2 replies to this topic

[blindwig]

I got tired of doing this:

If $i = 1 or $i = 3 or $i = 7 or $i = 12 ...

so I write a function to work like this:

If _StringInSet($i, '1,3,7,12,...')

And here it is:

CODE

Func _StringInSet($Value, $List, $Delim=',')
Dim $Result=0, $ListArray, $i
$ListArray=StringSplit($List,$Delim)
$i = 1
While $i<=$ListArray[0] And Not $Result
If $Value=$ListArray[$i] Then $Result=-1
$i = $i + 1
WEnd
Return $Result
EndFunc

It's smarter than just using StringInStr because it won't match partials.

[MHz]

Hi blindwig,

Yes, good idea. A good UDF, can save some extra typing.
I would have done it a little different.

Func _StringInSetOr($Value, $List, $Delim = '|')
    Dim $ListArray=StringSplit($List,$Delim)
    If @error Then Return 0
    For $i = 1 To $ListArray[0]
        If $Value = $ListArray[$i] Then Return 1
    Next
EndFunc

I try to avoid using comma's as delimiters. I have learnt my lesson with other languages, in the past. Accidently typing a period, somewhere in the string, makes fun debugging. But it is good that you chose as optional.

Or this is with error return

Func _StringInSetOr($Value, $List, $Delim = '|')
    Dim $ListArray=StringSplit($List,$Delim)
    If @error Then
        SetError(1)
        Return 0
    EndIf
    For $i = 1 To $ListArray[0]
        If $Value = $ListArray[$i] Then Return 1
    Next
EndFunc

Edited by MHz, 08 June 2005 - 11:38 AM.

[hou]

This is a very old thread, I know.
However this is my way doing it ...

Func StringInSet( $str, $set, $delim="|" );Tests wether string is in set of strings. Returns item no of str in set if found, 0 if not.
    $set= $delim & $set & $delim
    Local $res= StringInStr($set, $delim & $str & $delim)
    If $res>0 Then
        StringReplace( StringLeft($set,$res), $delim, " " )
        Return @extended    ;return replacement count (=item no)
    EndIf
    Return 0
EndFunc