I have a Contenttype with a custom Image with formatting and constraints for publishing -- StockImg
.
I have a rollup of all listitems with this Contenttype.
In my ItemStyle.xsl I have created a custom itemstyle
<xsl:template name="StockDisplay" match="Row[@Style='StockDisplay']" 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="SafeImageUrl">
<xsl:call-template name="OuterTemplate.GetSafeStaticUrl">
<xsl:with-param name="UrlColumnName" select="'ImageUrl'"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="DisplayTitle">
<xsl:call-template name="OuterTemplate.GetTitle">
<xsl:with-param name="Title" select="substring(@Title, 0 ,25)"/>
<xsl:with-param name="UrlColumnName" select="'LinkUrl'"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="StockImg">
<xsl:call-template name="OuterTemplate.GetSafeStaticUrl">
<xsl:with-param name="UrlColumnName" select="'ImageUrl'"/>
</xsl:call-template>
</xsl:variable>
...
<img class="image" src="{$StockImg}" title="" style="border: none;">
...
</xsl:template>
in which I would like to get the img url from my custom Publishing Image Field.
I know that by default, you'd use this:
<xsl:variable name="SafeImageUrl">
<xsl:call-template name="OuterTemplate.GetSafeStaticUrl">
<xsl:with-param name="UrlColumnName" select="'ImageUrl'"/>
</xsl:call-template>
</xsl:variable>
But since I want to retrieve the URL from a custom image field, this isn't doing the job for me.
How do I use the above variable to retrieve the URL from my custom image field?
Image with formatting and constraints for publishing
field type