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
答案 0 :(得分:3)
从不在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JOIN
和GROUP 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;
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