mysql数据库优化课程---12、mysql嵌套和链接查询(查询user表中存在的所有班级的信息?)
一、总结
一句话总结：
in：distinct：select * from class where id in(select distinct class_id from user);
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 |
+----+--------+------------+
2、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 |
+----+----------+-----+----------+
3、嵌套查询为什么用的少？
因为嵌套查询能做到的普通多表查询和链接查询一般都能做到
4、请统计每个班的总人数？
错误答案：select class_id,count(*) from user group by class_id;
left join：select class.name,count(user.id) tot from class left join user on class.id=user.class_id group by class.id;
right join：select class.name,count(user.id) tot from user right join class on class.id=user.class_id group by class.id;
需求6:请统计每个班的总人数?
1)错误答案
mysql> select class_id,count(*) from usergroup 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 |
+--------+-----+
5、嵌套查询和索引的关系？
嵌套查询里面或者外面的索引会失效，所以嵌套查询没有链接查询速度快
6、count(id)的用处在哪？
left连接之后统计班级人数，就是这个class4的0人，这里用count(*)就不行
需求6:请统计每个班的总人数?
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 |
+--------+-----+
7、链接查询的条件字段是什么？
on：select class.name,count(user.id) tot from class left join user on class.id=user.class_id group by class.id;
需求6:请统计每个班的总人数?
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 |
+--------+-----+
8、链接查询和普通多表查询的区别是什么(查询每个用户的用户名和对应的班级名称)？
内链接：select user.username,class.name from user inner join class on class.id=user.class_id;
普通多表查询：select user.username,class.name from user,class where user.class_id=class.id;
需求7: 查询每个用户的用户名和对应的班级名称?
1)正确答案(内链接)
mysql> select user.username,class.name from user inner join class on class.id=user.class_id;
+----------+--------+
| username | name |
+----------+--------+
| user1 | class1 |
| user2 | class1 |
| user3 | class1 |
| user4 | class1 |
| user5 | class2 |
| user6 | class3 |
| user7 | class2 |
| user8 | class1 |
| user9 | class2 |
| user10 | class3 |
+----------+--------+
2)正确答案(普通多表查询)
mysql> select user.username,class.name from user,class where user.class_id=class.id;
+----------+--------+
| username | name |
+----------+--------+
| user1 | class1 |
| user2 | class1 |
| user3 | class1 |
| user4 | class1 |
| user5 | class2 |
| user6 | class3 |
| user7 | class2 |
| user8 | class1 |
| user9 | class2 |
| user10 | class3 |
+----------+--------+
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 | -04-13 20:34:27 |
| user2 | 19 | class1 | -04-13 20:34:27 |
| user3 | 19 | class1 | -04-13 20:34:27 |
| user4 | 19 | class1 | -04-13 20:34:27 |
| user5 | 19 | class2 | -04-13 20:34:27 |
| user6 | 19 | class3 | -04-13 20:34:27 |
| user7 | 19 | class2 | -04-13 20:34:27 |
| user8 | 19 | class1 | -04-13 20:34:27 |
| user9 | 19 | class2 | -04-13 20:34:27 |
| user10 | 19 | class3 | -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 |
+--------+-----+
需求7: 查询每个用户的用户名和对应的班级名称?
1)正确答案(内链接)
mysql> select user.username,class.name from user inner join class on class.id=user.class_id;
+----------+--------+
| username | name |
+----------+--------+
| user1 | class1 |
| user2 | class1 |
| user3 | class1 |
| user4 | class1 |
| user5 | class2 |
| user6 | class3 |
| user7 | class2 |
| user8 | class1 |
| user9 | class2 |
| user10 | class3 |
+----------+--------+
2)正确答案(普通多表查询)
mysql> select user.username,class.name from user,class where user.class_id=class.id;
+----------+--------+
| username | name |
+----------+--------+
| user1 | class1 |
| user2 | class1 |
| user3 | class1 |
| user4 | class1 |
| user5 | class2 |
| user6 | class3 |
| user7 | class2 |
| user8 | class1 |
| user9 | class2 |
| user10 | class3 |
+----------+--------+
mysql所有班级名称和人数_mysql数据库优化课程---12 mysql嵌套和链接查询（查询user表中存在的所有班级的信息?）...
