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RE: sorting based on a variable

2002-11-12 08:31:53

try data[number($column)]

Perhaps your $column variable is a string or a result tree fragment, not
a number. You didn't show us its declaration.

Michael Kay
Software AG
home: Michael(_dot_)H(_dot_)Kay(_at_)ntlworld(_dot_)com
work: Michael(_dot_)Kay(_at_)softwareag(_dot_)com 

-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com 
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of 
Kasper Nielsen
Sent: 12 November 2002 14:34
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] sorting based on a variable


hi (another question)

lets say I want to some data for the $column data-column

<projects>
  <project name="a">
    <data> 1237</data>
    <data> 1234</data>
    <data> 1235</data>
    <data> 1236</data>
  </project>
  <project name="b">
    <data> 12</data>
    <data> 41234</data>
    <data> 51235</data>
    <data> 71236</data>
  </project>
  <project name="c">
    <data> 1</data>
    <data> 41234</data>
    <data> 51235</data>
    <data> 71236</data>
  </project>
</projects>

so if $column=1 then it would sort them into {project 
name="c", project name="b", project name="a"} however i've 
tried <xsl:for-each select="projects/project">
    <xsl:sort select="./data[$column]" data-type="number"/>
but it doesn't work?
however if just use something like
    <xsl:sort select="./data[1]" data-type="number"/> it works fine

anyone can tell me what im doing wrong?

regards
  Kasper Nielsen


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