Outputting SharePoint Hyperlink Column as URL - sharepoint

Question

I have some document URLs stored in a Sharepoint publishing column. When I output the info into a HTML page using:
<xsl:value-of select="#[ColumnName]" />
in ItemStyle.xml, I get [url], [document name] in the page. I would like to display this as a URL can anyone help with the XSL?

Answer 1

You could use:
<xsl:value-of select="substring-before(#[ColumnName],',')"/>
or whatever the separator is.

Answer 2

Thanks everyone, in the end I figured out the following based on a post at sguk
<xsl:variable name="Doc">
<xsl:call-template name="OuterTemplate.GetTitle">
<xsl:with-param name="Title" select="#DocumentLink1"/>
</xsl:call-template>
</xsl:variable>
with the following a tag code:
<a href="{substring-before($Doc,',')}">
<xsl:value-of select="substring-after($Doc,',')" />
</a>
or for an image:
<xsl:variable name="Image">
<xsl:call-template name="OuterTemplate.GetTitle">
<xsl:with-param name="Title" select="#img" />
</xsl:call-template>
</xsl:variable>
with the following img tag:
<img src="{substring-before($Image,',')}" alt="{substring-after($Image,',')}" />
I'm posting the solution back here as this proved ludicrously hard to figure out (probably my fault as I don't really 'get' XSL) but just in case anybody is looking for it, this code outputs images or links from the 'Hyperlink or Picture' column type in Sharepoint.

Answer 3

Another thing you can do is to take a list that shows URLs properly (like a Links list) and use SharePoint Designer to convert it to a DataView WebPart. In there will be the proper XSL for doing the conversion.

Answer 4

The easiest way to do this is with SharePoint designer:
click the field that shows "http://link, description"
a box with an > will appear, click it and you will get a "common xsl:value-of tasks" flyout.
It will have the field name and type, with the type set to "text".
Change the type to "hyperlink" and you will get a box allowing you to format the hyperlink. It will have all the necessary xsl already populated but you can input your own text or remove the link.

Answer 5

This hopefully helps. It shows "Project Site" when the hyperlink is entered and spaces when not.
<!--Project Site--><TD Class="{$IDAAO2UG}">
<xsl:variable name="Field" select="#Project_x0020_Site" />
<xsl:choose>
<xsl:when test="substring-before(#Project_x0020_Site, ', ')=''"><xsl:value-of select="substring-after(#Project_x0020_Site, ', ')" /></xsl:when>
<xsl:otherwise><A HREF="{substring-before(#Project_x0020_Site, ', ')}">
<xsl:choose>
<xsl:when test="substring-after(#Project_x0020_Site, ', ')=''"><xsl:value-of disable-output-escaping="no" select="substring-before(#Project_x0020_Site, ', ')" /></xsl:when>
<xsl:otherwise><xsl:value-of select="substring-after(#Project_x0020_Site, ', ')" /></xsl:otherwise>
</xsl:choose>
</A></xsl:otherwise>
</xsl:choose>
</TD>

Answer 6

In SharePoint 2013 you have to do things a bit differently because the #Url attribute is is no longer delimited with a comma. There is now a .desc sub property of #Url. Below is an example of how this works hopefully this saves someone else some time.
<xsl:template name="dvt_1.rowview">
<xsl:if test="string-length(#URL) > 0">
<div class="link-item item">
<a title="{#Comments}" target="_blank" href="{#URL}">
<xsl:value-of select="#URL.desc" />
</a>
</div>
</xsl:if>
</xsl:template>