Empleado con el salario más alto del departamento (empleados con los tres salarios más altos del departamento)

Estructura de la mesa

empleado

departamento

CREATE TABLE `department`  (
  `id` int(0) NOT NULL,
  `name` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_ai_ci NULL DEFAULT NULL,
  PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_0900_ai_ci ROW_FORMAT = Dynamic;

INSERT INTO `department` VALUES (1, 'IT 部门');
INSERT INTO `department` VALUES (2, '销售部门');

CREATE TABLE `employee`  (
  `id` int(0) NOT NULL,
  `name` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_ai_ci NULL DEFAULT NULL,
  `salary` int(0) NULL DEFAULT NULL,
  `departmentId` int(0) NULL DEFAULT NULL,
  PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_0900_ai_ci ROW_FORMAT = Dynamic;

INSERT INTO `employee` VALUES (1, 'joe', 70000, 1);
INSERT INTO `employee` VALUES (2, 'jim', 90000, 2);
INSERT INTO `employee` VALUES (3, 'heney', 80000, 2);
INSERT INTO `employee` VALUES (4, 'sam', 70000, 2);
INSERT INTO `employee` VALUES (5, 'kity', 60000, 1);
INSERT INTO `employee` VALUES (6, 'hellen', 68000, 1);
INSERT INTO `employee` VALUES (7, 'hym', 68000, 1);
INSERT INTO `employee` VALUES (8, 'liliy', 72000, 2);

1. Si solo consulta el nombre, puede utilizar el método más sencillo.

Consulta el ID del departamento y el salario del primer lugar en cada departamento.

1.查询各个部门第一名的部门Id和薪资
SELECT departmentId,MAX(salary) departmentId FROM employee GROUP BY departmentId;

+--------------+--------------+
| departmentId | departmentId |
+--------------+--------------+
|            1 |        70000 |
|            2 |        90000 |
+--------------+--------------+

2.使用in
SELECT * FROM employee JOIN department on employee.departmentId = department.id 
WHERE #双条件in
  (employee.departmentId,employee.salary) 
in #各部门第一名
	(SELECT departmentId,MAX(salary) departmentId FROM employee GROUP BY departmentId) ;

+----+------+--------+--------------+----+--------------+
| id | name | salary | departmentId | id | name         |
+----+------+--------+--------------+----+--------------+
|  1 | joe  |  70000 |            1 |  1 | IT 部门      |
|  2 | jim  |  90000 |            2 |  2 | 销售部门     |
+----+------+--------+--------------+----+--------------+

2. Consulta los tres primeros

1. Primero introduzca tres funciones, las funciones de clasificación rango (), densa_rank (), fila_número () en mysql

rango() se ordena como 1,2,2,4. (Uno de los segundos lugares empatados es el tercer lugar, por lo que se omite el tercer lugar. Los rangos en la siguiente figura son los rangos

nombre, consulte los rangos de id=6, 7 y 5 para obtener más detalles)

dividir por es equivalente a agrupar por en función de clasificación

SELECT *,rank() over(partition by departmentId ORDER BY salary desc) ranks FROM employee;

+----+--------+--------+--------------+-------+
| id | name   | salary | departmentId | ranks |
+----+--------+--------+--------------+-------+
|  1 | joe    |  70000 |            1 |     1 |
|  6 | hellen |  68000 |            1 |     2 |
|  7 | hym    |  68000 |            1 |     2 |
|  5 | kity   |  60000 |            1 |     4 |
|  2 | jim    |  90000 |            2 |     1 |
|  3 | heney  |  80000 |            2 |     2 |
|  8 | liliy  |  72000 |            2 |     3 |
|  4 | sam    |  70000 |            2 |     4 |
+----+--------+--------+--------------+-------+

row_number() está ordenado como 1,2,3,4 según el número de fila

SELECT *,row_number() over(partition by departmentId ORDER BY salary desc) ranks FROM employee;
| id | name   | salary | departmentId | ranks |
+----+--------+--------+--------------+-------+
|  1 | joe    |  70000 |            1 |     1 |
|  6 | hellen |  68000 |            1 |     2 |
|  7 | hym    |  68000 |            1 |     3 |
|  5 | kity   |  60000 |            1 |     4 |
|  2 | jim    |  90000 |            2 |     1 |
|  3 | heney  |  80000 |            2 |     2 |
|  8 | liliy  |  72000 |            2 |     3 |
|  4 | sam    |  70000 |            2 |     4 |

denso_rank() está ordenado como 1,2,2,3. (denso significa "denso, denso" en inglés. denso_rank() es que los números de clasificación son continuos e ininterrumpidos)

