How to convert this SQL query to XQuery? - sql

Question

How to convert this SQL query into the XQuery equivalent
SELECT e.essn AS one, f.essn AS two, e.pno, f.pno
FROM works_on AS e JOIN works_on AS f on e.pno = f.pno
AND e.essn < f.essn \n" . "ORDER BY `e`.`pno` ASC
I have tried this so far, but it does not give the desired results
let $prods := doc("../company/works_on.xml")//works_on[pno = $project/pnumber]
for $d in distinct-values($prods/essn),
$n in distinct-values($prods[essn = $d]/pno)
return <result essn="{$d}" pno="{$n}"/>
This is a sample of the XML file that I am using.
<dataroot>
<works_on>
<essn>123456789</essn>
<pno>1</pno>
<hours>32.5</hours>
</works_on>
<works_on>
<essn>123456789</essn>
<pno>2</pno>
<hours>7.5</hours>
</works_on>
<works_on>
<essn>333445555</essn>
<pno>2</pno>
<hours>10</hours>
</works_on>
<works_on>
<essn>333445555</essn>
<pno>3</pno>
<hours>10</hours>
</works_on>
This is the output so far
<results>
<project>
<pnumber>1</pnumber>
<employee>
<essn>123456789</essn>
<essn>453453453</essn>
</employee>
</project>
<project>
<pnumber>2</pnumber>
<employee>
<essn>123456789</essn>
<essn>333445555</essn>
<essn>453453453</essn>
</employee>
</project>
<project>
<pnumber>3</pnumber>
<employee>
<essn>333445555</essn>
<essn>666884444</essn>
</employee>
</project>
<project>
<pnumber>10</pnumber>
<employee>
<essn>333445555</essn>
<essn>987987987</essn>
<essn>999887777</essn>
</employee>
</project>
<project>
<pnumber>20</pnumber>
<employee>
<essn>333445555</essn>
<essn>888665555</essn>
<essn>987654321</essn>
</employee>
</project>
<project>
<pnumber>30</pnumber>
<employee>
<essn>987654321</essn>
<essn>987987987</essn>
<essn>999887777</essn>
</employee>
</project>
</results>
I'm trying to get the values for essn into pairs for example, on project 1, it will be as is, but for project 2, there will be 123456789 and 333445555 as one pair, 123456789 and 453453453 as another pair, and finally 333445555 and 453453453 as the final pair. They should be non repeating and non reversed pairs.

Answer 1

Your provided XML is cut for brevety (a good idea btw), but - which makes it a bit difficult to understand your issue - it is no more corresponding to your epected output.
At the moment I'm sitting in front of a SQL-Server, so the test code is T-SQL, but the idea of the XQuery should be the same with other engines:
--Your XML
DECLARE #xml XML=
N'<dataroot>
<works_on>
<essn>123456789</essn>
<pno>1</pno>
<hours>32.5</hours>
</works_on>
<works_on>
<essn>123456789</essn>
<pno>2</pno>
<hours>7.5</hours>
</works_on>
<works_on>
<essn>333445555</essn>
<pno>2</pno>
<hours>10</hours>
</works_on>
<works_on>
<essn>333445555</essn>
<pno>3</pno>
<hours>10</hours>
</works_on>
</dataroot>';
--The T-SQL-Query with the embedded XQuery
SELECT #xml.query
('<results>
{
for $p in distinct-values(/dataroot/works_on/pno/text())
order by $p
return <project><pnumber>{$p}</pnumber>
<employee>
{
for $e in distinct-values(/dataroot/works_on[pno=$p]/essn/text())
order by $e
return <essn>{$e}</essn>
}
</employee>
</project>
}
</results>')
The result (for the shortened XML)
<results>
<project>
<pnumber>1</pnumber>
<employee>
<essn>123456789</essn>
</employee>
</project>
<project>
<pnumber>2</pnumber>
<employee>
<essn>123456789</essn>
<essn>333445555</essn>
</employee>
</project>
<project>
<pnumber>3</pnumber>
<employee>
<essn>333445555</essn>
</employee>
</project>
</results>
The idea in short:
We find a distinct list of project numbers and write out the main output structure with the project numbers. Internally we create the <employee> element and fill it with the <essn> values found below any <works_on>, which has a <pno> as the current number.
Hope this helps...