sql在每个月查找最小日期的行

时间:2012-11-22 19:51:05

标签: sql postgresql min

我有一张桌子,让我们说“记录”结构:

id      date
--      ----
1       2012-08-30
2       2012-08-29
3       2012-07-25

我需要在PostgreSQL中编写一个SQL查询来获取每个月的MIN日期的record_id。

month    record_id
-----    ---------
8           2
7           3

我们看到2012-08-29< 2012-08-30,这是8个月,所以我们应该显示record_id = 2

我尝试过这样的事情,

SELECT
   EXTRACT(MONTH FROM date) as month,
   record_id,
   MIN(date) 
FROM Records
GROUP BY 1,2

但它显示了3条记录。

有人可以帮忙吗?

5 个答案:

答案 0 :(得分:3)

SELECT DISTINCT ON (EXTRACT(MONTH FROM date))
   id,
   date
FROM Records1
ORDER BY EXTRACT(MONTH FROM date),date

SQLFiddle http://sqlfiddle.com/#!12/76ca2/3

UPD:此查询:

1)按月份和日期对记录进行排序

2)每个月挑选第一条记录(由于订购,第一条记录有MIN(date)

此处详细信息http://www.postgresql.org/docs/current/static/sql-select.html#SQL-DISTINCT

答案 1 :(得分:2)

SQL Fiddle

select distinct on (date_trunc('month', date))
    date_trunc('month', date) as month,
    id,
    date
from records
order by 1, 3 desc

答案 2 :(得分:2)

如果您有最小重复日期,则会返回倍数:

Select
  minbymonth.Month,
  r.record_id
From (
  Select
    Extract(Month From date) As Month, 
    Min(date) As Date
  From
    records
  Group By 
    Extract(Month From date)
  ) minbymonth
    Inner Join
  records r
    On minbymonth.date = r.date
Order By
  1;

或者如果你有CTE

With MinByMonth As (
  Select
    Extract(Month From date) As Month, 
    Min(date) As Date
  From
    records
  Group By 
    Extract(Month From date)
  )
Select
  m.Month,
  r.record_id
From
  MinByMonth m
    Inner Join
  Records r
    On m.date = r.date
Order By
  1;

http://sqlfiddle.com/#!1/2a054/3

答案 3 :(得分:2)

select extract(month from date) 
, record_id
, date
from
(
    select
        record_id
       , date
       , rank() over (partition by extract(month from date) order by date asc) r
    from records
) x
where r=1
order by date

答案 4 :(得分:1)

我认为您需要使用子查询,如下所示:

SELECT
   EXTRACT(MONTH FROM r.date) as month,
   r.record_id 
FROM Records as r
INNER JOIN (
    SELECT
       EXTRACT(MONTH FROM date) as month,
       MIN(date) as mindate
    FROM Records
    GROUP BY EXTRACT(MONTH FROM date)
) as sub on EXTRACT(MONTH FROM r.date) = sub.month and r.date = sub.mindate