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There are a few options, with your current style use <xsl:value-of select="../Name"/>, the .. means use the parent, Customer, then choose the child Name. I think a better way would be to select only the Customer elements with a status: <xsl:for-each select="Customers/Customer[Status]"> to start with then just output Name and Status.
RE: [xsl] XSL Help
Subject: RE: [xsl] XSL Help From: "Joe Fawcett" <joefawcett@xxxxxxxxxxx> Date: Mon, 01 Oct 2007 16:37:22 +0100 |
There are a few options, with your current style use <xsl:value-of select="../Name"/>, the .. means use the parent, Customer, then choose the child Name. I think a better way would be to select only the Customer elements with a status: <xsl:for-each select="Customers/Customer[Status]"> to start with then just output Name and Status.
Joe http://joe.fawcett.name
From: "Waqar Ali" <waqarali_pk@xxxxxxxxxxx> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: [xsl] XSL Help Date: Mon, 01 Oct 2007 11:31:17 -0400
<Customers> <Customer> <Name>Joe Smith</Name> <Status>VIP</Status> </Customer> <Customer> <Name>Mary J</Name> </Customer> <Customer> <Name>John Hopkins</Name> <Status>VVIP</Status> </Customer> </Customers>
Generate a list of records with Status: <xsl:template name="SPECIAL" match="/"> <xsl:for-each select="Customers/Customer/Status"> <xsl:value-of select="/Name"/> <xsl:text>-</xsl:text> <xsl:value-of select="/Status"/> </xsl:for-each> </xsl:template>
The names/status are not getting printed. How can I move one level up to print
the names using value-of.
Desired Output: Joe Smith-VIP John Hopkins-VVIP
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