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Getting attribute Values using xslt

Posted on 2016-09-02
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Last Modified: 2016-09-08
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>

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However, I am not getting any results.

Can you please suggest what's gone wrong here?
Thanks,
Shail
0
Comment
Question by:Shailesh Shinde
  • 2
  • 2
4 Comments
 
LVL 60

Expert Comment

by:Geert Bormans
ID: 41781474
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>

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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
 
LVL 3

Author Comment

by:Shailesh Shinde
ID: 41781518
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)">

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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>

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Thanks,
Shail
0
 
LVL 60

Accepted Solution

by:
Geert Bormans earned 500 total points
ID: 41781544
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
 
LVL 3

Author Closing Comment

by:Shailesh Shinde
ID: 41789351
Yes, this works perfectly
0

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