Sql 按组聚合Postgres中的存储
我有这样一个hstore的数据:Sql 按组聚合Postgres中的存储,sql,postgresql,postgresql-9.5,Sql,Postgresql,Postgresql 9.5,我有这样一个hstore的数据: |brand|account|likes|views | |-----|-------|-----|----------------------| |Ford |ford_uk|1 |"3"=>"100" | |Ford |ford_us|2 |"3"=>"200", "5"=>"10" | |Jeep |jeep_uk|3 |"3"=>"300"
|brand|account|likes|views |
|-----|-------|-----|----------------------|
|Ford |ford_uk|1 |"3"=>"100" |
|Ford |ford_us|2 |"3"=>"200", "5"=>"10" |
|Jeep |jeep_uk|3 |"3"=>"300" |
|Jeep |jeep_us|4 |"3"=>"400", "5"=>"20" |
我希望能够按键、按品牌对商店进行汇总:
|brand|likes|views |
|-----|-----|----------------------|
|Ford |3 |"3"=>"300", "5"=>"10" |
|Jeep |7 |"3"=>"700", "5"=>"20" |
提供了一个很好的解决方案,说明如何在没有分组的情况下执行此操作。使其适应这种情况会产生如下结果:
SELECT
sum(likes) AS total_likes,
(SELECT hstore(array_agg(key), array_agg(value::text))
FROM (
SELECT s.key, sum(s.value::integer)
FROM (
SELECT((each(views)).*)
) AS s(key, value)
GROUP BY key
) x(key, value)) AS total_views
FROM my_table
GROUP BY brand
然而,这就产生了:
错误:子查询使用外部查询中未分组的列“my_table.views”
感谢您的帮助 这是因为在
分组依据
查询中使用的视图
列没有聚合函数。非常快速的解决方法:
with my_table(brand,account,likes,views) as (
values
('Ford', 'ford_uk', 1, '"3"=>"100"'::hstore),
('Ford', 'ford_uk', 2, '"3"=>"200", "5"=>"10"'),
('Jeep', 'jeep_uk', 3, '"3"=>"300"'::hstore),
('Jeep', 'jeep_uk', 4, '"3"=>"400", "5"=>"20"'))
SELECT
brand,
sum(likes) AS total_likes,
(SELECT hstore(array_agg(key), array_agg(value::text))
FROM (
SELECT s.key, sum(s.value::integer)
FROM
unnest(array_agg(views)) AS h, --<< aggregate views according to the group by, then unnest it into the table
each(h) as s(key,value)
GROUP BY key
) x(key, value)) AS total_views
FROM my_table
GROUP BY brand
之后,您的查询将更简单、更“规范”:
更新2
即使没有create aggregate
,函数hstore\u sum\u ffunc
也可能有用:
select
brand,
sum(likes) as total_likes,
hstore_sum_ffunc(array_agg(views)) as total_views
from my_table
group by brand;
这是因为在
分组依据
查询中使用的视图
列没有聚合函数。非常快速的解决方法:
with my_table(brand,account,likes,views) as (
values
('Ford', 'ford_uk', 1, '"3"=>"100"'::hstore),
('Ford', 'ford_uk', 2, '"3"=>"200", "5"=>"10"'),
('Jeep', 'jeep_uk', 3, '"3"=>"300"'::hstore),
('Jeep', 'jeep_uk', 4, '"3"=>"400", "5"=>"20"'))
SELECT
brand,
sum(likes) AS total_likes,
(SELECT hstore(array_agg(key), array_agg(value::text))
FROM (
SELECT s.key, sum(s.value::integer)
FROM
unnest(array_agg(views)) AS h, --<< aggregate views according to the group by, then unnest it into the table
each(h) as s(key,value)
GROUP BY key
) x(key, value)) AS total_views
FROM my_table
GROUP BY brand
之后,您的查询将更简单、更“规范”:
更新2
即使没有create aggregate
,函数hstore\u sum\u ffunc
也可能有用:
select
brand,
sum(likes) as total_likes,
hstore_sum_ffunc(array_agg(views)) as total_views
from my_table
group by brand;
如果为
hstore
创建一个聚合,这会变得更容易:
create aggregate hstore_agg(hstore)
(
sfunc = hs_concat(hstore, hstore),
stype = hstore
);
然后你可以这样做:
with totals as (
select t1.brand,
hstore(k, sum(v::int)::text) as views
from my_table t1, each(views) x(k,v)
group by brand, k
)
select brand,
(select sum(likes) from my_table t2 where t1.brand = t2.brand) as likes,
hstore_agg(views) as views
from totals t1
group by brand;
另一个选项是将可能较慢的相关子查询移动到CTE中:
with vals as (
select t1.brand,
hstore(k, sum(v::int)::text) as views
from my_table t1, each(views) x(k,v)
group by brand, k
), view_totals as (
select brand,
hstore_agg(views) as views
from vals
group by brand
), like_totals as (
select brand,
sum(likes) as likes
from my_table
group by brand
)
select vt.brand,
lt.likes,
vt.views
from view_totals vt
join like_totals lt on vt.brand = lt.brand
order by brand;
如果为
hstore
创建一个聚合,这会变得更容易:
create aggregate hstore_agg(hstore)
(
sfunc = hs_concat(hstore, hstore),
stype = hstore
);
然后你可以这样做:
with totals as (
select t1.brand,
hstore(k, sum(v::int)::text) as views
from my_table t1, each(views) x(k,v)
group by brand, k
)
select brand,
(select sum(likes) from my_table t2 where t1.brand = t2.brand) as likes,
hstore_agg(views) as views
from totals t1
group by brand;
另一个选项是将可能较慢的相关子查询移动到CTE中:
with vals as (
select t1.brand,
hstore(k, sum(v::int)::text) as views
from my_table t1, each(views) x(k,v)
group by brand, k
), view_totals as (
select brand,
hstore_agg(views) as views
from vals
group by brand
), like_totals as (
select brand,
sum(likes) as likes
from my_table
group by brand
)
select vt.brand,
lt.likes,
vt.views
from view_totals vt
join like_totals lt on vt.brand = lt.brand
order by brand;