由许多表组成

时间:2017-08-29 08:08:21

标签: sql postgresql group-by

我有下表:

performance
  --id
  --color
  --installs
  --date

performance_groups
  --id
  --performance_id
  --group_id

我想要一个像这样的SQL:

 SELECT color, targeting_id, SUM(installs) as installs
 FROM performance, performance_groups
GROUP BY color, group_id

但我希望所有小组都能完成分组。

例如:

performance
id     color     installs   date
1      Blue      5          2017-07-05
2      Red       10         2017-07-04
3      Blue      10         2017-07-04
4      Blue      10         2017-07-05

performance_groups
id   performance_id   group_id
1    1                1
2    1                2
3    2                3
4    3                1
5    3                2
6    4                1
7    4                3

我想得到这样的结果:

color group_ids installs
Blue  1,2     15
Red   3       10
Blue  1,3     10

4 个答案:

答案 0 :(得分:3)

从不FROM子句中使用逗号。始终使用正确的,明确的JOIN语法。

您的查询似乎是JOINGROUP BY

select p.color, string_agg(pg.group_id) as groups, 
       sum(installs) as installs
from performance p join
     performance_groups pg
     on pg.performance_id = p.id
group by color;

答案 1 :(得分:1)

使用array_agg并且不要忘记区别。联接可能会产生重复。

select p.color, array_agg(distinct pg.group_id) as groups, 
       sum(distinct installs) as installs
from performance p 
join performance_groups pg
     on pg.performance_id = p.id
group by color;

sqlfiddle demo

答案 2 :(得分:1)

感谢Gordon和Radium,我的答案得到了启发。

最终的解决方案是:

with pg as(
    select performance_id, array_agg(distinct group_id) as groups from performance_groups
    group by performance_id
) 
select groups, sum(installs) from performance join pg 
on pg.performance_id = performance.id
group by groups

见工作sqlfiddle

答案 3 :(得分:0)

We are using STUFF,XML PATH and GROUP BY in SQL 2008.

代码:

    select f.color,
    (SELECT STUFF(( SELECT ',' + convert(varchar(5),j.group_id) 
    FROM (select distinct g.group_id from performance_groups g 
    where g.performance_id in (select id from performance where color= 
    f.color)) j FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'), 1, 1, 
    '')) as group_ids,
    SUM(f.installs) installs
    from performance f group by color

输出:

color   group_ids   installs
Blue    1,2         15
Red     3           10