[MySQL]group by 与 having 结合函数 的统计技巧
group by 与 having 允许字段使用函数,根据函数运行的结果group by分组或having设置选择条件;
同时group by 与 having 也允许使用字段别名
示例表a:
| id | a | effective_date |
| 38 | 1 | 2018-03-23 09:55:58 |
| 39 | 9 | 2018-03-22 09:56:04 |
| 66 | 1 | 2018-02-21 09:56:09 |
| 68 | 6 | 2018-01-13 09:56:17 |
| 69 | 1 | 0000-00-00 00:00:00 |
示例表b:
| id | u_id | amount | t4_date |
| 1 | 39 | 30 | 2017-07-28 15:30:00 |
| 2 | 66 | 21 | 2018-03-08 10:07:40 |
| 3 | 69 | 13 | 2018-03-01 10:09:00 |
应用示例:
1.单表格式化分组
SELECT sum(a) FROM test_1 GROUP BY DATE_FORMAT(effective_date,"%Y-%m");
| sum(a) |
| 1 |
| 6 |
| 1 |
| 10 |
2.单表格式化分组(使用别名)
SELECT DATE_FORMAT(effective_date,"%Y-%m") AS date, sum(a) FROM test_1 GROUP BY date;
| date | sum(a) |
| 0000-00 | 1 |
| 2018-01 | 6 |
| 2018-02 | 1 |
| 2018-03 | 10 |
3.单表格式化分组,having选择输出的行(使用别名)
SELECT DATE_FORMAT(effective_date,"%Y-%m") AS date, sum(a) FROM test_1 GROUP BY date HAVING date = "2018-02";
| date | sum(a) |
| 2018-02 | 1 |
4.连表应用
SELECT DATE_FORMAT(b.t4_date,"%Y-%m") AS date, sum(a.a) FROM test_1 AS a LEFT JOIN test_4 AS b ON a.id = b.u_id GROUP BY date HAVING date = "2018-03";
| date | sum(a.a) |
| 2018-03 | 2 |

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