praveen

Sure this is possible, you'd need a double for-each to loop through all child nodes...

<xsl:for-each select="umbraco.library:GetMedia($MainFolder, true())">
<xsl:for-each select="umbraco.library:GetMedia(current()/@id, true())">
<xsl:if test="position() = 1">
<img src="./umbracoFile" />
</xsl:if>
</xsl:for-each>
</xsl:for-each>

First loop iterates folders of the MainFolder (either pass this is as a parameter on the xslt, or hard code this value in the xslt), inner loop fetches all child media nodes from the current node (Child1, Child2, ...) and gets the first image (using position() = 1)

Second param to GetMedia() (true()) is to make sure the GetMedia() call returns all child nodes instead of fetching just a media node (in which case you'd use false())

Hope this helps.

Regards,

/Dirk

Hey Praveen,

You could also give this a tr. Same basic loop as Dirk mentioned earlier.

<xsl:if test="$mediaFolder"> 
     <xsl:variable name="mediaItems" select="umbraco.library:GetMedia($mediaFolder/*/@id, 'true')" />     
     <xsl:for-each select="$mediaItems/Image">      
       <xsl:if test="position() = 1">          
         <img src="./umbracoFile" />     
     </xsl:if>       
    </xsl:for-each>     
</xsl:if>  

//fuji

Hi Dirk,

                      thanks for your reply, i am new in umbraco,  but it can not show the images ,

        please check the xslt, and  xslt show this error

    <xsl:variable name="imageRoot" select="$currentPage/imageFolder"/>

  
<xsl:for-each select="umbraco.library:GetMedia($imageRoot, true())">
<xsl:for-each select="umbraco.library:GetMedia(current()/@id, true())">
<xsl:if test="position() = 1">
<img src="./umbracoFile" />
xsl:if>
xsl:for-each>
xsl:for-each>

thanks,

praveen

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