MOSS Content Query Web part itemstyle.xsl
I have a Content Query Webpart (CQWP) pulling the URL and title from a News links list. The CQWP uses the XSLT style Orange.News.Links defined in ItemStyle.xsl.
I need to sort the title @Title0 field as commented out below because it causes an error.
Does anyone know whats causing this error? - Many Thanks. The XSLT code is below:
<xsl:template name="Orange.News.Links" match="Row[@Style='Orange.News.Links']" mode="itemstyle">
<xsl:param name="CurPos" />
<xsl:param name="Last" />
<xsl:variable开发者_开发知识库 name="SafeLinkUrl">
<xsl:call-template name="OuterTemplate.GetSafeLink">
<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="@URL"/>
<xsl:with-param name="UrlColumnName" select="'URL'"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="LinkTarget">
<xsl:if test="@OpenInNewWindow = 'True'" >_blank</xsl:if>
</xsl:variable>
<xsl:variable name="SafeImageUrl">
<xsl:call-template name="OuterTemplate.GetSafeStaticUrl">
<xsl:with-param name="UrlColumnName" select="'ImageUrl'"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="Header">
<xsl:if test="$CurPos = 1">
<![CDATA[<ul class="list_Links">]]>
</xsl:if>
</xsl:variable>
<xsl:variable name="Footer">
<xsl:if test="$Last = $CurPos">
<![CDATA[</ul>]]>
</xsl:if>
</xsl:variable>
<xsl:value-of select="$Header" disable-output-escaping="yes" />
<li>
<a>
<xsl:attribute name="href">
<xsl:value-of select="substring-before($DisplayTitle,', ')"></xsl:value-of>
</xsl:attribute>
<xsl:attribute name="title">
<xsl:value-of select="@Description"/>
</xsl:attribute>
<!-- <xsl:sort select="@Title0"/> -->
<xsl:value-of select="@Title0">
</xsl:value-of>
</a>
</li>
<xsl:value-of select="$Footer" disable-output-escaping="yes" />
</xsl:template>
I need to sort the title @Title0 field as commented out below because it causes an error.
Does anyone know whats causing this error?
Yes, <xsl:sort/>
can only be a child node of either <xsl:apply-templates>
or <xsl:for-each>
(and of <xsl:perform-sort>
in XSLT 2.0).
Recommendation: Take at least a mini-course in XSLT and XPath so that you at least grok the fundamental concepts.
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