xsl:sort - Help Required

[Andy Bowes]

Hi
I have got a section of XML that looks like:

<?xml version="1.0" encoding="UTF-8"?>
<report>
<clients>
<client id="1">
<name>Client1</name>
</client>
<client id="2">
<name>Client2</name>
</client>
<client id="3">
<name>Client3</name>
</client>
</clients>
<orders>
<order order-no="1" client-id="1">
<order-date>12-Sept-03</order-date>
</order>
<order order-no="2" client-id="1">
<order-date>12-Sept-03</order-date>
</order>
<order order-no="3" client-id="3">
<order-date>12-Sept-03</order-date>
</order>
<order order-no="4" client-id="2">
<order-date>12-Sept-03</order-date>
</order>
</orders>
</report>

I want to get a list of orders but sorted by client name and then order id.
i.e. something like :

<xsl:apply-templates select="order">
   <xsl:sort select="/report/clients/client[@id=current()/@client-id]/name"></xsl:sort>
   <xsl:sort data-type="number" select="@id"></xsl:sort>
</xsl:apply-templates>

I have tried all sorts of XPath statements but I have not found the solution, can anyone out there enlighten me.
Thanks

[Lasse Koskela]

XML:

<?xml version="1.0" encoding="UTF-8"?>
<report>
    <clients>
        <client id="1">
            <name>Charles</name>
        </client>
        <client id="2">
            <name>David</name>
        </client>
        <client id="3">
            <name>Bob</name>
        </client>
    </clients>
    <orders>
        <order order-no="2" client-id="1">
            <order-date>11-Sept-03</order-date>
        </order>
        <order order-no="1" client-id="1">
            <order-date>12-Sept-03</order-date>
        </order>
        <order order-no="3" client-id="3">
            <order-date>13-Sept-03</order-date>
        </order>
        <order order-no="4" client-id="2">
            <order-date>14-Sept-03</order-date>
        </order>
    </orders>
</report>

XSL:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl :o utput method="text"/>
    <xsl:template match="/">
        <xsl:apply-templates/>
    </xsl:template>
    <!-- ignore "clients" -->
    <xsl:template match="clients"/>
    <!-- sort "order" elements according to the client's name and call the "order" template -->
    <xsl:template match="orders">
        <xsl:apply-templates select="order">
            <xsl:sort select="/report/clients/client[@id=current()/@client-id]/name"></xsl:sort>
            <xsl:sort data-type="number" select="@order-no"></xsl:sort>
        </xsl:apply-templates>
    </xsl:template>
  
    <!-- print out the order number and client id -->
    <xsl:template match="order">
        order-no=<xsl:value-of select="@order-no"/> client-id=<xsl:value-of select="@client-id"/> client-name=<xsl:value-of select="/report/clients/client[@id=current()/@client-id]/name"/>
    </xsl:template>
</xsl:stylesheet>

Output:

order-no=3 client-id=3 client-name=Bob
order-no=1 client-id=1 client-name=Charles
order-no=2 client-id=1 client-name=Charles
order-no=4 client-id=2 client-name=David

[ October 03, 2003: Message edited by: Lasse Koskela ]

[Andy Bowes]

Thanks Lasse.
I also found another another alternative were I selected all clients in alphabetical order and then select the list of orders for each specific client.
I now need to see which one of these performs best.

[Lasse Koskela]

I now need to see which one of these performs best.

Might I ask to publish your findings here as well? Of course it's only one, very specific case in a specific environment, but I'm just curious...