Copied to clipboard

Flag this post as spam?

This post will be reported to the moderators as potential spam to be looked at


  • Haris 12 posts 32 karma points
    Jun 09, 2011 @ 02:46
    Haris
    0

    Picked content url

    Hi all,

    I have a content picker in my document type, and I want to retrieve the "Link to document" property of the picked content in XSLT.  Is this possible? If yes, how?

     

    Thanks

  • Tom Fulton 2030 posts 4996 karma points c-trib
    Jun 09, 2011 @ 03:48
    Tom Fulton
    0

    Hi,

    There's a function built into umbraco.library called NiceUrl that you can use to get a page's URL from it's ID (which the Content Picker stores).

    Ex:

    <xsl:if test="$currentPage/yourContentPickerPropertyAlias != ''">
      <a href="{umbraco.library:NiceUrl($currentPage/yourContentPickerPropertyAlias)}">Link text</a>
    </xsl:if>

    If you want to retrieve other properties from the picked page, like it's name, you can use GetXmlNodeById:

    <xsl:if test="$currentPage/yourContentPickerPropertyAlias != ''">
      <xsl:variable name="pickedPage" select="umbraco.library:GetXmlNodeById($currentPage/yourContentPickerPropertyAlias)"/>
        <a href="{umbraco.library:NiceUrl($pickedPage/@id)}">
          <xsl:value-of select="$pickedPage/@nodeName"/>
        </a>
    </xsl:if>

    Hope this helps,
    Tom

  • This forum is in read-only mode while we transition to the new forum.

    You can continue this topic on the new forum by tapping the "Continue discussion" link below.

Please Sign in or register to post replies