getmedia xslt on news list

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eddy 43 posts 64 karma points

Oct 09, 2011 @ 22:14

XSLT

I'm using 4.5.2 and the new Schema and want a list of news items with their images on the homepage. I've created the following xslt which displays the news title but I cannot get the associated media file to render. The Image is linked via the media picker and also cropped using the Image Cropper data type, therefore I need the image path followed by the suffix '_banner.jpg' which is the associated crop name.

I would appreciate any help.

Eddy

Xslt:

<ul id="newscarousel" >
<!-- This selects our data and spits it out in the following format once for each entry found  -->
  <xsl:for-each select="umbraco.library:GetXmlNodeById($source)/descendant::NewsItem [@isDoc and string(umbracoNaviHide) != '1']">
  <!-- This sorts the results by date decending -->
    <xsl:sort select="@createDate" order="descending" />    
    <!-- This limits the articles presented by the maxItem variable -->
    <xsl:if test="position() &lt;= $maxItems">
      <li>
      <!-- this prints all media nodes -->
        <img src="{umbraco.library:GetMedia(./bannerImage, false())/umbracoFile}" />
        <div id="carouseltitle">
          <h3><a href="{umbraco.library:NiceUrl(@id)}"><xsl:value-of select="@nodeName" /></a></h3>
        </div>
      </li>
    </xsl:if>
  </xsl:for-each>
</ul>

Here is the xml from one of the news items:

<Image id="1621" version="f098f151-e125-4b67-a507-8a2cc60ab8b4" parentID="1618" level="2" writerID="0" nodeType="1032" template="0" sortOrder="2" createDate="2011-10-06T09:48:24" updateDate="2011-10-06T09:48:25" nodeName="test2" urlName="test2" writerName="Administrator" nodeTypeAlias="Image" path="-1,1618,1621"><umbracoFile>/media/25270/penguins.jpg</umbracoFile><umbracoWidth>1024</umbracoWidth><umbracoHeight>768</umbracoHeight><umbracoBytes>777835</umbracoBytes><umbracoExtension>jpg</umbracoExtension><imagecropper><crops date="06/10/2011 09:48:37"><crop name="smallsquare" x="615" y="56" x2="805" y2="246" url="/media/25270/penguins_smallsquare.jpg" /><crop name="mediumsquare" x="128" y="59" x2="418" y2="334" url="/media/25270/penguins_mediumsquare.jpg" /><crop name="banner" x="199" y="50" x2="799" y2="300" url="/media/25270/penguins_banner.jpg" /></crops></imagecropper></Image>

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eddy 43 posts 64 karma points

Oct 09, 2011 @ 23:01

I'm one step closer. Thanks to some tips from Sascha here.

<xsl:if test="current()/bannerImage != ''">
    <xsl:variable name="imageSource" select="umbraco.library:GetMedia(current()/bannerImage, '0')" />  
    <img src="{$imageSource/umbracoFile}" />
</xsl:if>

Now I just need to insert the crop name '_banner' into the file path.

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Fuji Kusaka 2203 posts 4220 karma points

Oct 10, 2011 @ 04:52

Hi eddy,

You can try to concat your img tag

<xsl:variablename="imageSource"select="umbraco.library:GetMedia(current()/bannerImage, '0')"/>

<xsl:for-each select="$imageSource/Image">
   <img src="{concat(substring-before(umbracoFile,'.'),'_banner.jpg')}" />
</xsl:for-each>

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eddy 43 posts 64 karma points

Oct 10, 2011 @ 15:39

Thanks Fuji. I couldn't get the Concat to work but the Umbraco TV episode on the image cropper data type helped me out in the end. Here's what worked:

<xsl:choose>
          <xsl:when test="current()/bannerImage">
            <img>
              <xsl:attribute name="src">
                <xsl:value-of select="umbraco.library:GetMedia(current()/bannerImage, '0')/imagecropper//crop [@name='banner']/@url" />
              </xsl:attribute>
            </img>
          </xsl:when>
          <xsl:otherwise>
            <img src="media/25270/penguins_banner.jpg"></img>
          </xsl:otherwise>
        </xsl:choose>

Eddy

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Fuji Kusaka 2203 posts 4220 karma points

Oct 10, 2011 @ 21:28

0

Hey Eddy, at least you got it working.

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