I have list that contains two columns
Name: Single Line of Text
URL: Hyperlink or picture
Using XSLT I have shown Name columns value on my home page by using below custom style in Itemstyle.xsl file.
Custom style to show Link in PopUp window  itemstyle.xsl
- <xsl:template name=”OpenApplicationPagePopUp” match=”Row[@Style=’OpenApplicationPagePopUp‘]” mode=”itemstyle”>
- <xsl:variable 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=”@Title”/>
- <xsl:with-param name=”UrlColumnName” select=”‘LinkUrl'”/>
- </xsl:call-template>
- </xsl:variable>
- <div class=”item link-item”>
- <xsl:call-template name=”OuterTemplate.CallPresenceStatusIconTemplate”/>
- <a href=”{$SafeLinkUrl}” onclick=”ShowPopupDialog(GetGotoLinkUrl(this));return false;”>
- <xsl:if test=”$ItemsHaveStreams = ‘True'”>
- <xsl:attribute name=”onclick”>
- <xsl:value-of select=”@OnClickForWebRendering”/>
- </xsl:attribute>
- </xsl:if>
- <xsl:if test=”$ItemsHaveStreams != ‘True’ and @OpenInNewWindow = ‘True'”>
- <xsl:attribute name=”onclick”>
- <xsl:value-of disable-output-escaping=”yes” select=”$OnClickTargetAttribute”/>
- </xsl:attribute>
- </xsl:if>
- <xsl:value-of select=”$DisplayTitle”/>
- </a>
- </div>
- </xsl:template>
My Requirement to show only Suggestion Box in Popup window Other Links Should be open in new Tab So please suggest me what can change in above xslt…
Thanks
Deepak Chauhan
<xsl:template name=”OpenApplicationPagePopUp” match=”Row[@Style=’OpenApplicationPagePopUp’]” mode=”itemstyle”>
<xsl:variable name=”SafeLinkUrl”>
<xsl:call-template name=”OuterTemplate.GetSafeLink”>
<xsl:with-param name=”UrlColumnName” select=”‘LinkUrl'”/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name=”DisplayTitle”>
<xsl:call-template name=”OuterTemplate.GetTitle”>
<xsl:with-param name=”Title” select=”@Title”/>
<xsl:with-param name=”UrlColumnName” select=”‘LinkUrl'”/>
</xsl:call-template>
</xsl:variable>
<div class=”item link-item”>
<xsl:call-template name=”OuterTemplate.CallPresenceStatusIconTemplate”/>
<a href=”{$SafeLinkUrl}”>
<xsl:choose>
<xsl:when test=”$DisplayTitle=’Suggestion Box'”>
<xsl:attribute name=”onclick”>ShowPopupDialog(GetGotoLinkUrl(this));return false;</xsl:attribute>
<xsl:if test=”$ItemsHaveStreams = ‘True'”>
<xsl:attribute name=”onclick”>
<xsl:value-of select=”@OnClickForWebRendering”/>
</xsl:attribute>
</xsl:if>
<xsl:if test=”$ItemsHaveStreams != ‘True’ and @OpenInNewWindow = ‘True'”>
<xsl:attribute name=”onclick”>
<xsl:value-of disable-output-escaping=”yes” select=”$OnClickTargetAttribute”/>
</xsl:attribute>
</xsl:if>
</xsl:when>
<xsl:otherwise>
<xsl:attribute name=”target”>_blank</xsl:attribute>
</xsl:otherwise>
</xsl:choose>
<xsl:value-of select=”$DisplayTitle”/>
</a>
</div>
</xsl:template>
Thanks @Paul but when we put the condition
<xsl:if test=”$DisplayTitle = ‘Suggestion Box'”>
<xsl:attribute name=”target”>_blank</xsl:attribute>
</xsl:if>
so that Suggestion Box will be open in New Tab But My requirement is that Suggestion Box Should be open in popup window and other open in new Tab
Please suggest me regarding this
Thanks
Deepak Chauhan
<xsl:if test=”$DisplayTitle = ‘Suggestion Box'”>
<xsl:attribute name=”target”>_blank</xsl:attribute>
</xsl:if>
Put those lines between 25 and 26 in your source code.