我想做这样的事情：
select id,
count(*) as total,
FOR temp IN SELECT DISTINCT somerow FROM mytable ORDER BY somerow LOOP
sum(case when somerow = temp then 1 else 0 end) temp,
END LOOP;
from mytable
group by id
order by id
我创建了工作选择：
select id,
count(*) as total,
sum(case when somerow = 'a' then 1 else 0 end) somerow_a,
sum(case when somerow = 'b' then 1 else 0 end) somerow_b,
sum(case when somerow = 'c' then 1 else 0 end) somerow_c,
sum(case when somerow = 'd' then 1 else 0 end) somerow_d,
sum(case when somerow = 'e' then 1 else 0 end) somerow_e,
sum(case when somerow = 'f' then 1 else 0 end) somerow_f,
sum(case when somerow = 'g' then 1 else 0 end) somerow_g,
sum(case when somerow = 'h' then 1 else 0 end) somerow_h,
sum(case when somerow = 'i' then 1 else 0 end) somerow_i,
sum(case when somerow = 'j' then 1 else 0 end) somerow_j,
sum(case when somerow = 'k' then 1 else 0 end) somerow_k
from mytable
group by id
order by id
这是有效的，但它是'静态' - 如果将某些新值添加到'somerow'，我将不得不手动更改sql以从somerow列获取所有值，这就是为什么我想知道它是否可以用for循环做一些事情。
所以我想得到的是：
id somerow_a somerow_b ....
0 3 2 ....
1 2 10 ....
2 19 3 ....
. ... ...
. ... ...
. ... ...
所以我想做的是计算所有包含特定字母的行并按id分组（此id不是主键，但它重复 - 对于id大约有80种不同价值可能）。
答案 0 :(得分：3)
阵列对你有好处吗？ (SQL Fiddle)
select
id,
sum(totalcol) as total,
array_agg(somecol) as somecol,
array_agg(totalcol) as totalcol
from (
select id, somecol, count(*) as totalcol
from mytable
group by id, somecol
) s
group by id
;
id | total | somecol | totalcol
----+-------+---------+----------
1 | 6 | {b,a,c} | {2,1,3}
2 | 5 | {d,f} | {2,3}
在9.2中，可以有一组JSON对象(Fiddle)
select row_to_json(s)
from (
select
id,
sum(totalcol) as total,
array_agg(somecol) as somecol,
array_agg(totalcol) as totalcol
from (
select id, somecol, count(*) as totalcol
from mytable
group by id, somecol
) s
group by id
) s
;
row_to_json
---------------------------------------------------------------
{"id":1,"total":6,"somecol":["b","a","c"],"totalcol":[2,1,3]}
{"id":2,"total":5,"somecol":["d","f"],"totalcol":[2,3]}
在9.3中，添加了lateral，单个对象(Fiddle)
select to_json(format('{%s}', (string_agg(j, ','))))
from (
select format('%s:%s', to_json(id), to_json(c)) as j
from
(
select
id,
sum(totalcol) as total_sum,
array_agg(somecol) as somecol_array,
array_agg(totalcol) as totalcol_array
from (
select id, somecol, count(*) as totalcol
from mytable
group by id, somecol
) s
group by id
) s
cross join lateral
(
select
total_sum as total,
somecol_array as somecol,
totalcol_array as totalcol
) c
) s
;
to_json
---------------------------------------------------------------------------------------------------------------------------------------
"{1:{\"total\":6,\"somecol\":[\"b\",\"a\",\"c\"],\"totalcol\":[2,1,3]},2:{\"total\":5,\"somecol\":[\"d\",\"f\"],\"totalcol\":[2,3]}}"
在9.2中，也可以使用子查询而不是lateral
答案 1 :(得分：1)
SQL对返回类型非常严格。它需要事先知道要返回什么。
对于完全动态数量的结果值，您只能使用@Clodoaldo posted之类的数组。实际上，静态返回类型不会为每个值获取单独的列。
如果您知道通话时的列数（＆＃34;半动态＆＃34; ），您可以创建一个获取（并返回）多态参数的函数。与许多细节密切相关的答案：
（您还可以在那里找到@Clodoaldo数组的相关答案。）
您剩下的选择是对服务器使用两次往返。第一个用实际返回类型确定实际查询。第二个是根据第一次调用执行查询。
否则，您必须使用静态查询。在这样做的同时，我现在看到了两个更好的选项：
select id
, count(*) AS total
, count(somecol = 'a' OR NULL) AS somerow_a
, count(somecol = 'b' OR NULL) AS somerow_b
, ...
from mytable
group by id
order by id;
它是如何运作的？
crosstab() crosstab()起初更复杂，但用C语言编写，针对任务进行了优化，对于长列表则更短。您需要安装附加模块tablefunc。如果您不熟悉，请阅读此处的基础知识：
SELECT * FROM crosstab(
$$
SELECT id
, count(*) OVER (PARTITION BY id)::int AS total
, somecol
, count(*)::int AS ct -- casting to int, don't think you need bigint?
FROM mytable
GROUP BY 1,3
ORDER BY 1,3
$$
,
$$SELECT unnest('{a,b,c,d}'::text[])$$
) AS f (id int, total int, a int, b int, c int, d int);