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Traverse over children XSL

Posted on 2004-04-06
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Last Modified: 2006-11-17
I have an XSL transformation which works fine, but I need to traverse over all children as well and have no clue how to do that. The XSL looks like

**********************************
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">

<data>
<xsl:for-each select="/root/test">
<struct>
<var name="TITLE">
<string><xsl:value-of select="/root/test"/></string>
</var>
</struct>
</xsl:for-each>

</data>

</xsl:template>
</xsl:stylesheet>
**********************************

Which give me
**********************************
<?xml version="1.0" encoding="UTF-8"?>
<data>
<struct><var name="TITLE"><string>test 1yukkie!</string></var></struct>
<struct><var name="TITLE"><string>test 1yukkie!</string></var></struct>
<struct><var name="TITLE"><string>test 1yukkie!</string></var></struct>
</data>
**********************************

However the xml looks like

**********************************
<?xml version="1.0" encoding="iso-8859-1"?>
<root>
<test>test 1<anotherTest>yukkie!</anotherTest></test>
<test>test 1<anotherTest>yukkie!</anotherTest></test>
<test>test 1<anotherTest>yukkie!</anotherTest></test>
</root>
**********************************

What I am really after is

**********************************
<struct><var name="TITLE"><string>test 1
<struct><var name="TITLE"><string>yukkie!</string></var></struct>
</string></var></struct>
**********************************

In short I want to traverse through the tree and apply one style to all ellements. Then I'd also like to know how just traverse through them without knowing their names beforehand, and their name will become the <var name="Name here">
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Question by:Tacobell777
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3 Comments
 
LVL 26

Expert Comment

by:rdcpro
ID: 10770878
This:

<xsl:value-of select="/root/test"/>

will always return the value of the same node, because you're using an absolute path.  Use a path relative to the context node, if you want a value that's related to the context node.


As far as what you're trying to do, I don't quite follow your question.  

If the XML looks like:

<root>
<test>test 1<anotherTest>Foo 1</anotherTest></test>
<test>test 2<anotherTest>Foo 2</anotherTest></test>
<test>test 3<anotherTest>Foo 3</anotherTest></test>
</root>

what is the desired output?

Regards,
Mike Sharp
0
 
LVL 4

Accepted Solution

by:
maverick65 earned 2000 total points
ID: 10771152
I think you wanna have this:

<?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" version="1.0" encoding="UTF-8" indent="yes"/>
       <xsl:template match="/root">
                <data>
                      <!-- for all childs of root, whatever its name is -->
                      <xsl:apply-templates />
                </data>
      </xsl:template>
      
      
      <!-- template for all nodes which have no special rule -->
      <xsl:template match="*">
            <struct>
                  <var>
                        <!-- add the name attribute to the var tag -->
                        <xsl:attribute name="name">
                              <xsl:value-of select="name()"/>
                        </xsl:attribute>
                        <string>
                              <!-- further processing for all childs -->
                              <xsl:apply-templates/>
                        </string>
                  </var>
            </struct>
      </xsl:template>
      
      <!-- all tags with name test2 are differently processed -->
      <xsl:template match="test2">
            <test2>this is test 2</test2>
      </xsl:template>
</xsl:stylesheet>
0
 
LVL 5

Expert Comment

by:alambres
ID: 10772775
does the xml structure depend on you? or is it like this and no way to change it? you really can do you're aiming with xslt, but if you want to nest nodes, the first  I would do (if you are allowed to) is get the nesting already in the xml. that's:

example:

why got this:

<xmlRoot>
  <opt>a<desc>blabla</desc></opt>
  <opt>a1<desc>blablabla</desc></opt>
  <opt>a2<desc>blabla</desc></opt>
  <opt>b<desc>blabla</desc></opt>
  <opt>b1<desc>blablabla</desc></opt>
  <opt>b2<desc>blabla</desc></opt>
</xmlRoot>

much better, more logic:
<xmlRoot>
  <opt>a<desc>blabla</desc>
     <opt>a1<desc>blablabla</desc></opt>
     <opt>a2<desc>blabla</desc></opt>
  </opt>
  <opt>b<desc>blabla</desc>
     <opt>b1<desc>blablabla</desc></opt>
     <opt>b2<desc>blabla</desc></opt>
  </opt>
</xmlRoot>
0

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