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Subject:
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XSL newbie question
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Posted By:
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Someone2006
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Post Date:
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10/9/2006 9:13:18 PM
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Hi,
I'm a complete newbie to XSL and I was wondering if someone could help me with my first attempt of this technology.
Here's the XML:
<catalog xmlns="http://www.catalog.org/xml/export-0.3/">
<dir>
<id>user:XXX/home</id>
<text>my text 1</text>
<foo>zzz</foo>
</dir>
<dir>
<id>user:XXX/home/work</id>
<text>my text 2</text>
<foo>zzz</foo>
</dir>
<dir>
<id>user:YYY/home</id>
<text>my text 3</text>
<foo>zzz</foo>
</dir>
<dir>
<id>Something completely different</id>
<text>my text 4</text>
<foo>zzz</foo>
</dir>
</catalog>
I'd like to select the <dir> nodes who's <id> is something like "user:*/home". This is the expected output:
<new>
<dir>
<id>user:XXX/home</id>
<text>my text 1</text>
</dir>
<dir>
<id>user:YYY/home</id>
<text>my text 3</text>
</dir>
</new>
Thanks,
-S.
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Reply By:
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joefawcett
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Reply Date:
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10/10/2006 2:22:30 AM
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Assuming you're using version 1.0 then I think you need to combine the starts-with() and substring-after() string functions. So a possible XPath would be:/ns:catalog/ns:dir[starts-with(ns:id, 'user:') and substring-after(ns:id, '/') = 'home']
You will have to declare the ns prefix as bound to the default namespace. For links on this subject see http://p2p.wrox.com/topic.asp?TOPIC_ID=49630.
--
Joe (Microsoft MVP - XML)
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Reply By:
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mhkay
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Reply Date:
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10/10/2006 3:06:58 AM
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XSLT 2.0 supports regular expressions, so you can write
<xsl:template match="catalog">
<new>
<xsl:copy-of select="dir[matches(id, '^user:.*/work$')]"/>
</new>
</xsl:template>
Your example changes the namespace of all the elements, which the above transformation doesn't attempt: I wasn't sure that was an important part of your requirement or just something that happened by accident.
If you're constrained to use XSLT 1.0 it's a little more difficult because there are no regular expressions and no ends-with() function. You could change the predicate (the expression inside the square brackets) to:
starts-with(id, 'user:') and substring(id, string-length(id)-4) = '/work')
Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference
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Reply By:
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Someone2006
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Reply Date:
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10/10/2006 7:10:33 AM
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Hi,
I'm using Xalan-J and none of this is working :(
Any idea?
-S.
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Reply By:
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mhkay
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Reply Date:
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10/10/2006 7:20:10 AM
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If it's not working then you've done something wrong, and to tell you what you've done wrong, we'll need to see what you've done.
Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference
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