Reputation: 13
Hej. I am new at XML-query and have some problem finding a way to combine the result of two xquery into one.
Consider the XML:
<programs>
<program id="488">
<editor>Anna</editor>
<channel>132</channel>
<category>5</category>
</program>
<program id="178">
<editor>Olle</editor>
<channel>132</channel>
<category>68</category>
</program>
<program id="179">
<editor>Olle</editor>
<channel>132</channel>
<category>10</category>
</program>
</programs>
I want to extract list of editors along with the categories they have worked on which would be like this:
<li>Anna 5 </li>
<li>Olle 68 10</li>
Here is the xquery code I am using
let $editors :=
for $d in $sr/sr/programs/program
where $d/channel = "132"
return $d/editor
let $cat :=
for $a in $sr/sr/programs/program
where $a/editor = data($editors)
return concat($a/editor ,' ', $a/category)
for $result in distinct-values($cat)
return <li>{string($result)}</li>
Appreciate all the helps!
Upvotes: 1
Views: 259
Reputation: 66783
I would iterate over the distinct list of editors, obtained from an XPath with a predicate to restrict to those with channel
equal to 132, and then retrieve a distinct list of categories for those editors with a similar XPath with predicate filter, and then return the HTML with those values:
for $editor in distinct-values($sr/sr/programs/program[channel eq 132]/editor)
let $categories :=
distinct-values($sr/sr/programs/program[channel eq 132 and editor eq $editor]/category)
return <li>{$editor, $categories}</li>
Upvotes: 0