traversing external xml file - XSLT

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Claushingebjerg 939 posts 2574 karma points

Sep 01, 2010 @ 16:46

0

traversing external xml file

Im using an externale xml file as a data source, but having trouble getting the data i need in thye right order...

my source is as follows:

<xsl:variable name="kulturkalender">myfile.xml</xsl:variable>

<xsl:template match="/">
<xsl:variable name="kalender" select="umbraco.library:GetXmlDocumentByUrl($kulturkalender, 50000)" />

 <ul>
 <xsl:apply-templates select="$kalender//arrs" />
 </ul>

</xsl:template>

<xsl:template match="arrs">
<xsl:if test="position() <= 100">
<li >
<xsl:value-of select="@genre"/> 
 <xsl:value-of select="arr/kunstner"/>
 </li>
</xsl:if>
</xsl:template>

My xml is as follows:

<arrlist>
 <arrs genre="Musik">
 <arr ArrNr="4640717">
 <kunstner>Torsdagskoncerter på Østergade, Herning.</kunstner>
 </arr>
 <arr ArrNr="4743104">

 <kunstner>Jens Christian Wandt på Hjerl Hede.</kunstner>
</arr>

</arrs>
</arrlis>

What im looking to get is list the "@genre" on every "arr" so it lists like:

<ul>
<li>Musik Torsdagskoncerter på Østergade, Herning</li>
<li>Muisk Jens Christian Wandt på Hjerl Hede.</li>
</ul>

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Claushingebjerg 939 posts 2574 karma points

Sep 01, 2010 @ 16:46

0

im on 4.5.1 by the way

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Dirk De Grave 4541 posts 6021 karma points MVP 3x admin c-trib

Sep 01, 2010 @ 16:55
0
Have you tried using the full url to the xml file? Even if it's on the local server, I think you need to specify an url as the function does perform a WebRequest.

If that doesn't help, you can always output the error that gets returned from the function.

Error xml returned in case of exception is
```
<error url="url">error</error>
```
where url is the parameter fed to the function and the error is the error message returned from the exception (if it has occured)

Hope this helps.

Regards,

/Dirk
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Claushingebjerg 939 posts 2574 karma points

Sep 01, 2010 @ 17:22

0

Sorry for not being clear.

I typed my example as compact as possible...

my real url to the xml is a full url, i just didnt want to post it here. Im getting data out of it ok.

I'm just unable to get the "genre" attribute rendere in each <li>

Hope this makes sense.

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Steen Tøttrup 191 posts 291 karma points c-trib

Sep 01, 2010 @ 17:26
1
I'll try and beat Chriztian to it, and he'll probably be around later to come up with a much better solution.

I assume the problem isn't getting the xml, if it is, take a look at what Dirk has written.

So try this (this is done off the top of my head, so I haven't tested it):
```
 <xsl:apply-templates select="$kalender//arrlist/arrs" />
</xsl:template>

<xsl:template match="arrs">
 <xsl:if test="position() &lt;= 100">
  <ul>
   <xsl:apply-templates select="arr" />
  </ul>
 </xsl:if>
</xsl:template>

<xsl:template match="arr">
 <li><xsl:value-of select="../@genre" /> <xsl:value-of select="kunstner" /></li>
</xsl:template>
```
I'm not sure what sort order you want the result in ?!
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Claushingebjerg 939 posts 2574 karma points

Sep 01, 2010 @ 17:31

0

Works like a charm, Thanks

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Chriztian Steinmeier 2800 posts 8790 karma points MVP 8x admin c-trib

Sep 01, 2010 @ 22:12

0

Ha ha - good one, Steen :-)

/Chriztian

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