Subject: Re: how to set the pattern to get the node
From: que Li <queincanada@xxxxxxxx>
Date: Tue, 30 Nov 2004 11:24:26 -0500 (EST)
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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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