RE: [xsl] 2 Predicates in 1 for-each, possible?

Subject: RE: [xsl] 2 Predicates in 1 for-each, possible?
From: "Josh Canfield" <Josh.Canfield@xxxxxxxxxxxx>
Date: Tue, 10 Feb 2004 14:52:29 -0800
A predicate is used to filter a selection by converting the contents to a boolean value. 

Your selection is //communication/email, which translates to /descendant-or-self::node()/communication/email.
Alone this will return all email elements with a communication element as parent. 

Adding the predicate [emailaddress] specifies that you only want email nodes with an emailaddress node as a child. emailaddress will evaluate to true if there is an emailaddress child element of the selected node.

the predicate [//communication2[emailaddress2]] evaluates to true when there exists a communication2 element with a child emailaddress2 element anywhere in the document.

the predicate [emailaddress or //communication2[emailaddress2]] returns true when either of the two conditions described above are true.

It looks like you really want the union of all communication/email and communication2/emailaddress2 nodes.

<xsl:for-each select="//communication/email[emailaddress] | //communication2/emailaddress2">

Good luck,

-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Bert
Sent: Tuesday, February 10, 2004 12:29 PM
To: Xsl-List
Subject: [xsl] 2 Predicates in 1 for-each, possible?


Allow me to ask a question about 2 predicates in 1 for-each.

I have this xml-file:

<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="persons01.xsl"?>


I use this stylesheet to collect some information:

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
 <xsl:output method="xml" indent="yes"/>
 <xsl:template match="/">
   <xsl:for-each select="//communication/email[emailaddress or
    <xsl:sort select="."/>
    <xsl:value-of select="."/>
    <xsl:if test="position() != last()">,

I expected to get the following result:
e-mail@xxxxxxx(1), e-mail@xxxxxxx(2), e-mail@xxxxxx(1), e-mail@xxxxxx(2)

But what I get is this:
e-mail@xxxxxxx(1) , e-mail@xxxxxx(1)

It is obvious there is something wrong in my for-each-statement, but what?

Any help is this is welcome.

Kind regards,

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