mysql数据库优化课程---11、mysql普通多表查询

一、总结

一句话总结：select user.username,user.age,class.name,class.ctime from user,class where user.class_id=class.id;

1、sql中的注释是什么？

--

两个横杠

2、两表无条件查询的结果是什么？

条数乘积

select * from class,user;

1.两表数据进行所有组合.
2.数据量是两表条数乘积.

3、多表连接查询（有条件）的实质是什么？

排列组合

满足条件

在所有排列组合的情况下选出满足条件的记录

4、请查询出每一个学员的姓名、年龄、班级名称和班级创建时间？

select user.username,user.age,class.name,class.ctime from user,class where user.class_id=class.id;

5、请查询出每一个学员的姓名、年龄、班级名称和班级创建时间，并把时间戳转成正常日期显示出来？

select user.username,user.age,class.name,from_unixtime(class.ctime) ctime from user,class where user.class_id=class.id;

6、查询每个用户的用户名和对应的班级名称？

二、内容在总结中

mysql多表查询:
1.普通多表查询
2.嵌套查询或子查询
3.链接查询
1)左链接
2)右链接
3)内链接

查看class表数据:
mysql> select * from class;
+----+--------+------------+
| id | name | ctime |
+----+--------+------------+
| 1 | class1 | 1492086867 |
| 2 | class2 | 1492086867 |
| 3 | class3 | 1492086867 |
+----+--------+------------+

查看user表数据:
mysql> select * from user;
+----+----------+-----+
| id | username | age |
+----+----------+-----+
| 1 | user1    | 19 |
| 2 | user2    | 29 |
| 3 | user3    | 31 |
| 4 | user4    | 22 |
| 5 | user5    | 23 |
| 6 | user6    | 18 |
| 7 | user7    | 17 |
| 8 | user8    | 25 |
| 9 | user9    | 27 |
| 10 | user10   | 32 |
+----+----------+-----+

两表无条件查询:
mysql> select * from class,user;
+----+--------+------------+----+----------+-----+
| id | name   | ctime      | id | username | age |
+----+--------+------------+----+----------+-----+
| 1 | class1 | 1492086867 | 1 | user1    | 19 |
| 2 | class2 | 1492086867 | 1 | user1    | 19 |
| 3 | class3 | 1492086867 | 1 | user1    | 19 |
| 4 | class4 | 1492087405 | 1 | user1    | 19 |
| 1 | class1 | 1492086867 | 2 | user2    | 29 |
| 2 | class2 | 1492086867 | 2 | user2    | 29 |
| 3 | class3 | 1492086867 | 2 | user2    | 29 |
| 4 | class4 | 1492087405 | 2 | user2    | 29 |
| 1 | class1 | 1492086867 | 3 | user3    | 31 |
| 2 | class2 | 1492086867 | 3 | user3    | 31 |
| 3 | class3 | 1492086867 | 3 | user3    | 31 |
| 4 | class4 | 1492087405 | 3 | user3    | 31 |
| 1 | class1 | 1492086867 | 4 | user4    | 22 |
| 2 | class2 | 1492086867 | 4 | user4    | 22 |
| 3 | class3 | 1492086867 | 4 | user4    | 22 |
| 4 | class4 | 1492087405 | 4 | user4    | 22 |
| 1 | class1 | 1492086867 | 5 | user5    | 23 |
| 2 | class2 | 1492086867 | 5 | user5    | 23 |
| 3 | class3 | 1492086867 | 5 | user5    | 23 |
| 4 | class4 | 1492087405 | 5 | user5    | 23 |
| 1 | class1 | 1492086867 | 6 | user6    | 18 |
| 2 | class2 | 1492086867 | 6 | user6    | 18 |
| 3 | class3 | 1492086867 | 6 | user6    | 18 |
| 4 | class4 | 1492087405 | 6 | user6    | 18 |
| 1 | class1 | 1492086867 | 7 | user7    | 17 |
| 2 | class2 | 1492086867 | 7 | user7    | 17 |
| 3 | class3 | 1492086867 | 7 | user7    | 17 |
| 4 | class4 | 1492087405 | 7 | user7    | 17 |
| 1 | class1 | 1492086867 | 8 | user8    | 25 |
| 2 | class2 | 1492086867 | 8 | user8    | 25 |
| 3 | class3 | 1492086867 | 8 | user8    | 25 |
| 4 | class4 | 1492087405 | 8 | user8    | 25 |
| 1 | class1 | 1492086867 | 9 | user9    | 27 |
| 2 | class2 | 1492086867 | 9 | user9    | 27 |
| 3 | class3 | 1492086867 | 9 | user9    | 27 |
| 4 | class4 | 1492087405 | 9 | user9    | 27 |
| 1 | class1 | 1492086867 | 10 | user10   | 32 |
| 2 | class2 | 1492086867 | 10 | user10   | 32 |
| 3 | class3 | 1492086867 | 10 | user10   | 32 |
| 4 | class4 | 1492087405 | 10 | user10   | 32 |
+----+--------+------------+----+----------+-----+

结论:
1.两表数据进行所有组合.
2.数据量是两表条数乘积.

