SharePoint – Content Query Web Part open item in detail view

I wrote an article about opening an item with the CQWP in the edit form. I also found out how to open an item with the detail view. Just follow these instructions:

• Open SharePoint Designer
• Click at All Files
• Click at Style Library
• Click at XSL Style Sheets
• Make a back-up of ContentQueryMain and ItemStyle
• Check out ContentQueryMain and add the following code:

<xsl:template name=”OuterTemplate.GetDisplayLink”>
<xsl:param name=”UrlColumnName”/>
<xsl:if test=”$UseCopyUtil = ‘True'”>
<xsl:value-of select=”concat($RootSiteRef,’/_layouts/CopyUtil.aspx?Use=id&amp;Action=dispform&amp;ItemId=’,@ID,’&amp;ListId=’,@ListId,’&amp;WebId=’,@WebId,’&amp;SiteId=’,$SiteId,’&amp;Source=’,$Source)”/>
</xsl:if>
<xsl:if test=”$UseCopyUtil != ‘True'”>
<xsl:call-template name=”OuterTemplate.GetSafeStaticUrl”>
<xsl:with-param name=”UrlColumnName” select=”$UrlColumnName”/>
</xsl:call-template>
</xsl:if>
</xsl:template>

• Save and check in and publish
• Check out ItemStyle and add the following code:

<xsl:template name=”NameOfTheTemplate” match=”Row[@Style=’ NameOfTheTemplate’]” mode=”itemstyle”>
<xsl:variable name=”SafeLinkUrl”>
<xsl:call-template name=”OuterTemplate.GetDisplayLink”>
<xsl:with-param name=”UrlColumnName” select=”‘LinkUrl'”/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name=”DisplayTitle”>
<xsl:call-template name=”OuterTemplate.GetTitle”>
<xsl:with-param name=”Title” select=”@Title”/>
<xsl:with-param name=”UrlColumnName” select=”‘LinkUrl'”/>
</xsl:call-template>
</xsl:variable>
<div>
<xsl:call-template name=”OuterTemplate.CallPresenceStatusIconTemplate”/>
<a onclick=”javascript:SP.UI.ModalDialog.ShowPopupDialog(‘{$SafeLinkUrl}’); return false;” onmouseover=”javascript:this.style.cursor=’hand’;” title=”{@LinkToolTip}”>
<xsl:value-of select=”$DisplayTitle”/>
</a>
</div>
</xsl:template>

• Save, Check in and Publish.

The last thing you have to do is edit the CQWP and use the itemstyle. That should do the trick!