时间:2021-07-01 10:21:17 帮助过:20人阅读
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- JOIN语句的执行顺序
- INNER/LEFT/RIGHT/FULL JOIN的区别
- ON和WHERE的区别
一个完整的SQL语句中会被拆分成多个子句,子句的执行过程中会产生虚拟表(vt),但是结果只返回最后一张虚拟表。从这个思路出发,我们试着理解一下JOIN查询的执行过程并解答一些常见的问题。
如果之前对不同JOIN的执行结果没有概念,可以结合这篇文章往下看
以下是JOIN查询的通用结构
SELECT <row_list> FROM <left_table> <inner|left|right> JOIN <right_table> ON <join condition> WHERE <where_condition>
它的执行顺序如下(SQL语句里第一个被执行的总是FROM子句):
下面用一个例子介绍一下上述联表的过程(这个例子不是个好的实践,只是为了说明join语法)
创建一个用户信息表:
CREATE TABLE `user_info` ( `userid` int(11) NOT NULL, `name` varchar(255) NOT NULL, UNIQUE `userid` (`userid`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4
再创建一个用户余额表:
CREATE TABLE `user_account` ( `userid` int(11) NOT NULL, `money` bigint(20) NOT NULL, UNIQUE `userid` (`userid`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4
随便导入一些数据:
select * from user_info; +--------+------+ | userid | name | +--------+------+ | 1001 | x | | 1002 | y | | 1003 | z | | 1004 | a | | 1005 | b | | 1006 | c | | 1007 | d | | 1008 | e | +--------+------+ 8 rows in set (0.00 sec) select * from user_account; +--------+-------+ | userid | money | +--------+-------+ | 1001 | 22 | | 1002 | 30 | | 1003 | 8 | | 1009 | 11 | +--------+-------+ 4 rows in set (0.00 sec)
一共8个用户有用户名,4个用户的账户有余额。
取出userid为1003的用户姓名和余额,SQL如下:
SELECT i.name, a.money FROM user_info as i LEFT JOIN user_account as a ON i.userid = a.userid WHERE a.userid = 1003;
笛卡尔积操作后会返回两张表中所有行的组合,左表user_info有8行,右表user_account有4行,生成的虚拟表vt1就是8*4=32行:
SELECT * FROM user_info as i LEFT JOIN user_account as a ON 1; +--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1001 | 22 | | 1003 | z | 1001 | 22 | | 1004 | a | 1001 | 22 | | 1005 | b | 1001 | 22 | | 1006 | c | 1001 | 22 | | 1007 | d | 1001 | 22 | | 1008 | e | 1001 | 22 | | 1001 | x | 1002 | 30 | | 1002 | y | 1002 | 30 | | 1003 | z | 1002 | 30 | | 1004 | a | 1002 | 30 | | 1005 | b | 1002 | 30 | | 1006 | c | 1002 | 30 | | 1007 | d | 1002 | 30 | | 1008 | e | 1002 | 30 | | 1001 | x | 1003 | 8 | | 1002 | y | 1003 | 8 | | 1003 | z | 1003 | 8 | | 1004 | a | 1003 | 8 | | 1005 | b | 1003 | 8 | | 1006 | c | 1003 | 8 | | 1007 | d | 1003 | 8 | | 1008 | e | 1003 | 8 | | 1001 | x | 1009 | 11 | | 1002 | y | 1009 | 11 | | 1003 | z | 1009 | 11 | | 1004 | a | 1009 | 11 | | 1005 | b | 1009 | 11 | | 1006 | c | 1009 | 11 | | 1007 | d | 1009 | 11 | | 1008 | e | 1009 | 11 | +--------+------+--------+-------+ 32 rows in set (0.00 sec)
ON i.userid = a.userid 过滤之后vt2如下:
+--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1002 | 30 | | 1003 | z | 1003 | 8 | +--------+------+--------+-------+
LEFT JOIN会将左表未出现在vt2的行插入进vt2,每一行的剩余字段将被填充为NULL,RIGHT JOIN同理
本例中用的是LEFT JOIN,所以会将左表user_info剩下的行都添上 生成表vt3:
+--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1002 | 30 | | 1003 | z | 1003 | 8 | | 1004 | a | NULL | NULL | | 1005 | b | NULL | NULL | | 1006 | c | NULL | NULL | | 1007 | d | NULL | NULL | | 1008 | e | NULL | NULL | +--------+------+--------+-------+
WHERE a.userid = 1003 生成表vt4:
+--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1003 | z | 1003 | 8 | +--------+------+--------+-------+
SELECT i.name, a.money 生成vt5:
+------+-------+ | name | money | +------+-------+ | z | 8 | +------+-------+
虚拟表vt5作为最终结果返回给客户端
介绍完联表的过程之后,我们看看常用JOIN的区别
