Consider the XML and SQL:
declare @xml xml = '
<root>
<person id="11272">
<notes for="107">Some notes!</notes>
<item id="107" selected="1" />
</person>
<person id="77812">
<notes for="107"></notes>
<notes for="119">Hello</notes>
<item id="107" selected="0" />
<item id="119" selected="1" />
</person>
</root>'
select Row.Person.value('data(../@id)', 'int') as person_id,
Row.Person.value('data(@id)', 'int') as item_id,
Row.Person.value('data(../notes[@for=data(@id)][1])', 'varchar(max)') as notes
from @xml.nodes('/root/person/item') as Row(Person)
I end up with:
person_id item_id notes
----------- ----------- -------
77812 107 NULL
77812 119 NULL
11272 107 NULL
What I want is the ‘notes’ column to be pulled based on the @id attribute of the current item. If I replace [@for=data(@id)] in the selector with [@for=107] of course I get the value Some notes! in the last record. Is it possible to do this with XPath/XQuery, or am I barking up the wrong tree here? I think the problem is that
The XML is a bit awkward, yes, but I can’t really change it I’m afraid.
I found one solution that works, but it feels awfully heavy for something like this.
select Item.Person.value('data(../@id)', 'int') as person_id,
Item.Person.value('data(@id)', 'int') as item_id,
Notes.Person.value('text()[1]', 'varchar(max)') as notes
from @xml.nodes('/root/person/item') as Item(Person)
inner join @xml.nodes('/root/person/notes') as Notes(Person) on
Notes.Person.value('data(@for)', 'int') = Item.Person.value('data(@id)', 'int')
and
Notes.Person.value('data(../@id)', 'int') = Item.Person.value('data(../@id)', 'int')
Update!
I figured it out! I’m new at XQuery but this works, so I’m calling it job done 🙂 I changed the query for the notes to:
Item.Person.value('
let $id := data(@id)
return data(../notes[@for=$id])[1]
', 'varchar(max)') as notes
I would suggest that you do a
cross applyinstead of doing../to find a parent node. According to query plan it is a lot faster.You can even remove the
../in the flwor with one extra cross apply gaining a bit more.Comparing the queries against each other I got 67% on your query 17% on my first and 16% on the second. Note: these figures only give you a hint on what query will actually be faster in reality. Test the against your data to know for sure.