How do I produce an XML response from PostgreSQL 1:n tables? - sql

Question

I have the table "client" with:
id name registered_on status
-- ------- ------------- ------
1 Alice 2020-03-04 a
2 Vincent 2020-03-05 p
3 Anne 2020-03-06 a
And the table "account" with:
client_id account_number type balance
--------- -------------- ---- -------
1 300-1 CHK 100
2 307-5 SAV 24
2 307-6 CHK 350
I created them in DB Fiddle (for a similar question I asked before about producing JSON).
Now, I need a SQL query to produce the 1:n XML document:
<client id="1" name="Alice" registered_on="2020-03-04" status="a">
<account account_number="300-1" type="CHK" balance="100" />
</client>
<client id="2" name="Vincent" registered_on="2020-03-05" status="p">
<account account_number="307-5" type="SAV" balance="24" />
<account account_number="307-6" type="CHK" balance="350" />
</client>
<client id="3" name="Anne" registered_on="2020-03-06" status="a" />
There's a 1:n relationship between the tables and some clients may not have an account (such as "Anne"). The result is a simple join (probably an outer join) that I know how to do. I just don't get how to produce a XML document from it.
If it's makes it easier/shorter I'm open to an alternative XML result, as long as it represents the same data; using tags, instead of attributes, for example.

Answer 1

After trying a bunch of options I was able to find the answer(s).
Original Format: With Attributes
It's possible to produce the XML result using an outer join:
select
xmlserialize(content -- remove this line to keep as XML instead of VARCHAR
xmlagg(r)
as text) -- remove this line to keep as XML instead of VARCHAR
from (
select
xmlelement(name client,
xmlattributes(c.id, c.name, c.registered_on, c.status),
case when count(a.client_id) > 0 then
xmlagg(xmlelement(name account,
xmlattributes(a.account_number, a.type, a.balance) ))
end
) as r
from client c
left join account a on a.client_id = c.id
group by c.id
) s
Or using subqueries (shorter but less performant):
select
xmlserialize(content -- remove this line to keep as XML instead of VARCHAR
xmlagg(
xmlelement(name client, xmlattributes(id, name, registered_on, status),
( select xmlagg(xmlelement(name account,
xmlattributes(a.account_number, a.type, a.balance)
)) from account a where a.client_id = c.id
)
))
as text) -- remove this line to keep as XML instead of VARCHAR
from client c;
Result:
<client id="1" name="Alice" registered_on="2020-03-04" status="a">
<account account_number="300-1" type="CHK" balance="100.00" />
</client>
<client id="2" name="Vincent" registered_on="2020-03-05" status="p">
<account account_number="307-5" type="SAV" balance="24.00" />
<account account_number="307-6" type="CHK" balance="350.00" />
</client>
<client id="3" name="Anne" registered_on="2020-03-06" status="a" />
Alternative Format: Without Attributes
Some people prefer to avoid attributes altogether and always use tags. That can also be done, using:
select
xmlserialize(content -- remove this line to keep as XML instead of VARCHAR
xmlagg(xmlelement(name client,
xmlforest(id, name, registered_on, status),
( select xmlagg(xmlelement(name account,
xmlforest(a.account_number, a.type, a.balance)))
from account a where a.client_id = c.id
)
))
as text) -- remove this line to keep as XML instead of VARCHAR
from client c;
Result:
<client>
<id>1</id>
<name>Alice</name>
<registered_on>2020-03-04</registered_on>
<status>a</status>
<account>
<account_number>300-1</account_number>
<type>CHK</type>
<balance>100.00</balance>
</account>
</client>
<client>
<id>2</id>
<name>Vincent</name>
<registered_on>2020-03-05</registered_on>
<status>p</status>
<account>
<account_number>307-5</account_number>
<type>SAV</type>
<balance>24.00</balance>
</account>
<account>
<account_number>307-6</account_number>
<type>CHK</type>
<balance>350.00</balance>
</account>
</client>
<client>
<id>3</id>
<name>Anne</name>
<registered_on>2020-03-06</registered_on>
<status>a</status>
</client>