设计了关系的user表:
mysql> select * from user;
+----+----------+-----+----------+
| id | username | age | class_id |
+----+----------+-----+----------+
| 1 | user1    | 19 |        1 |
| 2 | user2    | 19 |        1 |
| 3 | user3    | 19 |        1 |
| 4 | user4    | 19 |        1 |
| 5 | user5    | 19 |        2 |
| 6 | user6    | 19 |        3 |
| 7 | user7    | 19 |        2 |
| 8 | user8    | 19 |        1 |
| 9 | user9    | 19 |        2 |
| 10 | user10   | 19 |        3 |
+----+----------+-----+----------+

需求1:请查询出每一个学员的姓名、年龄、班级名称和班级创建时间?
mysql> select user.username,user.age,class.name,class.ctime from user,class where user.class_id=class.id;
+----------+-----+--------+------------+
| username | age | name   | ctime      |
+----------+-----+--------+------------+
| user1    | 19 | class1 | 1492086867 |
| user2    | 19 | class1 | 1492086867 |
| user3    | 19 | class1 | 1492086867 |
| user4    | 19 | class1 | 1492086867 |
| user5    | 19 | class2 | 1492086867 |
| user6    | 19 | class3 | 1492086867 |
| user7    | 19 | class2 | 1492086867 |
| user8    | 19 | class1 | 1492086867 |
| user9    | 19 | class2 | 1492086867 |
| user10   | 19 | class3 | 1492086867 |
+----------+-----+--------+------------+

需求2:请查询出每一个学员的姓名、年龄、班级名称和班级创建时间，并把时间戳转成正常日期显示出来?
mysql> select user.username,user.age,class.name,from_unixtime(class.ctime) ctime from user,class where user.class_id=class.id;
+----------+-----+--------+---------------------+
| username | age | name   | ctime               |
+----------+-----+--------+---------------------+
| user1    | 19 | class1 | 2017-04-13 20:34:27 |
| user2    | 19 | class1 | 2017-04-13 20:34:27 |
| user3    | 19 | class1 | 2017-04-13 20:34:27 |
| user4    | 19 | class1 | 2017-04-13 20:34:27 |
| user5    | 19 | class2 | 2017-04-13 20:34:27 |
| user6    | 19 | class3 | 2017-04-13 20:34:27 |
| user7    | 19 | class2 | 2017-04-13 20:34:27 |
| user8    | 19 | class1 | 2017-04-13 20:34:27 |
| user9    | 19 | class2 | 2017-04-13 20:34:27 |
| user10   | 19 | class3 | 2017-04-13 20:34:27 |
+----------+-----+--------+---------------------+

需求3:请查询出每一个学员的姓名、年龄、班级名称和班级创建时间，并且只查询1班学员?
mysql> select user.username,user.age,class.name,class.ctime from user,class where user.class_id=class.id and class.id=1;
+----------+-----+--------+------------+
| username | age | name   | ctime      |
+----------+-----+--------+------------+
| user1    | 19 | class1 | 1492086867 |
| user2    | 19 | class1 | 1492086867 |
| user3    | 19 | class1 | 1492086867 |
| user4    | 19 | class1 | 1492086867 |
| user8    | 19 | class1 | 1492086867 |
+----------+-----+--------+------------+

需求4:查询user表中存在的所有班级的信息?
mysql> select * from class where id in(select distinct class_id from user);
+----+--------+------------+
| id | name | ctime |
+----+--------+------------+
| 1 | class1 | 1492086867 |
| 2 | class2 | 1492086867 |
| 3 | class3 | 1492086867 |
+----+--------+------------+

需求5:查询班级表中每个班的所有学员信息?
mysql> select * from user where class_id in(select id from class);
+----+----------+-----+----------+
| id | username | age | class_id |
+----+----------+-----+----------+
| 1 | user1    | 19 |        1 |
| 2 | user2    | 19 |        1 |
| 3 | user3    | 19 |        1 |
| 4 | user4    | 19 |        1 |
| 5 | user5    | 19 |        2 |
| 6 | user6    | 19 |        3 |
| 7 | user7    | 19 |        2 |
| 8 | user8    | 19 |        1 |
| 9 | user9    | 19 |        2 |
| 10 | user10   | 19 |        3 |
| 11 | user10   | 19 |        4 |
+----+----------+-----+----------+

左链接:
需求6:请统计每个班的总人数?

1)错误答案
mysql> select class_id,count(*) from user group by class_id;
+----------+----------+
| class_id | count(*) |
+----------+----------+
|        1 |        5 |
|        2 |        3 |
|        3 |        2 |
+----------+----------+

2)正确答案
mysql> select class.name,count(user.id) tot from class left join user on class.id=user.class_id group by class.id;
+--------+-----+
| name   | tot |
+--------+-----+
| class1 |   5 |
| class2 |   3 |
| class3 |   2 |
| class4 |   0 |
+--------+-----+

3)正确答案
mysql> select class.name,count(user.id) tot from user right join class on class.id=user.class_id group by class.id;
+--------+-----+
| name   | tot |
+--------+-----+
| class1 |   5 |
| class2 |   3 |
| class3 |   2 |
| class4 |   0 |
+--------+-----+