SELECT o.*,o.`payed`,(SELECT COUNT(s.payed)+1 FROM order_info s WHERE s.payed>o.payed) rank FROM order_info o ORDER BY o.payed DESC;
查询并排序1134方式

最新推荐文章于 2025-04-10 15:59:08 发布
SELECT o.*,o.`payed`,(SELECT COUNT(s.payed)+1 FROM order_info s WHERE s.payed>o.payed) rank FROM order_info o ORDER BY o.payed DESC;