找到每个键的最新合格行

Question

我的数据库中有这个 table：

我想得到 2 列：id_chantier 和 id_chef。
条件：date_fin not null并且有最后一个date_deb.
所以我想得到的行是数字 1 和 11.
我该怎么做？

Answer 1

你可以用 rank():

select id_chantier, id_chef
from (select t.*, rank() over (order by date_deb desc) as rnk
      from table t
     ) t
where date_fin is not null and rnk = 1;

Answer 2

SELECT DISTINCT ON (id_chef)
       id_chantier, id_chef
FROM   tbl
WHERE  date_fin IS NOT NULL
ORDER  BY id_chef, date_deb DESC NULLS LAST;

DISTINCT ON

的详细信息

Select first row in each GROUP BY group?

根据数据分布，可能有更快的解决方案：

Optimize GROUP BY query to retrieve latest record per user

Answer 3

-- 我想获取已关闭的网站 (chantier) 列表（date_fin 不为空）

SELECT *
FROM ztable t
WHERE date_fin IS NOT NULL
AND NOT EXISTS (
    SELECT * FROM ztable nx
    WHERE nx.id_chantier = t.id_chantier -- Same site
    AND nx.date_fin > t.date_fin         -- more recent
    );

找到每个键的最新合格行

Find the lastest qualifying row for each key

sql

postgresql

greatest-n-per-group