Regular Expressions and Replace Function

ttbhandari · #1 (**permalink**) May 17th, 2005, 03:50 PM

My valid characters for username are letters and numbers only.
so if the username is abc*67% i want to replace the invalid chars with nothiing. so the correct result should be abc67. i get abc67%.

I am able to replace only one invalid character. the others stay as it is . any input?

this is my code:-

    username = rtrim(rssqlusername("ag_username"))

    'Test the ag_username for valid chars
    ' Valid Characters Include Letters and Numbers only

    set Regexforinvaliddata = New RegExp
    Regexforinvaliddata.Pattern ="[^A-Za-z0-9 ]"
    set colMatches = Regexforinvaliddata.execute(username)

'Step through our matches
    for each Regexforinvaliddata in colmatches

            searchchar = Regexforinvaliddata.value

            username = replace(username, searchchar, "")
    Next

pgtips · #2 (**permalink**) May 18th, 2005, 05:45 AM

use the replace method of regexp instead of execute
Regexforinvaliddata.replace username, ""

ttbhandari · #3 (**permalink**) May 18th, 2005, 08:28 AM

I dont have to replace the entire username with nulls but
only the invalid characters --which i have to find with the execute method.

pgtips · #4 (**permalink**) May 18th, 2005, 02:00 PM

you misunderstand what the regexp replace method does. It replaces all characters that match the regular expression, not all characters in the string. Try it!

chipset · #5 (**permalink**) May 20th, 2005, 05:10 PM

why not write a function to perform the operation
for you
here is the one i have written;

Code:

Option Explicit
Const InvalideChar As String = "~`!@#$%^&*()_-+=|\{}[];':<>,.?/"


Private Function RemoveInvalideChar(Expression As String, _
                                sInvalideChar As String)


    Dim i As Integer            'counter
    Dim sChar As String         'hold single charater
    Dim iLenChar As Integer     'lenght of all character

    iLenChar = Len(sInvalideChar)
    For i = 1 To iLenChar
        sChar = Mid$(sInvalideChar, i, 1)
        Expression = Replace(Expression, sChar, vbNullString)
    Next i
    RemoveInvalideChar = Replace(Expression, Chr(34), vbNullString) ' chr(34) is "
End Function

hope it helps

Arowolo