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Hi I'm new with Xpath, so need a little help
I'm trying to get and count Contour records in a specific form submitted by specific members.
<xsl:variable name="recCountList" select="umbraco.contour:GetRecordsFromForm('b41befb1-549f-410a-babb-898a77f63759')[/uformrecord/memberkey=$profileMemberID]" /> Comments: <xsl:value-of select="count($recCountList/uformrecord)" />
Without the filter the recCountList returns all records from the form, with the filter it returns none.
How do I make the filter work?
Ok, changed to this and it works
<xsl:variable name="records" select="umbraco.contour:GetRecordsFromForm('b41befb1-549f-410a-babb-898a77f63759')" /> <xsl:variable name="recCountList" select="$records/uformrecord[memberkey=$profileMemberID]" /> <xsl:value-of select="count($recCountList)" /> comments
I am though a bit concerned with performance, if I count comments from many members. Or is there no reason to worry about the difference?
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List records with specific value
Hi I'm new with Xpath, so need a little help
I'm trying to get and count Contour records in a specific form submitted by specific members.
<xsl:variable name="recCountList" select="umbraco.contour:GetRecordsFromForm('b41befb1-549f-410a-babb-898a77f63759')[/uformrecord/memberkey=$profileMemberID]" />
Comments: <xsl:value-of select="count($recCountList/uformrecord)" />
Without the filter the recCountList returns all records from the form, with the filter it returns none.
How do I make the filter work?
Ok, changed to this and it works
<xsl:variable name="records" select="umbraco.contour:GetRecordsFromForm('b41befb1-549f-410a-babb-898a77f63759')" />
<xsl:variable name="recCountList" select="$records/uformrecord[memberkey=$profileMemberID]" />
<xsl:value-of select="count($recCountList)" /> comments
I am though a bit concerned with performance, if I count comments from many members. Or is there no reason to worry about the difference?
is working on a reply...
This forum is in read-only mode while we transition to the new forum.
You can continue this topic on the new forum by tapping the "Continue discussion" link below.