SQL子查询试图获得两个表中同一列的最大差异

Question

我正在寻找每个ID每天两个表之间的最大差异。我有一个MySQL数据库

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); 

在这里，下面的数据是我的查询：

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` 

我得到这个结果：

+---------------------+----+------+------+------------------------+ 
| 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) 

我想收到此结果： 我得到这个结果：

+---------------------+----+------+------+------------------------+ 
| 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 | <----- 
+---------------------+----+------+------+------------------------+ 

任何人都可以帮助我吗？我希望得到最大的区别，然后是与之相伴的路线。这个查询给了我正确的区别，但不是与它一起使用的id和总数。一位同事也建议按ID分组，但正如我所认为的那样，只是将结果放平，并且每个ID都列出来，而不是一天中具有最大差异的一个ID。

感谢很多提前

Answer 1

这一个应该为你工作。

它按降序对总和进行排序，给它们分配一个等级，然后只得到rank = 1的那些。以下查询

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 

Answer 2

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` 

在你的日期，但不是ID分组外部查询，因此你没有得到您所期望的ID。如果您想查找与最高差异相关的ID，那么首先您需要找到最高的差异性，然后使用另一个查询来确定哪些ID（或多个ID）与此相关联。你必须决定你想要做什么与重复。像这样的东西（未经测试）...

SELECT t3.`date`, t3.id, t3.diff 
    (SELECT t1.`date`, t1.id, t1.sums, t2.sums, max(t1.sums - t2.sums) as diff 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) as t3 
WHERE t3.diff = (correlated subquery to get maximum value of diff for each date) 

或使用单独的查询。

Answer 3

输出是：

+----------+- ---+------+------+-------+ 
|date  |id |sumf |sumb |maxdiff| 
+----------+- ---+------+------+-------+ 
|2010-01-10| 1| 30| 20|  10| 
|2010-01-10| 2| 40| 20|  20| 
|2010-01-10| 3| 60| 20|  40| 
|2010-01-11| 1| 20| 40|  20| 
|2010-01-11| 2| 20| 20|  0| 
|2010-01-11| 3| 20| 20|  0| 
+----------+- ---+------+------+-------+ 


select m.date, m.id, s.sumf, s.sumb, m.maxdiff 
from (
    --subquery2: get the maximum different absolute sum between foo and bar for each date/id combination 
    select s.date, s.id, max(abs(s.sumf - s.sumb)) as maxdiff 
    from (
     --subquery1: get the sum of values for each date/id combination 
     select f.date, f.id, sum(f.val) as sumf, sum(b.val) as sumb 
     from foo f 
     inner join bar b on f.date = b.date and f.id = b.id 
     group by f.date, f.id 
    ) s 
    group by s.date, s.id 
) m 
--join back against subquery1 to find out which sums gave us the max difference 
inner join (
    select f.date, f.id, sum(f.val) as sumf, sum(b.val) as sumb 
    from foo f 
    inner join bar b on f.date = b.date and f.id = b.id 
    group by f.date, f.id 
) s on m.date = s.date and m.id = s.id and m.maxdiff = abs(s.sumf - s.sumb) 

注：这将在被减去标的金额重复maxdiff S的情况下返回多行有不同的值。如果您必须返回sumf和sumb，我相信这是正确的行为，否则您不一定会获得创建maxdiff的值。