how to traverse and than print result

eruditionist · October 28th, 2008, 12:07 PM

<?xml version="1.0"?>
<root>
    <details>
        <list>A</list>
        <categ/>
        <title>Title A1</title>
        <link>linkA1</link>
    </details>
    <details>
        <list>A</list>
        <categ/>
        <title>Title A2</title>
        <link>linkA2</link>
    </details>
    <details>
        <list>B</list>
        <categ/>
        <title>TitleB1</title>
        <link>linkB1</link>
    </details>
    <details>
        <list>B</list>
        <categ/>
        <title>Title B2</title>
        <link>linkB2</link>
    </details>
        <details>
        <list>C</list>
        <categ>fan</categ>
        <title>Title C1</title>
        <link>linkC1</link>
    </details>
        <details>
        <list>C</list>
        <categ>fan</categ>
        <title>Title C2</title>
        <link>linkC2</link>
    </details>
    </details>
        <details>
        <list>C</list>
        <categ>pen</categ>
        <title>Title C3</title>
        <link>linkC2</link>
    </details>
</root>

based on <title> I need to first check to see if <categ> exists if it does than I need to print all title based on the current <categ>
else I need to print all the <title> based on the <list> so for instance user provides Title A1 which results in printing out Title A1 and Title A2 since they both have
<list>A</list>. If the user provides Title C2 than it should only print Title C2 and Title C1 but not Title C3 since it has a seperate <categ> I need to know what the statement would be based on this requirement?

mhkay · October 28th, 2008, 12:43 PM

Your requirements statement is terribly hard to read, but it seems to be saying "Select the <details> with the given <title>. If there is a <categ>, select all <details> elements with the same <categ>. Otherwise, select all <details> elements with the same <list>."

That is

<xsl:variable name="d" select="/root/details[title=$inputTitle]"/>

then

select="/root/details[($d/categ != '' and $d/categ = categ)
or ($d/categ = '' and $d/list = list]"/>

Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer's Reference

Martin Honnen · October 28th, 2008, 12:44 PM

Here is an example stylesheet, you simply need to use xsl:choose/xsl:when to check:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">

  <xsl:output method="text"/>

  <xsl:param name="title"/>

  <xsl:template match="/">
    <xsl:variable name="item" select="root/details[title = $title]"/>
    <xsl:choose>
      <xsl:when test="$item">
        <xsl:choose>
          <xsl:when test="$item/categ/text()">
            <xsl:apply-templates select="root/details[categ = $item/categ]/title"/>
          </xsl:when>
          <xsl:otherwise>
            <xsl:apply-templates select="root/details[list = $item/list]/title"/>
          </xsl:otherwise>
        </xsl:choose>
      </xsl:when>
      <xsl:otherwise>
        <xsl:text>No element found.</xsl:text>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>

  <xsl:template match="title">
    <xsl:value-of select="concat(., '#10;')"/>
  </xsl:template>

</xsl:stylesheet>

--
Martin Honnen
Microsoft MVP - XML

eruditionist · October 28th, 2008, 12:59 PM

Im trying this and it doesnt seem to be working.
Am I doing this right?

<xsl:variable name="d" select="/root/details[title='Title A1']"/>
<select="/root/details[($d/categ != '' and $d/categ = categ) or ($d/categ = '' and $d/list = list]"/>

mhkay · October 28th, 2008, 01:15 PM

There's a missing right bracket, but I can't see anything else wrong with it.

Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer's Reference