Re: [xsl] regular expression in replace()

[UlyLee <ulyleeka@xxxxxxxxx>]

i think i already solve my problem (i think :D) I used
a recursive function:

<xsl:function name="f:rep">
<xsl:param name="source" as="xs:string"/>
<xsl:param name="pattern" as="xs:string"/>
<xsl:param name="replace" as="xs:string"/>
<xsl:choose>
<xsl:when test="contains($source,$pattern)">
<xsl:value-of
select="substring-before($source,$pattern)"/>
<xsl:value-of select="$replace"/>
<xsl:value-of
select="f:rep(substring-after($source,$pattern),$pattern,$replace)"/>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$source"/>
</xsl:otherwise>
</xsl:choose>
</xsl:function>

this replaces the replace() function. so when i call

<xsl:sequence select="f:rep('Lucie et Suz. Beauvais
Beauvais, Suzanne', 'Suz.', 'Suzanne')"/>

i can get:

Lucie et Suzanne Beauvais Beauvais, Suzanne

I hope this is correct :D

--- UlyLee <ulyleeka@xxxxxxxxx> wrote:

> I have a string: "Lucie et Suz. Beauvais Suzanne"
> and
> i want to replace "Suz." with "Suzanne".
> 
> but when i use replace("Lucie et Suz. Beauvais
> Suzanne","Suz.","Suzanne") it gives me "Lucie et
> Suzanne Beauvais Suzannenne", i figured that this is
> because "." is treated as a regular expression thats
> why it replaced "Suza" with "Suzanne". I know i need
> to escape the "." to  "\." but what if my
> replace-pattern contains other regex characters like
> "?" "*" "+"?
> 
> Michael Kay suggested that i first make my
> replacement
> string to regelar expression or create a replace
> function that uses substring-before() and
> contains().
> How am i to go around this? I'm just starting out in
> XSL and the new features of XSLT 2.0 sometime
> confuses
> me.
> 
> My first alternative was to use replace($sourceStr,
> ".", "\.") but it says "\." is an invalid
> replacement string.
> 
> 
> 		
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