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.
__________________________________
Yahoo! Music Unlimited
Access over 1 million songs. Try it free.
http://music.yahoo.com/unlimited/
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--