《Mysql學習MySQL無GROUP BY直接HAVING返回空的問題分析》要點:
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有一張表,id是主鍵,這樣的寫法可以返回一條記錄:
MYSQL入門
代碼如下:
“SELECT * FROM t HAVING id=MIN(id);”
但是只是把MIN換成MAX,這樣返回就是空了:
代碼如下:
“SELECT * FROM t HAVING id=MAX(id);”
這是為什么呢?
我們先來做個試驗,驗證這種情況.
這是表結(jié)構(gòu),初始化兩條記錄,然后試驗:
代碼如下:
root@localhost : plx 10:25:10> show create table t2\G
*************************** 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,好像真的是這樣哎,怎么會這樣呢?我再試一下,把a字段改一個為10,然后試下a字段:MYSQL入門
代碼如下:
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)
我擦,這回MAX能返回,MIN不能了,這又是為啥呢?MYSQL入門
旁白
一般來說,HAVING子句是配合GROUP BY使用的,單獨使用HAVING本身是不符合規(guī)范的,
但是MySQL會做一個重寫,加上一個GROUP BY NULL,”SELECT * FROM t HAVING id=MIN(id)”會被重寫為”SELECT * FROM t GROUP BY NULL HAVING id=MIN(id)”,這樣語法就符合規(guī)范了.
繼續(xù)……
但是,這個 GROUP BY NULL 會產(chǎn)生什么結(jié)果呢?經(jīng)過查看代碼和試驗,可以證明,GROUP BY NULL 等價于 LIMIT 1:MYSQL入門
代碼如下:
root@localhost : plx 10:25:48> SELECT * FROM t2 GROUP BY NULL;
+------+----+
| a??? | id |
+------+----+
|?? 10 |? 1 |
+------+----+
1 row in set (0.00 sec)
也就是說,GROUP BY NULL 以后,只會有一個分組,里面就是第一行數(shù)據(jù).
但是如果這樣,MIN、MAX結(jié)果應(yīng)該是一致的,那也不應(yīng)該MAX和MIN一個有結(jié)果,一個沒結(jié)果啊,這是為什么呢,再做一個測試.
修改一下數(shù)據(jù),然后直接查看MIN/MAX的值:MYSQL入門
代碼如下:
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)
是不是發(fā)現(xiàn)問題了?
MAX/MIN函數(shù)取值是全局的,而不是LIMIT 1這個分組內(nèi)的.
因此,當GROUP BY NULL的時候,MAX/MIN函數(shù)是取所有數(shù)據(jù)里的最大和最小值!
所以啊,”SELECT * FROM t HAVING id=MIN(id)”本色上是”SELECT * FROM t HAVING id=1″, 就能返回一條記錄,而”SELECT * FROM t HAVING id=MAX(id)”本色上是”SELECT * FROM t HAVING id=3″,當然沒有返回記錄,這就是問題的根源.
測試一下GROUP BY a,這樣就對了,每個分組內(nèi)只有一行,所以MAX/MIN一樣大,這回是取得組內(nèi)最大和最小值.MYSQL入門
代碼如下:
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時MAX/MIN的行為,是這個問題的本色,所以啊,盡量使用標準語法,玩花樣SQL之前,一定要搞清楚它的行為是否與理解的一致.MYSQL入門
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