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Re: Replace Function - 5/22/2001 11:57:21 PM
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hennykoe
Posts: 9
Score: 0
Joined: 5/23/2001
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Status: offline
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Perhaps try doing the following function: Function ReplaceQuote(inStr) Dim inStrLen, quotChar inStrLen = Len(inStr) quotChar = Chr(34) Replace(inStr, quotChar, "'", 1, inStrLen) End Function Syntax: Replace(String, FindSTR, ReplaceSTR, Start, Count, Compare)
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