what is the best way to return xml from template

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AlexR 39 posts 61 karma points

May 05, 2010 @ 09:20

I created template that return teaserImage XML and is protected from empty numbers:

<xsl:template name="teaserImage">
  <xsl:param name="currentPage" />

  <xsl:if test="number($currentPage/data[@alias='teaserImage'])">
    <xsl:copy-of select="umbraco.library:GetMedia($currentPage/data[@alias='teaserImage'], 'false')" />
  </xsl:if>
</xsl:template>

<xsl:variable name="teaserImage">
  <xsl:call-template name="teaserImage">
    <xsl:with-param name="currentPage" select="$currentPage" />
  </xsl:call-template>
</xsl:variable>

Here I used copy-of function. And I think that using it is not the best way.

Can I return full XML without copy it?

For example, I can set variable without copy XML, but it is not protected:

<xsl:variable name="teaserImage" select="umbraco.library:GetMedia($currentPage/data[@alias='teaserImage'], 'false')" />

What is the best way to return XML from my template?

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

May 05, 2010 @ 09:37

0

Hi AlexR,

I just posted an answer to your other post about this.

/Chriztian

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

May 05, 2010 @ 09:42

1

Hi again,

Sorry, totally misunderstood your question I think.

If you need the media item XML, you need to use the copy-of instruction - that's how you do it.

/Chriztian

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AlexR 39 posts 61 karma points

May 05, 2010 @ 09:45

0

Thank you for answer. I asked because I just started to use XSLT.

If using copy-of is the only way to return XML I will use it.

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

May 05, 2010 @ 10:00
0
Just to clarify:

The XML for the media item is only available through the library extension, so that's why you need to use the copy-of instruction to actually get it back from a template call.

If you were to return the finished <img> (XML) tag from the template, you could just build it as a literal result, e.g.:
```
<xsl:template name="teaserImage">
  <xsl:param name="currentPage" />

  <xsl:if test="number($currentPage/data[@alias='teaserImage'])">
    <xsl:variable name="media" select="umbraco.library:GetMedia($currentPage/data[@alias='teaserImage'], false())" />
    <img src="{$media/data[@alias = 'umbracoFile']}" width="..." height="..." />
  </xsl:if>
</xsl:template>
```
Please note I've changed 'false' to false() - a common pitfall with XSLT :-)

Also note that you could do more checking for missing values etc. if you want to bulletproof it - Lee Kelleher has a blogpost on this, that you should check out.

/Chriztian
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AlexR 39 posts 61 karma points

May 05, 2010 @ 11:05

0

Chriztian, thank you for your note about false().

I returned XML for media because I want to work with image attributes in main template and so on.

I am creating one XSLT file with useful templates. I will include it into every usual XSLT files and I will use templates from included XSLT file.

I don't want write same functions in many templates.

So I need to know how to return node-sets from templates.

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