从一系列日期中获取所有前 5 个 id（一个数组）

Question

我需要从 tableA 中查询前 5 个值。比如下面

    select id, count(occurrence), date 
    from 
    (select id, unnest(value) as occurrence, date from tableA) as a
    group by id, occurrence, date 
    order by occurrence desc 
    limit 5

 id  | occurrence |   date   
-----+-------+----------
 330 |    11 | 20141015
 400 |    11 | 20141015
 390 |    10 | 20141015
 240 |    10 | 20141015
 501 |    10 | 20141015

收到ids后，查询同一个tableA，获取20140101到20141015其他日期的id的所有值。

expected result:

 id  | occurrence |   date   
-----+-------+----------
 330 |    11 | 20141015
 400 |    11 | 20141015
 390 |    10 | 20141015
 240 |    10 | 20141015
 501 |    10 | 20141015
 330 |    0  | 20141014
 400 |    1  | 20141014
 390 |    10 | 20141014
 240 |    15 | 20141014
 501 |    10 | 20141014
 330 |    11 | 20141013
 400 |    11 | 20141013
 390 |    11 | 20141013
 240 |    19 | 20141013
 501 |    10 | 20141013

但我究竟该怎么做呢？

我的 postgresql 版本是 8.1，我不能使用分区（如果你想建议的话）

---

编辑

select  id, count(value) as occurrence, date
from tableA
where id = ANY(
    select array(
        select id
        from (
            select date, unnest(id) as id
            from tableA where date>='20140101' and date<='20141015'
            )as a
        )
)
group by id, date

不返回任何内容。我的数组正确吗？

Answer 1

你可以尝试 select sth from table where dt > 20140101 and dt

Answer 2

select id, count(value) as occurrence, date
from t
where id in (
    select id
    from (
        select id, count(value) as occurrence, date 
        from t 
        group by id, date 
        order by occurrence desc 
        limit 5
    ) s
) q
group by id, date