[xquery-talk] Return Position()
Vallone, Philip Mr CTR USA AMC
Philip.Vallone at us.army.mil
Fri Jun 13 11:07:14 PDT 2008
Hi List,
I am trying to return the position of each <row> in the following xml
example:
Sample XML:
<table>
<title>
<pcnarr>Table 3. Response</pcnarr>
</title>
<tgroup cols="3">
<colspec colwidth="*"/>
<colspec colwidth="*"/>
<colspec colwidth="*"/>
<tbody>
<row>
<entry align="center">
<emphasis emph="bold">
<pcnarr>
Action</pcnarr>
</emphasis>
</entry>
<entry align="center">
<emphasis emph="bold">
<pcnarr>
Action</pcnarr>
</emphasis>
</entry>
<entry align="center">
<emphasis emph="bold">
<pcnarr>Action</pcnarr>
</emphasis>
</entry>
</row>
<row>
<entry align="center">
<emphasis emph="bold">
<pcnarr>
Action</pcnarr>
</emphasis>
</entry>
<entry align="center">
<emphasis emph="bold">
<pcnarr>
Action</pcnarr>
</emphasis>
</entry>
<entry align="center">
<emphasis emph="bold">
<pcnarr>Action</pcnarr>
</emphasis>
</entry>
</row>
</tbody>
</tgroup>
</table>
Here is my Xquery Expression:
for $rows in //table/tgroup/tbody
let $pos := $rows/row/position()
(: I want to return row position:)
return
<row num="{$pos}">Some Data</row>
Which returns:
<row num="1 2">Some Data</row>
Desired results:
<row num="1">Some Data</row>
<row num="2">Some Data</row>
Thanks for the help
Phil
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