xsl-list
[Top] [All Lists]

order problem

2003-06-27 01:02:41
Hi,

I have this element hierarchy

Source::
root
    body
        record1(*)
        record2(*)

Destination::
root1
    body1
        record3(*)
        record4(*)

Here occurence of record1 and record2 can occur any number of times.This
will be my source xml on which the transform is to be applied.Following is
the xsl

<xsl:template match="/" name="root">
<xsl:element name="root1">
 <xsl:element name="body1">
  <xsl:for-each select="/root/body/record1">
   <xsl:element name="record3"><xsl:value-of
select="concat(.,'1')"/></xsl:element>
  </xsl:for-each>
  <xsl:for-each select="/root/body/record2">
   <xsl:element name="record4"><xsl:value-of
select="concat(.,'2')"/></xsl:element>
  </xsl:for-each>
 </xsl:element>
</xsl:element>
</xsl:template>

Only problem would be that i need the order of records to be same in
output
as in input and input of records can be in random order,viz
<record1><record1><record2><record1><record2><record1>.But with the two
for-loops this will break the sequence.

How can this be done.

Any help on this would be deeply appreciated.

Thanks,
Shadab



 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list



<Prev in Thread] Current Thread [Next in Thread>