MySQL的UPDATE(更新数据)及语法详解

UPDATE [LOW_PRIORITY] [IGNORE] table_name
SET column1 = value1, column2 = value2, ...
[WHERE condition]
[ORDER BY ...]
[LIMIT row_count]

-- 更新表 students 中 id 为 1 的记录，将 name 字段设为 '张三'
UPDATE students
SET name = '张三'
WHERE id = 1;
-- 更新表 students 中所有记录，将 age 字段增加 1
UPDATE students
SET age = age + 1;

-- 将表 students 中所有记录的 age 字段增加 1
UPDATE students
SET age = age + 1;

-- 将表 students 中 name 为 '张三' 的记录的 class_id 更新为表 classes 中 name 为 '数学班' 的 class_id
UPDATE students
SET class_id = (SELECT id FROM classes WHERE name = '数学班')
WHERE name = '张三';

-- 更新表 orders 和 order_details，将订单总金额大于 1000 的订单状态设置为 '已完成'
UPDATE orders o
JOIN order_details od ON o.order_id = od.order_id
SET o.status = '已完成'
WHERE o.total_amount > 1000;

-- 根据学生的年龄更新他们的等级
UPDATE students
SET grade = CASE
    WHEN age < 18 THEN '初级'
    WHEN age BETWEEN 18 AND 25 THEN '中级'
    ELSE '高级'
END;

-- 根据学生的成绩更新他们的状态
UPDATE students
SET status = IF(score >= 60, '及格', '不及格');

-- 在学生的姓名后面添加 '同学'
UPDATE students
SET name = CONCAT(name, '同学');

-- 将学生的姓名中的 '张' 替换为 '李'
UPDATE students
SET name = REPLACE(name, '张', '李');

-- 如果学生的成绩为 NULL，则将其设为 0
UPDATE students
SET score = COALESCE(score, 0);

-- 假设 id 字段有索引
UPDATE students
SET name = '张三'
WHERE id = 1;

-- 批量更新多个记录
UPDATE employees
SET salary = CASE
    WHEN id = 1 THEN 50000
    WHEN id = 2 THEN 60000
    WHEN id = 3 THEN 70000
    ELSE salary
END
WHERE id IN (1, 2, 3);

-- 避免这种写法
UPDATE employees
SET salary = 50000;

-- 限制更新行数
UPDATE employees
SET salary = 50000
WHERE id > 1000
LIMIT 100;

START TRANSACTION;
UPDATE employees
SET salary = 50000
WHERE id BETWEEN 1 AND 1000;
COMMIT;
START TRANSACTION;
UPDATE employees
SET salary = 50000
WHERE id BETWEEN 1001 AND 2000;
COMMIT;

-- 将 id 为 1 的员工的工资设为 60000
UPDATE employees
SET salary = 60000
WHERE id = 1;

-- 将部门为 10 的所有员工的工资增加 10%
UPDATE employees
SET salary = salary * 1.1
WHERE department_id = 10;

-- 将 id 为 1 的员工的部门设为 20
UPDATE employees
SET department_id = 20
WHERE id = 1;

-- 将 id 为 1 的员工的部门设为 '研发部' 的部门 ID
UPDATE employees
SET department_id = (SELECT id FROM departments WHERE name = '研发部')
WHERE id = 1;

-- 将 id 为 1 的员工的名字设为 '李四'，工资设为 70000
UPDATE employees
SET name = '李四', salary = 70000
WHERE id = 1;