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有如下访客访问次数统计表 t_access_times
| 访客 |
月份 |
访问次数 |
| A |
2015-01 |
5 |
| A |
2015-01 |
15 |
| B
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|
2015-01 |
5 |
| A |
2015-01 |
8 |
| B |
2015-01 |
25 |
| A |
2015-01 |
5 |
| A |
2015-02 |
4 |
| A |
2015-02 |
6 |
| B |
2015-02 |
10 |
| B |
2015-02 |
5 |
| …… |
…… |
…… |
需要输出报表：t_access_times_accumulate
| 访客 |
月份 |
月访问总计 |
累计访问总计 |
| A |
2015-01 |
33 |
33 |
| A |
2015-02 |
10 |
43 |
| ……. |
……. |
……. |
……. |
| B |
2015-01 |
30 |
30 |
| B |
2015-02 |
15 |
45 |
| ……. |
……. |
……. |
……. |
实现步骤
可以用一个hql语句即可实现：
| select A.username,A.month,max(A.salary) as salary,sum(B.salary) as accumulate from (select username,month,sum(salary) as salary from t_access_times group by username,month) A inner join (select username,month,sum(salary) as salary from t_access_times group by username,month) B on A.username=B.username where B.month <= A.month group by A.username,A.month order by A.username,A.month; |