & b
MsgBox "'" & TrimNL(a) & "'", 0, "NewLine"
'MsgBox "'" & a & "'", 0, ""
End Sub
Sub TrimChr_test()
Dim a As String, b As String, c As Integer
b = "qQ"
For c = 1 To 4: b = b & b: Next
a = b & "_" & b
MsgBox "'" & TrimChr(a, "q", 1) & "'", 0, "TrimChr()"
'MsgBox a, 0, ""
End Sub
No comments have been posted about Various functions related to trim(). The difference is that you can trim a string for newline chara. Why not be the first to post a comment about Various functions related to trim(). The difference is that you can trim a string for newline chara.
