我有两张桌子:
person_id | name
1 name1
2 name2
3 name3
和第二张表:
person_id | date | balance
1 2016-03 1200 ---- \
1 2016-04 700 ---- > same person
1 2016-05 400 ---- /
3 2016-05 4000
考虑到person_id 1在第二个表上有三个记录,如何才能加入第一个记录? (即:余额400,对应日期:2016-05)。
例如:查询输出:
person_id | name | balance
1 name1 400
2 name2 ---
3 name3 4000
如果它可能更喜欢简单而不是解决方案的复杂性
答案 0 :(得分:5)
适用于所有数据库引擎的查询是
select t1.name, t2.person_id, t2.balance
from table1 t1
join table2 t2 on t1.person_id = t2.person_id
join
(
select person_id, max(date) as mdate
from table2
group by person_id
) t3 on t2.person_id = t3.person_id and t2.date = t3.mdate
答案 1 :(得分:1)
在任何支持ANSI标准窗口函数(大部分都是)的数据库中执行此操作的最佳方法是:
select t1.*, t2.balance
from table1 t1 left join
(select t2.*,
row_number() over (partition by person_id order by date desc) as seqnum
from table2 t2
) t2
on t1.person_id = t2.person_id and seqnum = 1;