[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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