I left something out. You forgot to close the "info" element, so you will have
to alter your source document to add a slash (/) before the right angle bracket
to get well-formed XML.
--
Charles Knell
cknell(_at_)onebox(_dot_)com - email
-----Original Message-----
From: drsystems(_at_)vsnl(_dot_)net
Sent: Mon, 30 Dec 2002 08:13:59 +0500 (IST)
To: XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Conditional branching on string attribute in IE5?
Hi,
This is for IE5. For an XML file like:
<xml>
<info path="abc.txt">
<info path="abc.jpg">
</xml>
I want to check the value of xml/info/path,
and if it contains a .jpg or .gif extension,
generate an img node, else generate an
href node.
I tried the following:
.
<xsl:for-each select="xml/info">
<xsl-if test="contains(@path,'.jpg')">
<xsl-element name="a">
..
and got an error, "Unknown method contains(@".
What is the right way to do this for IE5?
Upgrade to IE6 is not an option :-(
This is urgent, and I am stuck at this.
Please help out a new XSL enthusiast! :)
Thanks,
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list