[resolved] Wildcards

enoctis · 10-03-2005, 05:35 PM

I am trying to use the Replace() function to remove any parenthesised information in a string. For example:

X = Replace(X, "(" & * & ")", "")

Of course, this doesn't work as VB doesn't recognize * as a valid character. Is there any way to get around this; is there a function for wildcards in VB? Any help is greatly appreciated.

enoctis · 10-03-2005, 07:13 PM

Here is the code, but please don't just copy/paste it. Learn from it > Read the comments!

Code:

Dim X, A, B, C

X = "blah (because) blah"

If InStr(X, "(") > 0 Then '( was found in $X
A = InStr(X, "(") 'returns location of ( as integer
B = Mid(X, A) 'returns (* as string
    If InStr(B, ")") > 0 Then ') was found in $B
    C = InStr(X, ")") 'returns location of ) as integer
    C = (C - A) + 1 'returns length from ( to ) as integer
    B = Mid(X, A, C) 'returns (*)
    X = Replace(X, B, "") 'replaces (*) with blank
    Else ') was not found in $B, do nothing
    End If
End If
X = Replace(X, "  ", " ")

The reason for the last replacement is simple. Without that replacement $X would have two spaces between blah & blah, rather than one.

Hope this helps all of those out who are trying to use wildcards & can't use the Like function for their specific need.

10-03-2005, 05:35 PM	#1 (permalink)
enoctis Registered User Join Date: Aug 2005 Location: Ft. Hood, TX Posts: 59 OS: WinXP.pro My System	Wildcards I am trying to use the Replace() function to remove any parenthesised information in a string. For example: *X = Replace(X, "(" & & ")", "")** Of course, this doesn't work as VB doesn't recognize * as a valid character. Is there any way to get around this; is there a function for wildcards in VB? Any help is greatly appreciated.

10-03-2005, 07:13 PM	#2 (permalink)
enoctis Registered User Join Date: Aug 2005 Location: Ft. Hood, TX Posts: 59 OS: WinXP.pro My System	Work Around Here is the code, but please don't just copy/paste it. Learn from it > Read the comments! Code: Dim X, A, B, C X = "blah (because) blah" If InStr(X, "(") > 0 Then '( was found in $X A = InStr(X, "(") 'returns location of ( as integer B = Mid(X, A) 'returns (* as string If InStr(B, ")") > 0 Then ') was found in $B C = InStr(X, ")") 'returns location of ) as integer C = (C - A) + 1 'returns length from ( to ) as integer B = Mid(X, A, C) 'returns () X = Replace(X, B, "") 'replaces () with blank Else ') was not found in $B, do nothing End If End If X = Replace(X, " ", " ") The reason for the last replacement is simple. Without that replacement $X would have two spaces between blah & blah, rather than one. Hope this helps all of those out who are trying to use wildcards & can't use the Like function for their specific need.