accessing custom media type properties

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Alex Burr 77 posts 128 karma points

Dec 21, 2011 @ 17:10

XSLT

Hi everyone... I've built an image carousel macro (uses the Nivo Slider) and I'm very happy so far with getting images to spin. My macro takes a Media folder as a parameter, so that after dropping the carousel macro on the page, users can update the carousel by just adding/removing images from the Media folder.

I've added a link property (Content Picker) to the 'Image' Media Type in Umbraco, but I can't figure out the proper syntax to get my XSLT macro to actually insert a clickable link.

Here's what I have currently...

<xsl:param name="currentPage"/>

<!-- Define source of files in Media Library -->
<xsl:variable name="mediaNode" select="/macro/mediaSource" />
<xsl:variable name="id" select="$mediaNode/Gallery/@id" />
<xsl:variable name="width" select="/macro/width" />
<xsl:variable name="height" select="/macro/height" />

<!-- Displays all documents in a folder -->
<xsl:template match="/">
    <div id="slider" class="nivoSlider theme-default" style="width: {$width}px; height: {$height}px;">
        <xsl:if test="$id">
            <xsl:variable name="folderContent" select="umbraco.library:GetMedia($id,1)" />
            <xsl:for-each select="$folderContent/Image">
                    <img src="{umbracoFile}" alt="{@nodeName}" title="{@nodeName}" />
            </xsl:for-each>
        </xsl:if>
    </div>
</xsl:template>

This correctly adds/rotates images.
I was considering adding something like this within my <xsl:for-each> to output the link:

<xsl:element name="a">
       <xsl:attribute name="href">
            <xsl:value-of select="current()/link"/>
        </xsl:attribute>
       <img src="{umbracoFile}" alt="{@nodeName}" title="{@nodeName}" />
</xsl:element>

But no dice. Any help on the syntax would be great! Thanks in advance. :)

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Jan Skovgaard 11280 posts 23678 karma points MVP 10x admin c-trib

Dec 21, 2011 @ 18:53

0

Hi Alex

What have you called your link property?

I suppose you should be able to simply just write

<a href="{link}">
<img src="{umbracoFile}" alt="@nodeName" title="{@nodeName}" />
</a>

If this does not create the link then try adding <textarea><xsl:copy-of selec="." /></textarea> inside the for-each statement - this will return the XML for each image, which you can expect to see the structure of the XML and the element names.

Hope this helps.

/Jan

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Nigel Wilson 944 posts 2076 karma points

Dec 21, 2011 @ 19:38
1
Hi Alex

Maybe try
```
<xsl:attributename="href">
<xsl:value-ofselect="umbraco.library:NiceUrl(current()/link)"/>
</xsl:attribute>
```
It would pay to do a check to see if a content node had been selected as a link too, otherwise you may get a XSLT error.

Cheers

Nigel
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Alex Burr 77 posts 128 karma points

Dec 21, 2011 @ 20:19

Thanks so much! Here's what I ended up putting inside my for-each loop:

<xsl:choose>
  <xsl:when test="string(link)!=''">
    <a href="{umbraco.library:NiceUrl(current()/link)}">
      <img src="{umbracoFile}" alt="{@nodeName}" title="{@nodeName}" />
    </a>
  </xsl:when>
  <xsl:otherwise>
    <img src="{umbracoFile}" alt="{@nodeName}" title="{@nodeName}" />
  </xsl:otherwise>
</xsl:choose>

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