1. Introduction to knowledge points
What is a subquery? ?
It's very simple. It's actually a select statement nested within a select statement. The nested statement is called a subquery.
2. Classification
Where can subqueries appear?
In fact, subqueries can appear after select, from, and where statements. The structure is as follows:
select
....(select)
from
....(select)
where
....(select)
3. Example display
1.Subquery in where clause
Case: Use the table below to find the names and salaries of employees whose wages are higher than the minimum wage?
mysql> select * from emp;
+-------+--------+-----------+------+------------+---------+---------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+--------+-----------+------+------------+---------+---------+--------+
| 7369 | SMITH | CLERK | 7902 | 1980-12-17 | 800.00 | NULL | 20 |
| 7499 | ALLEN | SALESMAN | 7698 | 1981-02-20 | 1600.00 | 300.00 | 30 |
| 7521 | WARD | SALESMAN | 7698 | 1981-02-22 | 1250.00 | 500.00 | 30 |
| 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975.00 | NULL | 20 |
| 7654 | MARTIN | SALESMAN | 7698 | 1981-09-28 | 1250.00 | 1400.00 | 30 |
| 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850.00 | NULL | 30 |
| 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | NULL | 10 |
| 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 |
| 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000.00 | NULL | 10 |
| 7844 | TURNER | SALESMAN | 7698 | 1981-09-08 | 1500.00 | 0.00 | 30 |
| 7876 | ADAMS | CLERK | 7788 | 1987-05-23 | 1100.00 | NULL | 20 |
| 7900 | JAMES | CLERK | 7698 | 1981-12-03 | 950.00 | NULL | 30 |
| 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 |
| 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300.00 | NULL | 10 |
+-------+--------+-----------+------+------------+---------+---------+--------+
14 rows in set (0.01 sec)Implementation ideas:
Step 1: Find out what the minimum wage is
mysql> select min(sal) from emp;
+----------+
| min(sal) |
+----------+
| 800.00 |
+----------+
1 row in set (0.00 sec)Step 2: Find those whose salary is >800
mysql> select ename as '姓名',sal as '工资' from emp where sal >800;
+--------+---------+
| 姓名 | 工资 |
+--------+---------+
| ALLEN | 1600.00 |
| WARD | 1250.00 |
| JONES | 2975.00 |
| MARTIN | 1250.00 |
| BLAKE | 2850.00 |
| CLARK | 2450.00 |
| SCOTT | 3000.00 |
| KING | 5000.00 |
| TURNER | 1500.00 |
| ADAMS | 1100.00 |
| JAMES | 950.00 |
| FORD | 3000.00 |
| MILLER | 1300.00 |
+--------+---------+
13 rows in set (0.00 sec)Step 3: Merge
mysql> select ename as '姓名',sal as '工资' from emp where sal >(select min(sal) from emp );
+--------+---------+
| 姓名 | 工资 |
+--------+---------+
| ALLEN | 1600.00 |
| WARD | 1250.00 |
| JONES | 2975.00 |
| MARTIN | 1250.00 |
| BLAKE | 2850.00 |
| CLARK | 2450.00 |
| SCOTT | 3000.00 |
| KING | 5000.00 |
| TURNER | 1500.00 |
| ADAMS | 1100.00 |
| JAMES | 950.00 |
| FORD | 3000.00 |
| MILLER | 1300.00 |
+--------+---------+
13 rows in set (0.00 sec)2. Subquery in the from clause
Note : The subquery after from can treat the query result of the subquery as a temporary table . ( Trick )
Example: Use the table below to find the salary grade for the average salary for each position. (Employee table and salary registration table respectively)
mysql> select * from emp;
+-------+--------+-----------+------+------------+---------+---------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+--------+-----------+------+------------+---------+---------+--------+
| 7369 | SMITH | CLERK | 7902 | 1980-12-17 | 800.00 | NULL | 20 |
| 7499 | ALLEN | SALESMAN | 7698 | 1981-02-20 | 1600.00 | 300.00 | 30 |
| 7521 | WARD | SALESMAN | 7698 | 1981-02-22 | 1250.00 | 500.00 | 30 |
| 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975.00 | NULL | 20 |
