MySQL view exercise

Data preparation

CREATE TABLE dept  (
  deptno int NOT NULL,
  dname varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
  loc varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
  PRIMARY KEY (`deptno`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;

INSERT INTO dept VALUES (10, ' Teaching and Research Department ', ' Beijing ');
INSERT INTO dept VALUES (20, ' Department of science and Engineering ', ' Shanghai ');
INSERT INTO dept VALUES (30, ' The sales department ', ' Guangzhou ');
INSERT INTO dept VALUES (40, ' Finance Department ', ' wuhan ');

CREATE TABLE emp  (
  empno int NOT NULL,
  ename varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
  job varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
  mgr int NULL DEFAULT NULL,
  hiredate date NULL DEFAULT NULL,
  sal decimal(7, 2) NULL DEFAULT NULL,
  COMM decimal(7, 2) NULL DEFAULT NULL,
  deptno int NULL DEFAULT NULL,
  PRIMARY KEY (empno) USING BTREE,
  INDEX fk_emp(mgr) USING BTREE,
  CONSTRAINT fk_emp FOREIGN KEY (mgr) REFERENCES emp (empno) ON DELETE RESTRICT ON UPDATE RESTRICT
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;

INSERT INTO emp VALUES (1001, ' Gan Ning ', ' Clerk ', 1013, '2000-12-17', 8000.00, NULL, 20);
INSERT INTO emp VALUES (1002, ' Daisy ', ' Salesperson ', 1006, '2001-02-20', 16000.00, 3000.00, 30);
INSERT INTO emp VALUES (1003, ' Yin Tianzheng ', ' Salesperson ', 1006, '2001-02-22', 12500.00, 5000.00, 30);
INSERT INTO emp VALUES (1004, ' Liu bei ', ' The manager ', 1009, '2001-04-02', 29750.00, NULL, 20);
INSERT INTO emp VALUES (1005, ' Thank sun ', ' Salesperson ', 1006, '2001-09-28', 12500.00, 14000.00, 30);
INSERT INTO emp VALUES (1006, ' Guan yu ', ' The manager ', 1009, '2001-05-01', 28500.00, NULL, 30);
INSERT INTO emp VALUES (1007, ' Zhang Fei ', ' The manager ', 1009, '2001-09-01', 24500.00, NULL, 10);
INSERT INTO emp VALUES (1008, ' Zhugeliang ', ' analysts ', 1004, '2007-04-19', 30000.00, NULL, 20);
INSERT INTO emp VALUES (1009, ' Zeng a Niu ', ' Chairman of the board of directors ', NULL, '2001-11-17', 50000.00, NULL, 10);
INSERT INTO emp VALUES (1010, ' Xiangr ', ' Salesperson ', 1006, '2001-09-08', 15000.00, 0.00, 30);
INSERT INTO emp VALUES (1011, ' Zhou Tai ', ' Clerk ', 1008, '2007-05-23', 11000.00, NULL, 20);
INSERT INTO emp VALUES (1012, ' Cheng pu ', ' Clerk ', 1006, '2001-12-03', 9500.00, NULL, 30);
INSERT INTO emp VALUES (1013, ' Pang Tong ', ' analysts ', 1004, '2001-12-03', 30000.00, NULL, 20);
INSERT INTO emp VALUES (1014, ' Huang Gai ', ' Clerk ', 1007, '2002-01-23', 13000.00, NULL, 10);

CREATE TABLE salgrade  (
  grade int NOT NULL,
  losal int NULL DEFAULT NULL,
  hisal int NULL DEFAULT NULL,
  PRIMARY KEY (grade) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;

INSERT INTO salgrade VALUES (1, 7000, 12000);
INSERT INTO salgrade VALUES (2, 12010, 14000);
INSERT INTO salgrade VALUES (3, 14010, 20000);
INSERT INTO salgrade VALUES (4, 20010, 30000);
INSERT INTO salgrade VALUES (5, 30010, 99990);

Exercises

1： Query the name of the Department with the highest average salary in the Department

2： Query the Department name of the employee whose salary is higher than that of the leader 、 Employee name 、 Employee leader number

3： Query salary grade is 4 level ,2000 The place of employment after 2010 is the employee number in Shanghai 、 Name and salary , And find out the information of employees with the top three salaries

The answer to the exercise

-- 1： Query the name of the Department with the highest average salary in the Department 
SELECT
	a.deptno,
	a.dname,
	a.loc,
	avg_sal
FROM
	dept a,

(
SELECT
	* 
FROM
	( SELECT *, RANK ( ) ) OVER ( ORDER BY avg_sal DESC ) rn 
FROM
	( SELECT deptno, AVG( deptno ) avg_sal FROM emp GROUP BY deptno ) t 
	) tt 
WHERE
	rn = 1 
	) ttt 
WHERE
	a.deptno = ttt.deptno

--  Method 2  	
	CREATE VIEW test_view1 AS SELECT deptno, AVG( deptno ) avg_sal FROM emp GROUP BY deptno
	CREATE VIEW test_view2 AS SELECT *, RANK ( ) ) OVER ( ORDER BY avg_sal DESC) rn FROM test_view1
	CREATE VIEW test_view3 AS SELECT * FROM test_view2 tt WHERE rn=1

SELECT
	a.deptno,
	a.dname,
	a.loc,
	avg_sal
FROM
	dept a,
	test_view3 ttt
WHERE
	a.deptno = ttt.deptno
	

-- 2： Query the Department name of the employee whose salary is higher than that of the leader 、 Employee name 、 Employee leader number 
-- 2.1 Query the department number of the employee whose salary is higher than that of the leader 
CREATE view test_view4
AS
SELECT
	a.ename enane,
	a.sal sal,
	b.ename mgrname,
	b.sal msal,
	a.deptno 
FROM
	emp a,
	emp b 
WHERE
	a.mgr = b.empno 
	AND a.sal > b.sal;
	
	-- 2.2 Query the department number and Department list found in the first step with a linked list 
SELECT
	* 
FROM
	dept a
	JOIN test_view4 b ON a.deptno = b.deptno;
	
-- 3： Query salary grade is 4 level ,2000 The place of employment after 2010 is the employee number in Shanghai 、 Name and salary , And find out the information of employees with the top three salaries 
-- 3.1 demand 1： Query salary grade is 4 level ,2000 The place of employment after 2010 is the employee number in Shanghai 、 Name and salary 
CREATE VIEW test_view5 AS SELECT
a.deptno,
a.dname,
a.loc,
b.empno,
b.ename,
b.sal 
FROM
	dept a
	JOIN emp b ON a.deptno = b.deptno 
	AND YEAR ( hiredate ) > '2000' 
	AND a.loc = ' Shanghai '
	JOIN salgrade c ON c.grade = 4 
	AND ( b.sal BETWEEN c.losal AND c.hisal );
	
	SELECT * FROM
	(
	SELECT *,RANK() OVER(ORDER BY sal DESC)rn
	FROM
	test_view5
)t
WHERE rn<=3;