Question

我有一张桌子，如下表

id   user_id      bal      createdDate
1    001          100      2015-02-17 16:45:44
2    001          200      2015-02-18 18:45:44
3    002          300      2015-02-20 16:45:44
4    002          800      2015-02-18 18:45:44
5    001          300      2015-03-01 16:20:44
6    002          500      2015-03-17 16:45:44
7    002          200      2015-03-18 18:45:44
8    003          300      2015-03-10 16:45:44
9    003          80       2015-03-18 18:45:44
10   003          200      2015-03-21 16:20:44

我想要每个user_id的最新余额，并将它们全部求和。结果，我将从用户001,002,003中获得最新余额的总和

此表包含大约500万条记录
此表保存每个用户的帐户余额的历史记录。因此，最新日期是该用户的最新帐户余额。

下面是我的查询，但是由于MySQL工作台冻结，我没有得到任何结果。

SELECT (SUM(bal))
FROM hist_bal h1
WHERE h1.createDate = (SELECT MAX(h2.createDate)
FROM hist_bal h2
WHERE h2.user_id = h1.user_id  GROUP BY h2.user_id)

Answer 1

这应该很简单，不需要GROUP BY Clouse：

SELECT SUM(bal)
FROM hist_bal h1
WHERE h1.createDate = (SELECT MAX(h2.createDate)
FROM hist_bal h2
WHERE h2.user_id = h1.user_id)

Answer 2

SELECT SUM(h1.bal) AS Bal
FROM hist_bal h1 JOIN (SELECT user_id, MAX(h2.createDate) AS createDate
FROM hist_bal h2 GROUP BY h2.user_id) h2 ON h1.user_id = h2.user_id
AND h1.createDate = h2.createDate

Answer 3

选择 h1.user_id，SUM（h1.bal）从 hist_bal h1，hist_bal h2 哪里 h1.createDate =（选择最大值（h2.createDate）和h2.user_id = h1.user_id
通过...分组 h1.user_id

汇总每个用户的最新记录

3 个答案: