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

[que Li <queincanada@xxxxxxxx>]

Hi Geert:
     Thanks for help.
     If I do the way which you suggest, How I can
figure out the relative position which node is on the
needed node list. 

example : the node which I want is:
<List >
  <List_ID>10</List_ID>
   <Title>A</Title>
  <Parent_ID>1</Parent_ID>
 </List>
<List >
    <List_ID>12</List_ID>
   <Title>B</Title>
  <Parent_ID>20</Parent_ID>
 </List>
 <List >
   <List_ID>14</List_ID>
   <Title>C</Title>
    <Parent_ID>1</Parent_ID>
 </List>

and position() shoule be :
   List_ID=10 . position 1
   List_ID=12. position 2
   List_ID=14 . position 3


But if I do:
<xsl:when test="Parent_ID = 1">
        <!-- found a valid case -->
      </xsl:when>

  <xsl:when
 test="not(preceding-sibling::List[List_ID =
 $self/Parent_ID]
                          or
 following-sibling::List[List_ID =
 $self/Parent_ID])">

Then I can't figure out how to tell node is first one
or the last one during the node which I need list.
their position will  be:
   List_ID=10 . position 1
   List_ID=12. position 3
   List_ID=14 . position 4



My xml file:

<Lists>
<List >
  <List_ID>10</List_ID>
   <Title>A</Title>
  <Parent_ID>1</Parent_ID>
 </List>
 <List>
   <List_ID>11</List_ID>
   <Title>A1</Title>
   <Parent_ID>10</Parent_ID>
 </List>
<List >
    <List_ID>12</List_ID>
   <Title>B</Title>
  <Parent_ID>20</Parent_ID>
 </List>
 <List >
   <List_ID>14</List_ID>
   <Title>C</Title>
    <Parent_ID>1</Parent_ID>
 </List>
  </Lists>


any idea?

Thanks
Helena




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