Subject: [xsl] regular expression in replace() From: UlyLee <ulyleeka@xxxxxxxxx> Date: Tue, 18 Oct 2005 04:51:30 -0700 (PDT) |
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/
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