How to perform a 'join' with a sub element in XQuery - xquery

Question

Im not sure how to word what I am trying to do, but I am trying to get all touples of FID, BID, and Something. Consider the following XML:
<FOO>
<FID>f1</FID>
<NAME>f1</NAME>
<BAR>
<BID>b1</BID>
<SOMETHING>15</SOMETHING>
</BAR>
<BAR>
<BID>b2</BID>
<SOMETHING>25</SOMETHING>
</BAR>
</FOO>
<FOO>
<FID>f2</FID>
<NAME>f2</NAME>
<BAR>
<BID>b1</BID>
<SOMETHING>35</SOMETHING>
</BAR>
<BAR>
<BID>b3</BID>
<SOMETHING>0</SOMETHING>
</BAR>
</FOO>
What I need is:
b1 f1 15
b1 f2 35
b2 f1 25
b3 f2 0
Anyone know the syntax that I would use?
I tried:
for $foo in /root/FOO
for $bar in /root/FOO/BAR
let $fid := $foo/FID/text() where $foo/BAR/BID/text()=$bar/BID/text()
let $bid := $foo/BAR/BID/text() where $foo/BAR/BID/text()=$bar/BID/text()
let $something := $foo/BAR/SOMETHING/text() where $foo/BAR/BID/text()=$bar/BID/text()

Answer 1

If you'd wanted to order by the <FID/> elements first, it'd be as easy as looping over the <FOO/>s, for each of them over its <BAR/>s and dumping a string each time:
for $foo in /root/FOO
for $bar in /BAR
return string-join(($foo/FID, $bar/BID, $bar/SOMETHING), ' ')
For grouping by <BID/>s, you have to loop over those first and collect the other information relatively:
for $bar in //BAR
order by $bar/BID
return string-join(($bar/BID, $bar/../FID, $bar/SOMETHING), ' ')
A small remark (as I wasn't able to run your code without major cleanup): $foo and $FOO are not the same, XQuery is case sensitive. Furthermore, you're missing a return clause.