[xsl] regular expression in replace()

[UlyLee <ulyleeka@xxxxxxxxx>]

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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