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Luke Ambrogio wrote:
could be rewritten more simply as:
<xsl:apply-templates select="tblMIMCompanies[tblMIMSubSectors[@subsectorID=$subsector]]"/>
Unless there's more than one tblMIMSubSectors with the same subsectorID inside one tblMIMCompanies, that would give the same output, or am I wrong? (and if there are such duplicate subsectorIDs,
the first one would output multiple but identical tblMIMCompanies, whereas the second one would'nt)
Re: [xsl] xsl:apply-template and xsl:for-ech confusion
Subject: Re: [xsl] xsl:apply-template and xsl:for-ech confusion From: Anton Triest <anton@xxxxxxxx> Date: Fri, 01 Oct 2004 15:37:34 +0200 |
Luke Ambrogio wrote:
ok the solution was infact usingOK but this:
<xsl:apply templates select=".."/>
since there was the child element tblMIMSubsector being processed
<xsl:for-each select="tblMIMCompanies/tblMIMSubSectors[@subsectorID=$subsector]"> <xsl:apply-templates select=".."/> </xsl:for-each>
could be rewritten more simply as:
<xsl:apply-templates select="tblMIMCompanies[tblMIMSubSectors[@subsectorID=$subsector]]"/>
Unless there's more than one tblMIMSubSectors with the same subsectorID inside one tblMIMCompanies, that would give the same output, or am I wrong? (and if there are such duplicate subsectorIDs,
the first one would output multiple but identical tblMIMCompanies, whereas the second one would'nt)
Cheers, Anton
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