| 7654 | MARTIN | SALESMAN | 7698 | 1981-09-28 | 1250.00 | 1400.00 | 30 |
| 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850.00 | NULL | 30 |
| 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | NULL | 10 |
| 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 |
| 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000.00 | NULL | 10 |
| 7844 | TURNER | SALESMAN | 7698 | 1981-09-08 | 1500.00 | 0.00 | 30 |
| 7876 | ADAMS | CLERK | 7788 | 1987-05-23 | 1100.00 | NULL | 20 |
| 7900 | JAMES | CLERK | 7698 | 1981-12-03 | 950.00 | NULL | 30 |
| 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 |
| 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300.00 | NULL | 10 |
+-------+--------+-----------+------+------------+---------+---------+--------+
14 rows in set (0.01 sec)
mysql> select * from salgrade;
+-------+-------+-------+
| GRADE | LOSAL | HISAL |
+-------+-------+-------+
| 1 | 700 | 1200 |
| 2 | 1201 | 1400 |
| 3 | 1401 | 2000 |
| 4 | 2001 | 3000 |
| 5 | 3001 | 9999 |
+-------+-------+-------+
5 rows in set (0.00 sec)Implementation ideas:
Step 1: Find the average salary for each position (average according to position grouping)
mysql> select job,avg(sal) avgsal from emp group by job;
+-----------+-------------+
| job | avgsal |
+-----------+-------------+
| CLERK | 1037.500000 |
| SALESMAN | 1400.000000 |
| MANAGER | 2758.333333 |
| ANALYST | 3000.000000 |
| PRESIDENT | 5000.000000 |
+-----------+-------------+
5 rows in set (0.00 sec)Step 2: Overcome the psychological barrier and treat the above query results as a real table, which can be named 't'
Table connection between t table and salgrade table, condition: t table avg(sal) between losal and hisal
Step 3: Merge
(Note: If there is an aggregate function in the subquery after from, you must give it an alias, otherwise an error will be reported)
mysql> select t.*,s.grade
-> from (select job,avg(sal) avgsal from emp group by job) t
-> join salgrade s
-> on t.avgsal between s.losal and s.hisal;
+-----------+-------------+-------+
| job | avgsal | grade |
+-----------+-------------+-------+
| CLERK | 1037.500000 | 1 |
| SALESMAN | 1400.000000 | 2 |
| MANAGER | 2758.333333 | 4 |
| ANALYST | 3000.000000 | 4 |
| PRESIDENT | 5000.000000 | 5 |
+-----------+-------------+-------+
5 rows in set (0.00 sec)3. The subquery that appears after select
Case: Use the method to find out the department name of each employee and ask to display the employee name and department name?
Step 1: First check the employee’s employee name and department number
mysql> select e.ename,e.deptno from emp e;
+--------+--------+
| ename | deptno |
+--------+--------+
| SMITH | 20 |
| ALLEN | 30 |
| WARD | 30 |
| JONES | 20 |
| MARTIN | 30 |
| BLAKE | 30 |
| CLARK | 10 |
| SCOTT | 20 |
| KING | 10 |
| TURNER | 30 |
| ADAMS | 20 |
| JAMES | 30 |
| FORD | 20 |
| MILLER | 10 |
+--------+--------+
14 rows in set (0.00 sec)Step 2: Check the department name of each employee
mysql> select e.ename,e.deptno,(select d.dname from dept d where e.deptno=d.deptno) as dname from emp e;
+--------+--------+------------+
| ename | deptno | dname |
+--------+--------+------------+
| SMITH | 20 | RESEARCH |
| ALLEN | 30 | SALES |
| WARD | 30 | SALES |
| JONES | 20 | RESEARCH |
| MARTIN | 30 | SALES |
| BLAKE | 30 | SALES |
| CLARK | 10 | ACCOUNTING |
| SCOTT | 20 | RESEARCH |
| KING | 10 | ACCOUNTING |
| TURNER | 30 | SALES |
| ADAMS | 20 | RESEARCH |
| JAMES | 30 | SALES |
| FORD | 20 | RESEARCH |
| MILLER | 10 | ACCOUNTING |
+--------+--------+------------+
14 rows in set (0.00 sec)Note: For the subquery after select, only one result can be returned at a time. If there is more than one record, an error will be reported!