SELECT *,dense_rank() over(partition by departmentId ORDER BY salary desc) ranks FROM employee;

| id | name   | salary | departmentId | ranks |
+----+--------+--------+--------------+-------+
|  1 | joe    |  70000 |            1 |     1 |
|  6 | hellen |  68000 |            1 |     2 |
|  7 | hym    |  68000 |            1 |     2 |
|  5 | kity   |  60000 |            1 |     3 |
|  2 | jim    |  90000 |            2 |     1 |
|  3 | heney  |  80000 |            2 |     2 |
|  8 | liliy  |  72000 |            2 |     3 |
|  4 | sam    |  70000 |            2 |     4 |
+----+--------+--------+--------------+-------+

2. Utilice la función mysql para consultar a los tres empleados principales.

1.先查询出薪资排名
SELECT *,rank() over(partition by departmentId ORDER BY salary desc) ranks FROM employee;
| id | name   | salary | departmentId | ranks |
+----+--------+--------+--------------+-------+
|  1 | joe    |  70000 |            1 |     1 |
|  6 | hellen |  68000 |            1 |     2 |
|  7 | hym    |  68000 |            1 |     2 |
|  5 | kity   |  60000 |            1 |     4 |
|  2 | jim    |  90000 |            2 |     1 |
|  3 | heney  |  80000 |            2 |     2 |
|  8 | liliy  |  72000 |            2 |     3 |
|  4 | sam    |  70000 |            2 |     4 |

2.再根据已查出的薪资排名，筛选出前三名员工
SELECT * FROM department 
JOIN (
	SELECT *,rank() over(partition by departmentId ORDER BY salary desc) ranks FROM employee) a 
on department.id = a.departmentId WHERE a.ranks <=3 ;

| id | name         | id | name   | salary | departmentId | ranks |
+----+--------------+----+--------+--------+--------------+-------+
|  1 | IT 部门      |  1 | joe    |  70000 |            1 |     1 |
|  1 | IT 部门      |  6 | hellen |  68000 |            1 |     2 |
|  1 | IT 部门      |  7 | hym    |  68000 |            1 |     2 |
|  2 | 销售部门     |  2 | jim    |  90000 |            2 |     1 |
|  2 | 销售部门     |  3 | heney  |  80000 |            2 |     2 |
|  2 | 销售部门     |  8 | liliy  |  72000 |            2 |     3 |

3. Utilice el método tradicional para consultar los tres primeros.

1.查询出每个薪资数，比其它薪资数高的数量
select * from employee a,employee b where a.departmentId=b.departmentId and a.salary<b.salary;

El significado de esta declaración, cuando se traduce, es,

Hay 3 registros para salarios inferiores a 70000

Hay 2 registros para salario inferior a 68000

. . . . . ¿Por qué no obtuve 1 registro, porque hay dos 68000?

Salario inferior a 60000 tiene 0 registros

select * from employee a JOIN  department ON a.departmentId = department.id 
where (
	select count(1) from employee where a.departmentId=departmentId and a.salary<salary) <3 
order by a.departmentId, a.salary desc;


# where中select count(1) from employee where a.departmentId=departmentId and a.salary<salary) <3

整体查询时a.salary<salary 的意思是，employee的薪资比a中的高(注意:和之前的单独查询是意思不一样哦)
那么

当a中是70000是，count出来薪资比它高的数量是0条，小于3条  满足
当a中是68000是，count出来薪资比它高的数量是1条，小于3条  满足
当a中是60000是，count出来薪资比它高的数量是3条，等于3条  不满足


| id | name   | salary | departmentId | id | name         |
+----+--------+--------+--------------+----+--------------+
|  1 | joe    |  70000 |            1 |  1 | IT 部门      |
|  6 | hellen |  68000 |            1 |  1 | IT 部门      |
|  7 | hym    |  68000 |            1 |  1 | IT 部门      |
|  2 | jim    |  90000 |            2 |  2 | 销售部门     |
|  3 | heney  |  80000 |            2 |  2 | 销售部门     |
|  8 | liliy  |  72000 |            2 |  2 | 销售部门     |