Changing an XPath query from a value within for-each

Question

I'm trying to do a for each on each page:

<xsl:for-each select="/root/page">
   <xsl:value-of select="/root/titles/@page/"/>
</xsl:for-each>

Sample XML;

<root>
  <titles>
   <en>A title</en>
   <de>Ein Titel</de> 
 </titles>
 <page title="de">
    ....
 </page>
</root>

The problem I am having is that the XPath isn't resolving to the /root/titles/de/...
How can I get it to work?

I realize the question is quite specific, however I'm not able to come up with the right words to google it. — monksy
– monksy, Commented May 31, 2011 at 3:27
I don't think that any of the answers provided are correct. See mine if it fits. — Emiliano Poggi
– Emiliano Poggi, Commented May 31, 2011 at 4:17

danyolgiax · Accepted Answer · 2011-05-31 04:38:40Z

1

updated

 <xsl:for-each select="/root/page">
   <xsl:variable name="language" select="@title"/>
   <xsl:value-of select="/root/titles/*[name()=$language]"/>
</xsl:for-each>

now it works for me!

edited May 31, 2011 at 4:38

answered May 31, 2011 at 3:32

danyolgiax

13.2k10 gold badges71 silver badges118 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

danyolgiax Over a year ago

I cannot try at the moment but I think it's a good starting point.

monksy Over a year ago

I'm getting this error: Error loading stylesheet: XPath parse failure: Name or Nodetype test expected:

monksy Over a year ago

The result of the value-of comes back blank for me but the path prints out. EDIT: Nevermind... its a completely unrelated issue

danyolgiax Over a year ago

ah ok! I was investigating further! np ;-)

Emiliano Poggi · Accepted Answer · 2011-05-31 04:57:14Z

1

Use xsl:key. Because is not clear the output you need I've provided just an example for you showing how to get the title from titles for each page according to the page language.

XSLT 1.0 tested under Saxon 6.5.5

<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output omit-xml-declaration="yes" indent="yes"/>
    <xsl:strip-space elements="*"/>

    <xsl:key name="titles" match="titles/*" use="local-name()"/>

    <xsl:template match="/root">
     <xsl:for-each select="page">
             <xsl:value-of select="concat('title-',@title,key('titles',@title),'&#xA;')"/>
     </xsl:for-each>
 </xsl:template>

</xsl:stylesheet>

Applied on this input

 <?xml version="1.0" encoding="UTF-8"?>
<root>
  <titles>
   <en>A title</en>
   <de>Ein Titel</de> 
 </titles>
 <page title="de">
    ....
    </page>
     <page title="en">
    ....
 </page>
</root>

produces:

title-de:Ein Titel
title-en:A title

Tested on Mozilla Firefox 3.6.17 as follows:

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="test_trans.xsl"?>
<root>
  <titles>
   <en>A title</en>
   <de>Ein Titel</de> 
 </titles>
 <page title="de">
    ....
    </page>
     <page title="en">
    ....
 </page>
</root>

Produces

 title-deEin Titeltitle-enA title

Obviously without new lines, because we are displaying it in a browser and the transform should be changed to produce HTML.

edited May 31, 2011 at 4:57

answered May 31, 2011 at 4:16

Emiliano Poggi

24.9k8 gold badges58 silver badges66 bronze badges

2 Comments

monksy Over a year ago

What about for Firefox? I'm getting a "Error loading stylesheet: Parsing an XSLT stylesheet failed. " only on the key.

Emiliano Poggi Over a year ago

Tried on Mozilla Firefox 3.6.17 and works fine as expected. See my answer edit.

harpo · Accepted Answer · 2011-05-31 05:18:19Z

1

It's not really clear what you want, but it looks like you should be aware that xsl:for-each changes the context to the selected node, just like a template does. So this might get you somewhere:

<xsl:for-each select="/root/page">
   <xsl:value-of select="/root/titles/*[name()=current()/@title]"/>
</xsl:for-each>

Hope this helps.

edited May 31, 2011 at 5:18

answered May 31, 2011 at 3:53

harpo

43.5k15 gold badges102 silver badges133 bronze badges

2 Comments

monksy Over a year ago

I'm wanting the XSL to grab /root/title/de because de was an attribute within <page>..</page>

harpo Over a year ago

@monksy, sorry, misread, updated my answer. Basically like danyolgiax's only without the extra variable. You can also use local-name, but it's only needed if you're dealing with namespaces.

Andrew Cooper · Accepted Answer · 2011-05-31 03:46:17Z

0

<xsl:for-each select="/root/page">
    <xsl:variable name="language" select="@title"\>
    <xsl:value-of select="/root/titles/$language"/>
</xsl:for-each>

edited May 31, 2011 at 3:46

answered May 31, 2011 at 3:30

Andrew Cooper

32.7k5 gold badges86 silver badges124 bronze badges

1 Comment

monksy Over a year ago

I updated the code... I just thought I'd give you fair warning

Collectives™ on Stack Overflow

Changing an XPath query from a value within for-each

4 Answers 4

4 Comments

2 Comments

2 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

4 Comments

2 Comments

2 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related