通过 MySQL 查询优化组

Question

我有一个 orders table 可以连续保存一个用户订单的一个项目。因此每个用户都可以出现在多行中。每个订单行都有一个状态 - PENDING、APPROVED 等等。我想为每个用户计算每个状态的订单数量，即有多少订单已获批准，有多少仍待处理等等。我提出了以下查询：

SELECT 
    u.email, u.id,
    (SELECT count(status) FROM orders WHERE user_id = o.user_id AND status = 'PENDING') pending,
    (SELECT count(status) FROM orders WHERE user_id = o.user_id AND status = 'APPROVED') approved,
    (SELECT count(status) FROM orders WHERE user_id = o.user_id AND status = 'REJECTED') rejected,
    (SELECT count(status) FROM orders WHERE user_id = o.user_id AND status = 'ERROR') error
FROM orders o INNER JOIN users u ON o.user_id = u.id
GROUP BY o.user_id, u.email

问题是速度太慢了！我在 orders table 中有大约 5000 条记录，执行起来只需几分钟。请教如何优化。

Answer 1

您可以使用条件聚合摆脱子查询：

SELECT 
    u.email, u.id,
    COUNT(CASE WHEN status = 'PENDING' THEN 1 END) pending,
    COUNT(CASE WHEN status = 'APPROVED' THEN 1 END) approved,
    COUNT(CASE WHEN status = 'REJECTED' THEN 1 END) rejected,
    COUNT(CASE WHEN status = 'ERROR' THEN 1 END) error
FROM orders o INNER JOIN users u ON o.user_id = u.id
GROUP BY o.user_id, u.email

Answer 2

实际上不确定您是否需要分组依据：

SELECT u.email, u.id,o.pending,o.approved,o.rejected,o.error 
FROM users u INNER JOIN 
(
    SELECT
    user_id,
    COUNT(CASE WHEN status = 'PENDING' THEN 1 ELSE 0 END) pending,
    COUNT(CASE WHEN status = 'APPROVED' THEN 1 ELSE 0 END) approved,
    COUNT(CASE WHEN status = 'REJECTED' THEN 1 ELSE 0 END) rejected,
    COUNT(CASE WHEN status = 'ERROR' THEN 1 ELSE 0 END) error
    FROM orders
) o
ON u.id = o.user_id

通过 MySQL 查询优化组

Optimizing a group by MySQL query

mysql

sql-optimization