Getting attribute Values using xslt

Hi All,

I am trying to get the attribute value from 100's of xml files using below code..
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
	version="2.0" xmlns:saxon="http://sf.net/saxon" xmlns:xs="http://www.w3.org/2001/XMLSchema"
	exclude-result-prefixes="saxon xs">
	<xsl:output indent="yes" method="html" encoding="UTF-8" />
	<xsl:strip-space elements="*" />
	<xsl:template match="/">
		<html>
			<body>
				<ol>
					<xsl:for-each select="//fileref/@href">
						<xsl:variable name="file" select="concat('../input.files/',.)" />
						<xsl:message>Processing ...<xsl:value-of select="." /></xsl:message>
						<xsl:for-each select="document($file)/distinct-values(//trans-unit/@id)">
							<xsl:sort order="ascending" select="." />
							<li>
								<xsl:value-of select="." />
							</li>
						</xsl:for-each>
					</xsl:for-each>
				</ol>
			</body>
		</html>
	</xsl:template>
</xsl:stylesheet>

Open in new window

However, I am not getting any results.

Can you please suggest what's gone wrong here?
Thanks,
Shail
LVL 3
Shailesh ShindeLocalization Engineering & AutomationAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Geert BormansInformation ArchitectCommented:
technically I don't see a direct issue
but it is very hard to comment on this... and it is really easy for you to debug

1. first test the outer loop
				<ol>
					<xsl:for-each select="//fileref/@href">
						<xsl:variable name="file" select="concat('../input.files/',.)" />
							<li>
								<xsl:value-of select="$file" />
							</li>
						</xsl:for-each>
					</xsl:for-each>
				</ol>

Open in new window


So then you know whether the source XML file already gives you the correct file urls
(you have not added the source XML so I can't test that myself)
0
Shailesh ShindeLocalization Engineering & AutomationAuthor Commented:
Hi Geert,

Yes, I did that and when I changed the xpath something like below it has shown the results.
<xsl:for-each select="document($file)/distinct-values(//@id)">

Open in new window

This is strange as xml structure is quite simple...
<file>
<body>
<group>
<trans-unit id="m1"></trans-unit>
<trans-unit id="m1"></trans-unit>
<trans-unit id="m2"></trans-unit>
<trans-unit id="m2"></trans-unit>
</group>
</body>
</file>

Open in new window

Thanks,
Shail
0
Geert BormansInformation ArchitectCommented:
so... you are proving you question was not detailed enough
Please, in future give ALL information, because only now it is apparent what you should be looking at

if this is working
            <xsl:for-each select="document($file)/distinct-values(//@id)">
                        
and this is not
            <xsl:for-each select="document($file)/distinct-values(//trans-unit/@id)">

it simply means that there are no @id in a <trans-unit> element
I am pretty certain that the XML example you show is not complete
I bet there is an xml namespace declaration binding the default namespace to a namespace different from the NULL namespace

I assume this would work
            <xsl:for-each select="document($file)/distinct-values(//*:trans-unit/@id)">
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Shailesh ShindeLocalization Engineering & AutomationAuthor Commented:
Yes, this works perfectly
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Web Languages and Standards

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.