【发布时间】:2020-03-31 09:36:53
【问题描述】:
我的桌子
mysql> desc attendance;
+--------+------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+------------+------+-----+---------+----------------+
| aid | bigint(20) | NO | PRI | NULL | auto_increment |
| sid | int(10) | YES | MUL | NULL | |
| cid | bigint(20) | YES | MUL | NULL | |
| ttid | bigint(20) | YES | MUL | NULL | |
| did | int(3) | YES | MUL | NULL | |
| date | date | YES | | NULL | |
| hour | varchar(3) | YES | | NULL | |
| stuid | bigint(20) | YES | MUL | NULL | |
| status | varchar(8) | YES | | NULL | |
+--------+------------+------+-----+---------+----------------+
9 rows in set (0.47 sec)
它的价值是
mysql> select * from attendance;
+-----+------+------+------+------+------------+------+-------+---------+
| aid | sid | cid | ttid | did | date | hour | stuid | status |
+-----+------+------+------+------+------------+------+-------+---------+
| 1 | 2 | 13 | 4 | 3 | 2020-03-25 | p1 | 16 | present |
| 2 | 2 | 13 | 4 | 3 | 2020-03-25 | p1 | 17 | absent |
| 3 | 2 | 13 | 4 | 3 | 2020-03-25 | p1 | 18 | present |
| 4 | 2 | 13 | 4 | 3 | 2020-03-25 | p1 | 19 | absent |
| 5 | 2 | 13 | 4 | 3 | 2020-03-25 | p1 | 20 | present |
| 6 | 2 | 13 | 4 | 3 | 2020-03-25 | p1 | 15 | absent |
| 7 | 2 | 13 | 2 | 2 | 2020-03-17 | p6 | 16 | present |
| 8 | 2 | 13 | 2 | 2 | 2020-03-17 | p6 | 17 | absent |
| 9 | 2 | 13 | 2 | 2 | 2020-03-17 | p6 | 18 | present |
| 10 | 2 | 13 | 2 | 2 | 2020-03-17 | p6 | 19 | absent |
| 11 | 2 | 13 | 2 | 2 | 2020-03-17 | p6 | 20 | present |
| 12 | 2 | 13 | 2 | 2 | 2020-03-17 | p6 | 15 | absent |
| 13 | 2 | 13 | 4 | 3 | 2020-03-25 | p2 | 16 | present |
| 14 | 2 | 13 | 4 | 3 | 2020-03-25 | p2 | 17 | present |
| 15 | 2 | 13 | 4 | 3 | 2020-03-25 | p2 | 18 | present |
| 16 | 2 | 13 | 4 | 3 | 2020-03-25 | p2 | 19 | present |
| 17 | 2 | 13 | 4 | 3 | 2020-03-25 | p2 | 20 | present |
| 18 | 2 | 13 | 4 | 3 | 2020-03-25 | p2 | 15 | present |
| 19 | 2 | 13 | 4 | 3 | 2020-03-25 | p8 | 16 | absent |
| 20 | 2 | 13 | 4 | 3 | 2020-03-25 | p8 | 17 | absent |
| 21 | 2 | 13 | 4 | 3 | 2020-03-25 | p8 | 18 | present |
| 22 | 2 | 13 | 4 | 3 | 2020-03-25 | p8 | 19 | present |
| 23 | 2 | 13 | 4 | 3 | 2020-03-25 | p8 | 20 | present |
| 24 | 2 | 13 | 4 | 3 | 2020-03-25 | p8 | 15 | absent |
+-----+------+------+------+------+------------+------+-------+---------+
24 rows in set (0.00 sec)
现在,我要对表进行分组,但返回值为 0 在对表进行分组时,它返回的每一列值都返回 0。 为什么会这样。我需要在场或不在场,而不是 0。 谁能解决这个逻辑错误?
【问题讨论】:
-
不要不将代码作为图片发布!将文本粘贴到问题中。
-
对于 SQL 问题,如果您以一种我们可以创建表的方式显示架构,如果可能的话,在本地使用测试数据来测试答案,这将非常有用。所以尝试
SHOW CREATE TABLE YourTableName;并将输出复制/粘贴到您的问题中 -
您的查询仍然无效。
GROUP BY中没有很多列,也没有对它们应用聚合函数。旧的 MySQL 版本或配置错误的版本允许这样做,但可能会产生随机的、不需要的结果。