How do I output a link's description within ItemStyle.xsl?

Question

I have a custom list which contains a column of Hyperlink or Picture and URL formatted as Hyperlink. I then have a Content Query Web part on a page that uses this list as the source (Show items from the following list) and have selected a custom item style.

When creating a new item the Link URL's address and description are both populated.

However, I'm having difficulty finding a way to pull the link description when working within ItemStyle.xsl. I've seen LinkToolTip referred to, but that doesn't seem to be returning anything:

<a href="{$SafeLinkUrl}" title="{@LinkToolTip}">
    More <xsl:value-of select="@LinkToolTip" />
</a>

Unfortunately I can't seem to find anything online that shows how to do this in ItemStyle.xsl - just programmatically within code/CAML/views, or they use set text for all links.

What am I missing?

Answer 1

This is only a work-around, but as I fear it may be what I need to do if I'm only using the ItemStyle.xsl ...

I added a new text column to my custom list called Link Text.

I then updated the ItemStyle.xsl from the version in the original question to the following:

<a href="{$SafeLinkUrl}">
    <xsl:value-of select="@LinkText" />
</a>

Upon saving the ItemStyle.xsl file I was able to edit my content query Web part so that the new LinkText field under Presentation > Fields to display pulled from Link Text;.

My Web part is now correctly pulling that field's value into the Web part once output.

It feels like a hack, especially since we're already populating a description when we add a link URL, so I would hope there's a better way. I think what's important is that adding @LinkToolTip to the template in ItemStyle.xsl does not display a new field under Presentation > Fields to display, while others added like this do. (This field name is also used in the default item style.)

Answer 2

I had almost the same problem, the simplest, shortest solution to show a link from a Hyperlink or Picture field in the CQWP that I've found is the following:

I have a list which contains two columns of interest for this example: Title and SiteURL. The SiteURL column is of type "Hyperlink or Picture".

When that column enters the ItemStyle.xsl the value looks like "http://url.of/link, Url title", so to make a link of if and display we need to get everything before the ',' and after the ',' and make a link of it.

This is my resulting template in ItemStyles.xsl, without any other modifications.

<xsl:template name="ThaBomb" match="Row[@Style='ThaBomb']" mode="itemstyle">
    <xsl:variable name="MySiteLink" select="substring-before(@SiteURL, ',')"/>
    <xsl:variable name="MySiteTitle" select="substring-after(@SiteURL, ',')"/>
    <a href="{$MySiteLink}">
        <xsl:value-of select="@Title"/>
    </a>
</xsl:template>

When that's done, I went to edit the CQWP on the page, under Presentation->Item style I selected my template "ThaBomb" instead of Default. Also under Presentation->Fields to display I found my column SiteURL empty, in there I linked it to the actual column name (from the list), which in my case is also SiteURL.

As for getting all the field names, I had to modify (and add to) the xsl from your link (http://www.heathersolomon.com/blog/articles/CustomItemStyle.aspx) like so:

<xsl:for-each select="@*">
    Field: <xsl:value-of select="local-name()"/> : <xsl:value-of select="."/><br/>
</xsl:for-each>

to get the names and also add the values (which I thought I needed).

I believe the properties will display one of these fields under "Fields to display" for any field referenced in the template, but not necessarily bind it correctly. Anyway, after bound. Just hit OK and that did it for me, although in SharePoint 2013.

Hope it helps. / Richard