painless way to display hyperlink as url?

Question

I've got a list that makes use of the "Hyperlink or Picture" field. The list is used to hold URLs.

What I really want to do is to display those URLs, not the pretty descriptions for them.

I searched here for a solution and found http://msdn.microsoft.com/en-us/library/ff604024.aspx

Man, that looks like a lotta work to display a URL, when it's right there.

So, I am appealing to the SP community here, hoping and praying that somebody has a painless means of displaying the URL instead of the description. Does anybody got something?

Answer 1

Customization the Rendering of a Field on a List View using XSLT is preferable option here.

Solution

Note: the method described below represents one of the possible ways how it could be accomplished with XSLT, i.e.:

a)SharePoint Designer (SPD) is used for customization

b)XSLT text is embedded using Xsl property in XSLListViewWebPart

In SharePoint Designer (SPD) open view page in Designer Mode. Select list item in List View that have to be customized (Hyperlink field) and then click Customize Item as shown on picture below

Switch to Code mode and find generated code for rendering hyperlink field:

<xsl:template name="FieldRef_Hyperlink_URL_body.Link" ddwrt:dvt_mode="body" match="FieldRef[(@Format='Hyperlink') and @Name='Link']" mode="URL_body" ddwrt:ghost="show" xmlns:ddwrt2="urn:frontpage:internal">
  <xsl:param name="thisNode" select="."/>
  <xsl:variable name="url" select="$thisNode/@*[name()=current()/@Name]" />
  <xsl:variable name="desc" select="$thisNode/@*[name()=concat(current()/@Name, '.desc')]" />
  <xsl:choose>
    <xsl:when test="$url=''">
      <xsl:if test="$desc=''">
        <xsl:value-of select="$desc"/>
      </xsl:if>
    </xsl:when>
    <xsl:otherwise>
      <a href="{$url}" >
        <xsl:choose>
          <xsl:when test="$desc=''">
            <xsl:value-of select="$url"/>
          </xsl:when>
          <xsl:otherwise>
            <xsl:value-of select="$desc"/>
          </xsl:otherwise>
        </xsl:choose>
      </a>
    </xsl:otherwise>
  </xsl:choose>
</xsl:template>

Note:

a) In my case match attribute has following value: "FieldRef[(@Format='Hyperlink') and @Name='Link']", where Name corresponds to field internal name

b) By default for Hyperlink field if description value is empty, then url is displayed

So, if you need for force it to display url instead of description replace it with the following template:

<xsl:template name="FieldRef_Hyperlink_URL_body.Link"  match="FieldRef[(@Format='Hyperlink') and @Name='Link']" mode="URL_body" ddwrt:ghost="">
  <xsl:param name="thisNode" select="."/>
  <xsl:variable name="url" select="$thisNode/@*[name()=current()/@Name]" />
  <a href="{$url}" >
    <xsl:value-of select="$url"/>
  </a>
</xsl:template>

Result page