Mysql - 将具有不同日期值的两个表与分组依据组合
Mysql - Combine two tables with different date value with group by
我正在做项目,
我需要显示用户每天获得的积分和交换的积分。
我有两个不同的 table 来存储数据。 tables(a,b)
如下:
table一个:
id user_id earned created_at
--------------------------------------------
1 1 1 2016-12-14
2 2 2 2016-12-14
3 2 1 2016-12-14
4 3 1 2016-12-15
table b:
id user_id exchanged created_at
--------------------------------------------
1 1 1 2016-12-14
2 1 2 2016-12-14
3 2 1 2016-12-14
4 4 1 2016-12-15
5 3 3 2016-12-16
我想合并两个 table 日期如下
user_id earned exchanged created_at
-------------------------------------------------
1 1 1 2016-12-14
2 3 1 2016-12-14
3 1 0 2016-12-15
4 0 1 2016-12-15
3 0 3 2016-12-16
我试过搜索 SO,最后得到以下查询 (sqlfiddle):
select user_id, created_at, sum(earned) as earned, sum(exchanged) as exchanged from (
SELECT
a.user_id,
DATE_FORMAT(a.created_at, '%d-%m-%Y') AS created_at,
a.earned,
0 AS exchanged
FROM
a
LEFT JOIN
b ON DATE_FORMAT(a.created_at, '%y%m%d') = DATE_FORMAT(b.created_at, '%y%m%d')
UNION SELECT
b.user_id,
DATE_FORMAT(b.created_at, '%d-%m-%Y') AS created_at,
0 AS earned,
b.exchanged
FROM
a
RIGHT JOIN
b ON DATE_FORMAT(a.created_at, '%y%m%d') = DATE_FORMAT(b.created_at, '%y%m%d')
) as tbl group by tbl.created_at, tbl.user_id
但是显示的兑换点数不正确
更新:仅当每个用户每天的两个表中都有条目时,此查询才会获取结果。
select ex.user_id, ex.created_at, sum(earned) as earned, sum(exchanged) as exchanged
from exchanged ex, earned ea
where ea.created_at = ex.created_at
and ea.user_id = ex.user_id
group by user_id, created_at
尝试...
select user_id, created_at, sum(earned), sum(exchanged) from
((select id, user_id, earned, 0 as exchanged, created_at from a)
union all
(select id, user_id, 0 as earned, exchanged, created_at from b)) combined
group by user_id, created_at
我正在做项目, 我需要显示用户每天获得的积分和交换的积分。
我有两个不同的 table 来存储数据。 tables(a,b)
如下:
table一个:
id user_id earned created_at
--------------------------------------------
1 1 1 2016-12-14
2 2 2 2016-12-14
3 2 1 2016-12-14
4 3 1 2016-12-15
table b:
id user_id exchanged created_at
--------------------------------------------
1 1 1 2016-12-14
2 1 2 2016-12-14
3 2 1 2016-12-14
4 4 1 2016-12-15
5 3 3 2016-12-16
我想合并两个 table 日期如下
user_id earned exchanged created_at
-------------------------------------------------
1 1 1 2016-12-14
2 3 1 2016-12-14
3 1 0 2016-12-15
4 0 1 2016-12-15
3 0 3 2016-12-16
我试过搜索 SO,最后得到以下查询 (sqlfiddle):
select user_id, created_at, sum(earned) as earned, sum(exchanged) as exchanged from (
SELECT
a.user_id,
DATE_FORMAT(a.created_at, '%d-%m-%Y') AS created_at,
a.earned,
0 AS exchanged
FROM
a
LEFT JOIN
b ON DATE_FORMAT(a.created_at, '%y%m%d') = DATE_FORMAT(b.created_at, '%y%m%d')
UNION SELECT
b.user_id,
DATE_FORMAT(b.created_at, '%d-%m-%Y') AS created_at,
0 AS earned,
b.exchanged
FROM
a
RIGHT JOIN
b ON DATE_FORMAT(a.created_at, '%y%m%d') = DATE_FORMAT(b.created_at, '%y%m%d')
) as tbl group by tbl.created_at, tbl.user_id
但是显示的兑换点数不正确
更新:仅当每个用户每天的两个表中都有条目时,此查询才会获取结果。
select ex.user_id, ex.created_at, sum(earned) as earned, sum(exchanged) as exchanged
from exchanged ex, earned ea
where ea.created_at = ex.created_at
and ea.user_id = ex.user_id
group by user_id, created_at
尝试...
select user_id, created_at, sum(earned), sum(exchanged) from
((select id, user_id, earned, 0 as exchanged, created_at from a)
union all
(select id, user_id, 0 as earned, exchanged, created_at from b)) combined
group by user_id, created_at