Schema Part 2 says:
A metacharacter is either ., \, ?, *, +, {, } (, ), [ or ]. These characters
have special meanings in .regular expression.s, but can be escaped to form
.atom.s that denote the sets of strings containing only themselves, i.e., an
escaped .metacharacter. behaves like a .normal character..
and XPath adds ^ and $ to the list.
So you can turn your replacement string into a regex using
replace($in, "[.\\?*+{}()\[\]^$]", "\\$0")
Note that in both the regex and the replacement string, \ is represented as
\\.
Not tested.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: UlyLee [mailto:ulyleeka(_at_)yahoo(_dot_)com]
Sent: 18 October 2005 12:52
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] regular expression in replace()
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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