Max value in a many-to-many relationship
I'm using SQL Server 2008 and I have 3 tables, x
, y
and z
. y
exists to create a many-to-many relationship between x
and z
.
x y z
-- -- --
id xid id
zid sort
All of the above fields are int
.
I want to find the best-performing method (excluding denormalising) of finding the z
with the highest sort
for any x
, and return all fields from all three tables.
Sample data:
x: id
--
1
2
y: xid zid
--- ---
1 1
1 2
1 开发者_运维百科 3
2 2
z: id sort
-- ----
1 5
2 10
3 25
Result set should be
xid zid
--- ---
1 3
2 2
Note that if more than one z
exists with the same highest sort
value, then I still only want one row per x
.
Note also that in my real-world situation, there are other fields in all three tables which I will need in my result set.
One method is with a sub query. This however is only good for getting the ID of Z. If you need more/all columns from both x and z tables then this is not the best solution.
SELECT
x.id,
(
SELECT TOP 1
z.zid
FROM
y
INNER JOIN
z
ON
z.id = y.zid
WHERE
y.xid = x.id
ORDER BY
z.sort DESC
)
FROM
x
This is how you can do it and return all the data from all the tables.
SELECT
*
FROM
x
INNER JOIN
y
ON
y.xid = x.id
AND
y.zid =
(
SELECT TOP 1
z2.zid
FROM
y y2
INNER JOIN
z z2
ON
z2.id = y2.zid
WHERE
y2.xid = x.id
ORDER BY
z2.sort DESC
)
INNER JOIN
z
ON
z.id = y.zid
select xid,max(zid) as zid from y
group by xid
select xid, zid /* columns from x; and columns from y or z taken from q */
from (select y.xid, y.zid, /* columns from y or z */
row_number() over(partition by y.xid order by z.sort desc) r
from y
join z on z.id = y.zid
) q
join x on x.id = q.xid
where r = 1
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