Subject: RE: xsl filter problem
From: "Yuval" <yuvalkro@xxxxxxxxxxxx>
Date: Thu, 26 Dec 2002 12:10:01 +0200
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Hi again,
My xml looks like this :
<customers>
<customer>
<city>a</city>
<name>Jhon<name>
</customer>
<customer>
<city>b</city>
<name>Don<name>
</customer>
<customer>
<city>c</city>
<name>Ron<name>
</customer>
</customers>
The data is to presented in a table and I want to filter it according to
the "city" - meaning I have 4 options :
1.a
2.b
3.c
4.all the cities = no filter
Im using a html select box to let the user choose which city to filter
and the first option is "all cities".for this parameter I give the value
*.
The xsl looks like this :
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output method="xml" encoding="windows-1255" />
<xsl:param name="city" />
<xsl:template match="/">
<customers>
<xsl:for-each
select="customers/customer[city=$city]">
<customer>
<xsl:for-each select="*">
<xsl:element
name='{name()}'>
<xsl:value-of
select="." />
</xsl:element>
</xsl:for-each>
</customer>
</xsl:for-each>
</customers>
</xsl:template>
</xsl:stylesheet>
When choosing "all cities" - the result is an empty table instead of the
list of all cities with the customers names.
I tried to giva the param "city" a diect value of "*" or " '*' " and it
still didn't work.
How can I pass a value to this param which will act like : <xsl:for-each
select="customers/customer[city=*]"> ??
Thanks in advance
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
| Current Thread |
- xsl filter problem
- Yuval - Wed, 25 Dec 2002 18:50:02 -0500 (EST)
- Mike Brown - Wed, 25 Dec 2002 19:59:21 -0500 (EST)
- Yuval - Thu, 26 Dec 2002 05:10:00 -0500 (EST) <=
- Mike Brown - Thu, 26 Dec 2002 06:05:23 -0500 (EST)
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