Question

I’m looking to get the max discrepancy between two tables per day, per id. I have the following data in a mysql database

insert into test.foo values ('2010-01-10', 1, 10);
insert into test.foo values ('2010-01-10', 1, 5);
insert into test.foo values ('2010-01-10', 2, 10);
insert into test.foo values ('2010-01-10', 2, 10);
insert into test.foo values ('2010-01-10', 3, 15);
insert into test.foo values ('2010-01-10', 3, 15);
insert into test.foo values ('2010-01-11', 1, 5);
insert into test.foo values ('2010-01-11', 1, 5);
insert into test.foo values ('2010-01-11', 2, 5);
insert into test.foo values ('2010-01-11', 2, 5);
insert into test.foo values ('2010-01-11', 3, 5);
insert into test.foo values ('2010-01-11', 3, 5);

insert into test.bar values ('2010-01-10', 1, 5);
insert into test.bar values ('2010-01-10', 1, 5);
insert into test.bar values ('2010-01-10', 2, 5);
insert into test.bar values ('2010-01-10', 2, 5);
insert into test.bar values ('2010-01-10', 3, 5);
insert into test.bar values ('2010-01-10', 3, 5);
insert into test.bar values ('2010-01-11', 1, 10);
insert into test.bar values ('2010-01-11', 1, 10);
insert into test.bar values ('2010-01-11', 2, 5);
insert into test.bar values ('2010-01-11', 2, 5);
insert into test.bar values ('2010-01-11', 3, 5);
insert into test.bar values ('2010-01-11', 3, 5);

Here is my query:

SELECT t1.`date`, t1.id, t1.sums, t2.sums, max(t1.sums - t2.sums) FROM
  (select `date`, id, sum(val) sums
   from test.foo
   group by `date`, id) as t1,
  (select `date`, id, sum(val) sums
   from test.bar
   group by `date`, id) as t2
WHERE t1.`date` = t2.`date` AND t1.id = t2.id
group by t1.`date`

I’m getting this result:

+---------------------+----+------+------+------------------------+
| date                | id | sums | sums | max(t1.sums - t2.sums) |
+---------------------+----+------+------+------------------------+
| 2010-01-10 00:00:00 |  1 |   15 |   10 |                     20 |
| 2010-01-11 00:00:00 |  1 |   10 |   20 |                      0 |
+---------------------+----+------+------+------------------------+
2 rows in set (0.00 sec)

I’d like to be getting this result:
I’m getting this result:

+---------------------+----+------+------+------------------------+
| date                | id | sums | sums | max(t1.sums - t2.sums) |
+---------------------+----+------+------+------------------------+
| 2010-01-10 00:00:00 |  1 |   15 |   10 |                     20 |
| 2010-01-11 00:00:00 |  2 |   10 |   10 |                      0 |  <----- 
+---------------------+----+------+------+------------------------+

Can anyone help me? I’d like to be getting the max difference, and then the line that went along with it. This query gives me the correct difference, but not the id and sums that go with it. A colleague suggested also grouping by id, but as I thought, that just flattened out the result and each id was listed instead of the one id for the day that had the max difference.

Thanks much in advance

Answer 1

This one should work for you.

It sorts the sums in descending order, assigning them a rank, and then only get those with rank=1.

SELECT id, `date`, sums FROM (
  SELECT id, `date`, sums,
  CASE
    WHEN @d != `date` THEN @rownum := 1 
    ELSE @rownum := @rownum + 1
  END AS rank,
  @d := `date`
FROM
(
  SELECT t1.`date`, t1.id, t1.sums t1_sums, t2.sums t2_sums, (t1.sums - t2.sums) sums
  FROM
    (select `date`, id, sum(val) sums
     from foo
     group by `date`, id) as t1,
    (select `date`, id, sum(val) sums
     from bar
     group by `date`, id) as t2,
     (SELECT @rownum := 0, @d := NULL) r
  WHERE t1.`date` = t2.`date` AND t1.id = t2.id
  GROUP BY t1.`date`, t1.id, t2.`date`, t2.id
  ORDER BY t1.`date`, (t1.sums - t2.sums) DESC, t1.id
  ) x
) y
WHERE rank = 1