Sql 如何选择最近日期项目
我有一个表用户:Sql 如何选择最近日期项目,sql,oracle,greatest-n-per-group,Sql,Oracle,Greatest N Per Group,我有一个表用户: select t.* from user t; uid uname 1 tom 2 jim 3 bob 4 lily 和桌上玩具 select t.* from toys t; tid uid tdate 1 1 7/12/15 2 1 6/12/15 3 2 9/12/15 4 2 10/12/15 5 3 12/12/15 现在我想要 uid tid uname tdate
select t.* from user t;
uid uname
1 tom
2 jim
3 bob
4 lily
和桌上玩具
select t.* from toys t;
tid uid tdate
1 1 7/12/15
2 1 6/12/15
3 2 9/12/15
4 2 10/12/15
5 3 12/12/15
现在我想要
uid tid uname tdate
1 2 tom 6/12/15
2 3 jim 9/12/15
3 5 bob 12/12/15
4 lily
我该怎么办?(我使用oracle数据库) 每个玩具和用户的最新日期:
SELECT a.uid, b.tid, a.uname, MAX(b.tdate) AS tdate
FROM user a LEFT JOIN toys b ON a.uid = b.uid
GROUP BY a.uid, b.tid, a.uname
或每个用户最近的玩具(如有):
您可以为每个用户生成一个序列,并仅返回每个玩具的最小日期:
SELECT user.uid, toys.tid, user.name, toys.tdate
FROM
(SELECT tid, uid, tdate,
ROW_NUMBER() OVER (PARTITION BY uid ORDER BY tdate asc)
AS SEQ
FROM toys
)table1
LEFT JOIN user on toys.uid = user.uid
WHERE SEQ = 1
您希望玩具的用户按日期排序,还是只希望玩具的用户按最新日期排序?使用最短日期,您认为是正确的,但写入错误。此sql无法执行
SELECT user.uid, toys.tid, user.name, toys.tdate
FROM
(SELECT tid, uid, tdate,
ROW_NUMBER() OVER (PARTITION BY uid ORDER BY tdate asc)
AS SEQ
FROM toys
)table1
LEFT JOIN user on toys.uid = user.uid
WHERE SEQ = 1