Structure du tableau
employé
département
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 vous interrogez uniquement le prénom, vous pouvez utiliser la méthode la plus simple
Rechercher l'identifiant du département et le salaire de la première place dans chaque département
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. Interrogez les trois premiers
1. Introduisez d'abord trois fonctions, les fonctions de classement Rank(), dense_rank(), row_number() dans MySQL
Rank() est trié comme 1,2,2,4. (L'une des 2èmes places à égalité est la troisième place, donc la troisième place est ignorée. Les rangs dans la figure ci-dessous sont les rangs
nom, voir les rangs de id=6, 7 et 5 pour plus de détails)
partitionner par équivaut à regrouper par dans le tri des fonctions
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() est trié comme 1,2,3,4 trié en fonction du numéro de ligne
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 |
dense_rank() est trié comme 1,2,2,3. (dense signifie "dense, dense" en anglais. dense_rank() signifie que les numéros de tri sont continus et ininterrompus)
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. Utilisez la fonction MySQL pour interroger les trois meilleurs employés
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. Utilisez la méthode traditionnelle pour interroger les trois premiers
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;La signification de cette déclaration, une fois traduite, est la suivante :
Il existe 3 enregistrements pour des salaires inférieurs à 70 000
Il y a 2 enregistrements pour salaire inférieur à 68000
. . . . . Pourquoi n'ai-je pas obtenu 1 enregistrement, car il y a deux 68 000
Un salaire inférieur à 60 000 a 0 enregistrement
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 |