[xsl] how to set the pattern to get the node

Subject: [xsl] how to set the pattern to get the node
From: que Li <queincanada@xxxxxxxx>
Date: Mon, 29 Nov 2004 18:42:56 -0500 (EST)
Hi:
 I can't figure out this problem for several days.

I just wonder is it possible for me to filter node bu
the select ?

xml:
<Lists>
  <List>
   <List_ID>11</List_ID>
   <Title>A1</Title>
    <Parent_ID>10</Parent_ID>
 </List>
<List>
  <List_ID>10</List_ID>
   <Title>A</Title>
  <Parent_ID>1</Parent_ID>
 </List>
 <List>
   <List_ID>12</List_ID>
   <Title>A2</Title>
   <Parent_ID>10</Parent_ID>
 </List>
<List>
    <List_ID>13</List_ID>
   <Title>C</Title>
  <Parent_ID>16</Parent_ID>
 </List>
  </Lists>


How I can get the node which parent_ID =1 or parent_ID
is not 1 but No other sibling node List_ID equal to
current node parent_ID

I try to use:
   <xsl:template match="List[Parent_ID=1 or
not(../List[List_ID = current()/Parent_ID])] ">

But don't work 

Thanks

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