SQL速查1

 

SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------

-- Table structure for employees
-- ----------------------------
DROP TABLE IF EXISTS `employees`;
CREATE TABLE `employees`  (
  `id` int(0) NULL DEFAULT NULL,
  `name` varchar(30) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_bin NULL DEFAULT NULL,
  `department` varchar(30) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_bin NULL DEFAULT NULL,
  `salary` int(0) NULL DEFAULT NULL,
  `manager_id` int(0) NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_0900_bin ROW_FORMAT = Dynamic;

-- ----------------------------

-- Records of employees
-- ----------------------------
INSERT INTO `employees` VALUES (1, 'Aditi', 'HR', 30000, 5);
INSERT INTO `employees` VALUES (2, 'Rahul', 'IT', 50000, 6);
INSERT INTO `employees` VALUES (3, 'Neha', 'IT', 60000, 6);
INSERT INTO `employees` VALUES (4, 'Aman', 'Sales', 40000, 7);
INSERT INTO `employees` VALUES (5, 'Kiran', 'HR', 70000, 0);
INSERT INTO `employees` VALUES (6, 'Mohit', 'IT', 80000, 0);
INSERT INTO `employees` VALUES (7, 'Suresh', 'Sales', 65000, 0);
INSERT INTO `employees` VALUES (8, 'Pooja', 'HR', 30000, 5);

SET FOREIGN_KEY_CHECKS = 1;

1.SELECT

SELECT name, salary FROM employees;

2.FROM

SELECT * FROM employees;

3.WHERE

SELECT * FROM employees WHERE salary > 30000;

4.ORDER BY

SELECT * FROM employees ORDER BY salary DESC;

5.COUNT()

SELECT COUNT(*) FROM employees;