Subject: Re: [xsl] how to set the pattern to get the node From: que Li <queincanada@xxxxxxxx> Date: Tue, 30 Nov 2004 11:24:26 -0500 (EST) |
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 ______________________________________________________________________ Post your free ad now! http://personals.yahoo.ca
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