How to get the parent node from child using XSLT

Question

I'm having trouble to retrieve all the parents for repeating child name.

<NCAAScores>
   <levels>
      <level>
         <name>Western Conference</name>
         <teams>
            <team>
               <name>Dallas Stars</name>
               <scorable>
                  <win>60</win>
                  <lose>35</lose>
               </scorable>
            </team>
            <team>
               <name>Chicago Blackhawks</name>
               <scorable>
                  <win>60</win>
                  <lose>23</lose>
               </scorable>
            </team>
            <team>
               <name>Edmonton Oilers</name>
               <scorable>
                  <win>55</win>
                  <lose>9</lose>
               </scorable>
            </team>
            <team>
               <name>Philadelphia Flyers</name>
               <scorable>
                  <win>5</win>
                  <lose>9</lose>
               </scorable>
            </team>
         </teams>
      </level>
      <level>
         <name>Eastern Conference</name>
         <teams>
            <team>
               <name>Dallas Stars</name>
               <scorable>
                  <win>1</win>
                  <lose>34</lose>
               </scorable>
            </team>
<!---And so on, you get the idea-->  
         </teams>
      </level>
  </levels>
</NCAAScores>

If I want to retrieve all of the parent level having team "Dallas Stars" what would be the approach?

I tried with below

./levels[/level/teams/team/name = 'Dallas Stars']

, which didn't help.

How about ./levels[level/teams/team/name = 'Dallas Stars'] — Joel M. Lamsen
– Joel M. Lamsen, Commented May 21, 2018 at 4:01
Thanks @JoelM.Lamsen , with that I'm getting "Type error at char 1 in xsl:value-of/@select on line 78 column 81 XPTY0019: Required item type of first operand of '/' is node(); supplied value has item type xs:string" — sahaS
– sahaS, Commented May 21, 2018 at 13:14

Aniket V · Accepted Answer · 2018-05-22 04:53:45Z

3

Please try the following template. This uses the ancestor:: axis to look up for level/name from the current node context.

<xsl:template match="level">
    <xsl:for-each select="teams/team[name='Dallas Stars']">
        <name>
            <xsl:value-of select="ancestor::level/name" />
        </name>
    </xsl:for-each>
</xsl:template>

Alternately, you can also use

<xsl:value-of select="../../name" />

to go up the levels from the current node context, but I prefer using ancestor for ease of understanding.

Output

<name>Western Conference</name>
<name>Eastern Conference</name>

edited May 22, 2018 at 4:53

answered May 22, 2018 at 4:45

Aniket V

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1 Comment

sahaS Over a year ago

Thanks @Ankit V , that did help

fafl · Accepted Answer · 2018-05-22 11:43:46Z

1

The type error you mentioned comes from comparing a node with a string. You have to first extract the text from the node. Try this:

//levels/level[teams/team/name/text()='Dallas Stars']/name

I prepared it as a fiddle here: https://xsltfiddle.liberty-development.net/6qVRKw2/1

answered May 22, 2018 at 11:43

fafl

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1 Comment

sahaS Over a year ago

Thanks @fafl , that helps, however , while I'm calling that from another template , I'm always getting Required item type of the context item for the child axis is node(); supplied value has item type xs:string Please see the fiddle Line 79 xsltfiddle.liberty-development.net/bFDb2C6

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