XSL to find/replace a string?

[Blaeze]

I am using XSL to display my Twitter feed on my info page:[url="http://zonkzone.net/xls_test/mobile_info.html"]http://zonkzone.net/xls_test/mobile_info.html[/url]I've followed the tutorials on w3schools and it's all working very well but I am getting stuck trying to detect/replace strings with XSL.Basically I need to detect any string starting with 'http://...' and replace with '<a href="http://...">http://...</a>' to make them active links in HTML.The XLS file I am using is located here:[url="http://zonkzone.net/xls_test/xsl/twitter_mobile.xsl"]http://zonkzone.net/xls_test/xsl/twitter_mobile.xsl[/url](Just hit view source)Is that something that can be easily achieved with XSL and if not what is the alternative?Any suggestions well appreciated.

[boen_robot]

Not in XSLT 1.0. Maybe with regular expressions in XSLT 2.0 or with EXSLT's reges:replace() if your XSLT processor suppors it.Either way, doing this is just as hard in XSLT 2.0 as it would be in any language. Doing it in XSLT 1.0 is harder.

[Blaeze]

Not in XSLT 1.0. Maybe with regular expressions in XSLT 2.0 or with EXSLT's reges:replace() if your XSLT processor suppors it.Either way, doing this is just as hard in XSLT 2.0 as it would be in any language. Doing it in XSLT 1.0 is harder.

Thanks for checking out. So for an XSL novice like me probably no easy way round?

[Martin Honnen]

Do you use an XSLT 2.0 or 1.0 processor? With XSLT 2.0 you could use analyze-string http://www.w3.org/TR/xslt20/#analyze-string

[rsundare]

Do you use an XSLT 2.0 or 1.0 processor? With XSLT 2.0 you could use analyze-string http://www.w3.org/TR/xslt20/#analyze-string

I am also a novice in xsl. Can't we use the contains function. I have used it to enclose the value with double quotes like below.<xsl:when test="contains(current(),',')">"<xsl:value-of select="."/>"</xsl:when>