浅谈MySQL 中 null 值的那些坑

CREATE TABLE users (
    id INT PRIMARY KEY,
    name VARCHAR(50),
    age INT,
    email VARCHAR(100)
);
 
INSERT INTO users VALUES 
(1, 'Alice', 25, 'alice@example.com'), 
(2, 'Bob', 30, NULL),
(3, 'Charlie', NULL, 'charlie@example.com'); 

SELECT * FROM users WHERE email = NULL; -- 查不到任何结果 
SELECT * FROM users WHERE email <> NULL; -- 同样查不到任何结果 

SELECT * FROM users WHERE email IS NULL; -- 正确的结果：Bob 的记录 
SELECT * FROM users WHERE age IS NOT NULL; -- 正确的结果：Alice 和 Charlie 的记录 

SELECT * FROM users WHERE email = NULL; -- 错误！返回空结果 

CREATE TABLE orders (
    order_id INT PRIMARY KEY,
    user_id INT,
    amount DECIMAL(10, 2)
);
 
CREATE TABLE users (
    user_id INT PRIMARY KEY,
    name VARCHAR(50),
    email VARCHAR(100)
);
 
INSERT INTO orders VALUES 
(1, 1, 100.00),
(2, 2, 200.00),
(3, 3, 300.00);
 
INSERT INTO users VALUES 
(1, 'Alice', 'alice@example.com'), 
(2, 'Bob', NULL),
(3, 'Charlie', 'charlie@example.com'); 

SELECT * FROM orders 
WHERE user_id IN (SELECT user_id FROM users WHERE email IS NULL); -- 正确的结果：Bob 的订单 

SELECT * FROM orders 
WHERE user_id IN (SELECT user_id FROM users WHERE email = NULL); -- 错误！返回空结果 

-- 查询 email 为 null 的用户 
SELECT * FROM users WHERE email IS NULL;
 
-- 查询 age 不为 null 的用户 
SELECT * FROM users WHERE age IS NOT NULL;

-- 正确的写法：使用 COALESCE 将 null 转换为一个不存在的值 
SELECT * FROM orders 
WHERE user_id IN (
    SELECT COALESCE(user_id, -1) FROM users WHERE email IS NULL 
);

-- 插入 null 值 
INSERT INTO users (id, name, age, email) 
VALUES (4, 'David', NULL, 'david@example.com'); 
 
-- 更新 null 值 
UPDATE users SET email = NULL WHERE id = 4;

-- 查询 email 为 null 的用户 
SELECT * FROM users WHERE email IS NULL;
 
-- 查询 email 为空字符串的用户 
SELECT * FROM users WHERE email = '';

-- 按 age 升序排列，null 排在最前面 
SELECT * FROM users ORDER BY age ASC;
 
-- 按 age 降序排列，null 排在最后面 
SELECT * FROM users ORDER BY age DESC;

-- 计算所有用户的平均年龄（忽略 null 值）
SELECT AVG(age) FROM users;