MySQL中五大基础聚合函数的区别与应用

SELECT COUNT(*) FROM orders;
SELECT COUNT(1) FROM orders;
SELECT COUNT(status) FROM orders;

EXPLAIN SELECT COUNT(*) FROM orders;
EXPLAIN SELECT COUNT(1) FROM orders;
-- 两条语句的 Extra、rows、key 完全一致

-- 假设 orders 表有 100 行，其中 status 列有 5 个 NULL
SELECT COUNT(*) FROM orders;          -- 结果：100
SELECT COUNT(1) FROM orders;          -- 结果：100
SELECT COUNT(status) FROM orders;     -- 结果：95

-- 有多少个不同的用户下过单
SELECT COUNT(DISTINCT user_id) FROM orders;

-- 快速获取近似行数
SELECT TABLE_ROWS
FROM information_schema.TABLES
WHERE TABLE_SCHEMA = 'your_db' AND TABLE_NAME = 'orders';

-- 所有订单的总金额
SELECT SUM(amount) FROM orders;

-- 所有订单的平均金额
SELECT AVG(amount) FROM orders;

-- 假设 amount 列有值：100, 200, NULL, 300
-- SUM(amount) = 600（NULL 被忽略）
-- AVG(amount) = 600 / 3 = 200（不是 600 / 4 = 150）

SELECT AVG(IFNULL(amount, 0)) FROM orders;
-- 现在 NULL 被当作 0，结果是 600 / 4 = 150

-- 最大的订单金额
SELECT MAX(amount) FROM orders;

-- 最早的订单时间
SELECT MIN(create_time) FROM orders;

-- 找出金额最大的那条订单的完整信息
SELECT * FROM orders
WHERE amount = (SELECT MAX(amount) FROM orders);

-- 每个用户的订单数
SELECT user_id, COUNT(*) AS order_count
FROM orders
GROUP BY user_id;

-- 每个月的销售总额
SELECT
    DATE_FORMAT(create_time, '%Y-%m') AS month,
    SUM(amount) AS monthly_total
FROM orders
GROUP BY DATE_FORMAT(create_time, '%Y-%m');

-- 错误：status 没有出现在 GROUP BY 中，也不是聚合函数的参数
SELECT user_id, status, COUNT(*)
FROM orders
GROUP BY user_id;

-- 正确：status 也加入分组
SELECT user_id, status, COUNT(*)
FROM orders
GROUP BY user_id, status;

-- WHERE：在分组前过滤原始行
SELECT user_id, COUNT(*) AS order_count
FROM orders
WHERE status = 'paid'          -- 先过滤掉未支付的订单
GROUP BY user_id;

-- HAVING：在分组后过滤聚合结果
SELECT user_id, COUNT(*) AS order_count
FROM orders
GROUP BY user_id
HAVING COUNT(*) > 10;           -- 再过滤掉订单数不超过 10 的组

-- 完整的执行顺序：WHERE → GROUP BY → HAVING → ORDER BY → LIMIT
SELECT user_id, SUM(amount) AS total
FROM orders
WHERE status = 'paid'           -- 第一步：过滤原始行
GROUP BY user_id                -- 第二步：分组
HAVING total > 1000             -- 第三步：过滤分组结果
ORDER BY total DESC             -- 第四步：排序
LIMIT 10;                       -- 第五步：取前 10 条

-- 错误：WHERE 还没到分组阶段，不能用聚合函数
SELECT user_id, COUNT(*)
FROM orders
WHERE COUNT(*) > 10
GROUP BY user_id;

-- 正确：用 HAVING
SELECT user_id, COUNT(*)
FROM orders
GROUP BY user_id
HAVING COUNT(*) > 10;

-- 错误：MySQL 不允许在 WHERE 中直接用聚合函数
SELECT * FROM orders WHERE amount > AVG(amount);

-- 正确：用子查询包一层
SELECT * FROM orders
WHERE amount > (SELECT AVG(amount) FROM orders);

-- 找出每个用户消费最高的那笔订单
SELECT o.*
FROM orders o
INNER JOIN (
    SELECT user_id, MAX(amount) AS max_amount
    FROM orders
    GROUP BY user_id
) t ON o.user_id = t.user_id AND o.amount = t.max_amount;