拿上文的第三步添加外部行来举例,若LEFT JOIN替换成INNER JOIN,则会跳过这一步,生成的表vt3与vt2一模一样:
+--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1002 | 30 | | 1003 | z | 1003 | 8 | +--------+------+--------+-------+
若LEFT JOIN替换成RIGHT JOIN,则生成的表vt3如下:
+--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1002 | 30 | | 1003 | z | 1003 | 8 | | NULL | NULL | 1009 | 11 | +--------+------+--------+-------+
因为user_account(右表)里存在userid=1009这一行,而user_info(左表)里却找不到这一行的记录,所以会在第三步插入以下一行:
| NULL | NULL | 1009 | 11 |
上文引用的文章中提到了标准SQL定义的FULL JOIN,这在mysql里是不支持的,不过我们可以通过LEFT JOIN + UNION + RIGHT JOIN 来实现FULL JOIN:
SELECT * FROM user_info as i RIGHT JOIN user_account as a ON a.userid=i.userid union SELECT * FROM user_info as i LEFT JOIN user_account as a ON a.userid=i.userid;
他会返回如下结果:
+--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1002 | 30 | | 1003 | z | 1003 | 8 | | NULL | NULL | 1009 | 11 | | 1004 | a | NULL | NULL | | 1005 | b | NULL | NULL | | 1006 | c | NULL | NULL | | 1007 | d | NULL | NULL | | 1008 | e | NULL | NULL | +--------+------+--------+-------+
ps:其实我们从语义上就能看出LEFT JOIN和RIGHT JOIN没什么差别,两者的结果差异取决于左右表的放置顺序,以下内容摘自mysql官方文档:
RIGHT JOIN works analogously to LEFT JOIN. To keep code portable across databases, it is recommended that you use LEFT JOIN instead of RIGHT JOIN.
所以当你纠结使用LEFT JOIN还是RIGHT JOIN时,尽可能只使用LEFT JOIN吧
上文把JOIN的执行顺序了解清楚之后,ON和WHERE的区别也就很好理解了。
举例说明:
SELECT * FROM user_info as i LEFT JOIN user_account as a ON i.userid = a.userid and i.userid = 1003;
SELECT * FROM user_info as i LEFT JOIN user_account as a ON i.userid = a.userid where i.userid = 1003;
第一种情况LEFT JOIN在执行完第二步ON子句后,筛选出满足i.userid = a.userid and i.userid = 1003的行,生成表vt2,然后执行第三步JOIN子句,将外部行添加进虚拟表生成vt3即最终结果:
vt2: +--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1003 | z | 1003 | 8 | +--------+------+--------+-------+ vt3: +--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | NULL | NULL | | 1002 | y | NULL | NULL | | 1003 | z | 1003 | 8 | | 1004 | a | NULL | NULL | | 1005 | b | NULL | NULL | | 1006 | c | NULL | NULL | | 1007 | d | NULL | NULL | | 1008 | e | NULL | NULL | +--------+------+--------+-------+
而第二种情况LEFT JOIN在执行完第二步ON子句后,筛选出满足i.userid = a.userid的行,生成表vt2;再执行第三步JOIN子句添加外部行生成表vt3;然后执行第四步WHERE子句,再对vt3表进行过滤生成vt4,得的最终结果:
vt2: +--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1002 | 30 | | 1003 | z | 1003 | 8 | +--------+------+--------+-------+ vt3: +--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1001 | x | 1001 | 22 | | 1002 | y | 1002 | 30 | | 1003 | z | 1003 | 8 | | 1004 | a | NULL | NULL | | 1005 | b | NULL | NULL | | 1006 | c | NULL | NULL | | 1007 | d | NULL | NULL | | 1008 | e | NULL | NULL | +--------+------+--------+-------+ vt4: +--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1003 | z | 1003 | 8 | +--------+------+--------+-------+
如果将上例的LEFT JOIN替换成INNER JOIN,不论将条件过滤放到ON还是WHERE里,结果都是一样的,因为INNER JOIN不会执行第三步添加外部行
SELECT * FROM user_info as i INNER JOIN user_account as a ON i.userid = a.userid and i.userid = 1003;
SELECT * FROM user_info as i INNER JOIN user_account as a ON i.userid = a.userid where i.userid = 1003;
返回结果都是:
+--------+------+--------+-------+ | userid | name | userid | money | +--------+------+--------+-------+ | 1003 | z | 1003 | 8 | +--------+------+--------+-------+
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