Links List description column

How to get the description value of the hyperlink field in the link list for xslt list view webpart ? i.e how do I get the value of description in the XSLT that I provide for XslLink?

Hi,
If there is a hyperlink field in your list view, then there would be a variable “$desc” in the xslt code, you can call it like this:
<xsl:value-of select="$desc"/>
Here are two links with code samples about how to customize the the rendering of a hyperlink field for your reference:
https://social.msdn.microsoft.com/Forums/sharepoint/en-US/68b05f40-9b62-485b-ba8e-4684a7e34899/truncate-url-in-xslt-view
http://sharepoint.stackexchange.com/questions/58672/painless-way-to-display-hyperlink-as-url
More information about XSLT language:
http://www.w3schools.com/xsl/
Thanks
Patrick Liang
Forum Support
Please remember to mark the replies as answers if they help and unmark them if they provide no help. If you have feedback for TechNet Subscriber Support,
contact [email protected]
Patrick Liang
TechNet Community Support

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