文章给大家介绍关于解决MySQL中无GROUP BY直接HAVING的问题,如果你不想使用group by而直接使用having碰到问题可参考此文章.
今天有同学给我反应,有一张表,id是主键,这样的写法可以返回一条记录:
“SELECT * FROM t HAVING id=MIN(id);”
但是只是把MIN换成MAX,这样返回就是空了,代码如下"
“SELECT * FROM t HAVING id=MAX(id);”
这是为什么呢？我们先来做个试验,验证这种情况,这是表结构,初始化两条记录,然后试验,代码如下:
- root@localhost : plx 10:25:10> show create table t2G
- *************************** 1. row ***************************
- Table: t2
- Create Table: CREATE TABLE `t2` (
- `a` int(11) DEFAULT NULL,
- `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
- PRIMARY KEY (`id`)
- ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8
-
- root@localhost : plx 10:25:15> select * from t2;
- +
- | a | id |
- +
- | 1 | 1 |
- | 1 | 3 |
- +
- 2 rows in set (0.00 sec)
-
- root@localhost : plx 10:25:20> SELECT * FROM t2 HAVING id=MIN(id);
- +
- | a | id |
- +
- | 1 | 1 |
- +
- 1 row in set (0.00 sec)
-
- root@localhost : plx 10:25:30> SELECT * FROM t2 HAVING id=MAX(id);
- Empty set (0.00 sec)
初看之下,好像真的是这样哎,怎么会这样呢？我再试一下,把a字段改一个为10,然后试下a字段,代码如下:
- root@localhost : plx 10:26:58> select * from t2;
- +------+----+
- | a | id |
- +------+----+
- | 10 | 1 |
- | 1 | 3 |
- +------+----+
- 2 rows in set (0.00 sec)
-
- root@localhost : plx 10:28:20> SELECT * FROM t2 HAVING a=MAX(a);
- +------+----+
- | a | id |
- +------+----+
- | 10 | 1 |
- +------+----+
- 1 row in set (0.00 sec)
-
- root@localhost : plx 10:28:28> SELECT * FROM t2 HAVING a=MIN(a);
- Empty set (0.00 sec)
这回MAX能返回,MIN不能了,这又是为啥呢？一般来说,HAVING子句是配合GROUP BY使用的,单独使用HAVING本身是不符合规范的,但是MySQL会做一个重写,加上一个GROUP BY NULL，”SELECT * FROM t HAVING id=MIN(id)”会被重写为”SELECT * FROM t GROUP BY NULL HAVING id=MIN(id)”,这样语法就符合规范了.
继续……但是,这个 GROUP BY NULL 会产生什么结果呢？经过查看代码和试验,可以证明,GROUP BY NULL 等价于 LIMIT 1,代码如下:
- root@localhost : plx 10:25:48> SELECT * FROM t2 GROUP BY NULL;
- +
- | a | id |
- +
- | 10 | 1 |
- +
- 1 row in set (0.00 sec)
也就是说,GROUP BY NULL 以后,只会有一个分组,里面就是第一行数据,但是如果这样,MIN、MAX结果应该是一致的,那也不应该MAX和MIN一个有结果,一个没结果啊,这是为什么呢,再做一个测试.
修改一下数据,然后直接查看MIN/MAX的值,代码如下:
- root@localhost : plx 10:26:58> select * from t2;
- +------+----+
- | a | id |
- +------+----+
- | 10 | 1 |
- | 1 | 3 |
- +------+----+
- 2 rows in set (0.00 sec)
-
- root@localhost : plx 10:27:04> SELECT * FROM t2 GROUP BY NULL;
- +------+----+
- | a | id |
- +------+----+
- | 10 | 1 |
- +------+----+
- 1 row in set (0.00 sec)
-
- root@localhost : plx 10:30:21> SELECT MAX(a),MIN(a),MAX(id),MIN(id) FROM t2 GROUP BY NULL;
- +--------+--------+---------+---------+
- | MAX(a) | MIN(a) | MAX(id) | MIN(id) |
- +--------+--------+---------+---------+
- | 10 | 1 | 3 | 1 |
- +--------+--------+---------+---------+
- 1 row in set (0.00 sec)
是不是发现问题了？MAX/MIN函数取值是全局的,而不是LIMIT 1这个分组内的.
因此,当GROUP BY NULL的时候,MAX/MIN函数是取所有数据里的最大和最小值,所以啊,”SELECT * FROM t HAVING id=MIN(id)”本质上是”SELECT * FROM t HAVING id=1″,就能返回一条记录,而”SELECT * FROM t HAVING id=MAX(id)”本质上是”SELECT * FROM t HAVING id=3″,当然没有返回记录,这就是问题的根源.
测试一下GROUP BY a,这样就对了,每个分组内只有一行,所以MAX/MIN一样大,这回是取得组内最大和最小值,代码如下:
- root@localhost : plx 11:29:49> SELECT MAX(a),MIN(a),MAX(id),MIN(id) FROM t2 GROUP BY a;
- +--------+--------+---------+---------+
- | MAX(a) | MIN(a) | MAX(id) | MIN(id) |
- +--------+--------+---------+---------+
- | 1 | 1 | 3 | 3 |
- | 10 | 10 | 5 | 5 |
- +--------+--------+---------+---------+
- 2 rows in set (0.00 sec)
-
GROUP BY NULL时MAX/MIN的行为,是这个问题的本质,所以啊,尽量使用标准语法,玩花样SQL之前,一定要搞清楚它的行为是否与理解的一致. |
