Hello,
I have an issue with transforming an xml, the goal is to exclude <Source> element if the value of the attribute @guid has 'badguid', the element <client> and its attribute @guid are optional, meaning it may not be there all the time, here is my example:

XML:
----

<?xml version="1.0" encoding="utf-8"?>
<Feed>
    <Source>
        <BlackListedFor />
        <Link href="link1" />
    </Source>
    <Source>
        <BlackListedFor>
            <client guid="badguid" />
        </BlackListedFor>
        <Link href="link2" />
    </Source>
    <Source>
        <BlackListedFor />
        <Link href="link3" />
    </Source>
</Feed>

, and here is the XSLT
--------------------------

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" indent="yes"/>
    <xsl:template match="/Feed">
        <xsl:copy>
            <xsl:apply-templates select="Source[BlackListedFor/client[@guid!='badguid']]"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>


I'm not able to cover all cases, what should I use here? conditional processing? a special predicate??

thanks.

Try using the not() function instead of != operator:

<xsl:apply-templates select="Source[BlackListedFor/client[not(@guid='badguid')]]"/>

/- Sam Judson : Wrox Technical Editor -/

Try replacing

select="Source[BlackListedFor/client[@guid!='badguid']]"

by

select="Source[not(BlackListedFor/client/@guid='badguid')]"

It's the difference between something that has an attribute A that is not equal to B, and something that does not have an attribute A that is equal to B.


Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

thanks mhkay and samjudson !! mhkay solution worked :), and your comment does make sence now.

-Tamer