How to add 'join' xml files using xslt?

Question

I have found several similar answers but I can't quite get why this is not working; This is a stripped down version of my real problem, but I have tested it and get the same result;

test1.xml

<?xml version="1.0" encoding="UTF8"?>
<fieldList>
<sourceField name="SourceTime">
<fid>REC_TIME</fid>
</sourceField>
</fieldList>

lookup.xml

<?xml version="1.0" encoding="UTF-8"?>
<data>
<field  name="REC_TIME" fid_type="DATE"/>
</data>

joinTest.xsml

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="UTF-8" media-type="text/plain"/>
<xsl:variable name="fieldDict" select="'lookup.xml'" />
<xsl:template match="/fieldList/*[fid]">
<xsl:variable name="id" select="current()"/>
<xsl:value-of select="$id"/>
<xsl:for-each select="document($fieldDict)/data/field[@name = $id]">
<xsl:value-of select="@type"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>

Now if I change the xpath to [@name] I get

REC_TIME
MONKEY

which is the result I want, but obviously the real file has much more than one entry so I need the filter to actually work!

I am testing this with xsltproc on Linux

xsltproc --version
Using libxml 20705, libxslt 10124 and libexslt 813
xsltproc was compiled against libxml 20632, libxslt 10124 and libexslt 813
libxslt 10124 was compiled against libxml 20632
libexslt 813 was compiled against libxml 20632

Many thanks.

Good question, +1. See my answer for a complete and short solution. :) — Dimitre Novatchev
– Dimitre Novatchev, Commented Nov 5, 2010 at 13:18
@Martin I sort of did, "REC_TIME MONKEY... which is what I would expect with the $id bit", it could be clearer, I'll edit it. — Chris Huang-Leaver
– Chris Huang-Leaver, Commented Nov 7, 2010 at 11:43

Dimitre Novatchev · Accepted Answer · 2010-11-05 13:16:00Z

2

This transformation is the simplest correction of your XSLT code, that produces the wanted result:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:my="my:my"
>
    <xsl:output method="xml" encoding="UTF-8" media-type="text/plain"/>

    <my:lookup>
        <data>
          <field  name="REC_TIME" fid_type="DATE"/>
        </data>
    </my:lookup>

    <xsl:variable name="fieldDict" select="document('')/*/my:lookup" />

    <xsl:template match="/fieldList/*[fid]">
        <xsl:variable name="id" select="fid"/>
        <xsl:value-of select="$id"/>
        <xsl:text> </xsl:text>
        <xsl:for-each select="$fieldDict/data/field[@name = $id]">
            <xsl:value-of select="@fid_type"/>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the provided XML document:

<fieldList>
    <sourceField name="SourceTime">
        <fid>REC_TIME</fid>
    </sourceField>
</fieldList>

the wanted result is produced:

REC_TIME DATE

Do note:

The only significant change to your code was the definition of $id.
Usually for such operations it is recommended to use keys in order to speedup processing.

answered Nov 5, 2010 at 13:16

Dimitre Novatchev

244k27 gold badges308 silver badges438 bronze badges

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

2 Comments

Chris Huang-Leaver Over a year ago

BOOOM! yes that worked. I'm curious as to why though. Is it because current() actually points to some other part of the document? When I print out the value using value-of, it looks the same as your version.

Dimitre Novatchev Over a year ago

@Chris-Huang-Leaver: The string value of current() is the concatenation of all descendent text nodes -- this includes the white-space-only text nodes. Maybe using <xsl:strip-space elements="*"/> could help.

Collectives™ on Stack Overflow

How to add 'join' xml files using xslt?

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related