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Am stuggling with this Xslt and cant figure a way out.
Here is the situation and am trying to display only Child Nodes instead of Parent node by using GetXmlNodeById
Folder Content (id= 1755 ) Folder News 1 News Item 1 News Item 2 Folder News 2 News Item 1 News Item 2 Folder News 3 News Item 1 News Item 2
I only want to display the Child nodes of Folder News 1, Folder News 2 and Folder News 3.
<xsl:for-each select="umbraco.library:GetXmlNodeById($level)//* [@isDoc]"> <xsl:sort select="@createDate" order="descending" /> <li> <xsl:value-of select="@nodeName"/> </li> <li> <xsl:call-template name="ImageGallery"/> </li> </xsl:for-each>
Instead am getting both the Folder and item Name. Any help on this please?
xpath:
"/" Selects children from the root node"//" Selects descendants from the root node
So without any knowledge about xslt, i think you could do : umbraco.library:GetXmlNodeById($level)/*[@isDoc]
also, you might wanna check these examples : http://our.umbraco.org/wiki/reference/xslt/45-xml-schema/xslt-examples-updated-to-new-schema
wait... did you only want to show the items or the folders ? if items - it would be umbraco.library:GetXmlNodeById($level)/*[@isDoc]/*[@isDoc]
*could/should be replaces by document types...
i think... again not an xslt guy, but that xpath should match.
Hi Steffen,
I got this working what i did is
<xsl:for-each select="umbraco.library:GetXmlNodeById($level)//* [@isDoc][not(umbracoNaviHide = 1)]"> <xsl:sort select="@createDate" order="descending" /> <xsl:call-template name="linkItems"/> <a href="{umbraco.library:NiceUrl(@id)}"><xsl:call-template name="ImageGallery"/> </a> </xsl:for-each> </xsl:template> <!-- Template for the image --> <xsl:template name="ImageGallery"> <xsl:variable name="mediapickerproperty" select="./galleryPicker"/> <xsl:variable name="numMedia" select="4"/> <xsl:if test="$mediapickerproperty > 0"> <xsl:for-each select="umbraco.library:GetMedia($mediapickerproperty, 1)/PhotoGallery[1]"> <xsl:if test="position() <= $numMedia" > <img src="{umbraco.library:Replace(umbraco.library:GetMedia(@id, 0)/umbracoFile, '.', '_Thumbnail.')}"/> </xsl:if> </xsl:for-each> </xsl:if> </xsl:template> <!-- Display only Nodes with Document Type "Album" --> <xsl:template name=" linkItems "> <xsl:if test="self::Album"> <a href="{umbraco.library:NiceUrl(@id)}"> <xsl:value-of select="@nodeName"/> </a> </xsl:if> </xsl:template>
So from my current page i get all the attributes from the Child Nodes then i just make a sorting to get the most current @createDate which will go through all Parents nodes checking for new documents created
//fuji
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Display Only Child Nodes
Am stuggling with this Xslt and cant figure a way out.
Here is the situation and am trying to display only Child Nodes instead of Parent node by using GetXmlNodeById
Folder Content (id= 1755 )
Folder News 1
News Item 1
News Item 2
Folder News 2
News Item 1
News Item 2
Folder News 3
News Item 1
News Item 2
I only want to display the Child nodes of Folder News 1, Folder News 2 and Folder News 3.
Instead am getting both the Folder and item Name. Any help on this please?
xpath:
"/" Selects children from the root node
"//" Selects descendants from the root node
So without any knowledge about xslt, i think you could do : umbraco.library:GetXmlNodeById($level)/*[@isDoc]
also, you might wanna check these examples : http://our.umbraco.org/wiki/reference/xslt/45-xml-schema/xslt-examples-updated-to-new-schema
wait... did you only want to show the items or the folders ? if items - it would be umbraco.library:GetXmlNodeById($level)/*[@isDoc]/*[@isDoc]
*could/should be replaces by document types...
i think... again not an xslt guy, but that xpath should match.
Hi Steffen,
I got this working what i did is
So from my current page i get all the attributes from the Child Nodes then i just make a sorting to get the most current @createDate which will go through all Parents nodes checking for new documents created
//fuji
is working on a reply...