php - MySQL Join并排除count = 0的行

我有以下简单查询，它是垃圾回收脚本的一部分。该脚本应删除未使用的购物车。超过24小时前更新的购物车尚未使用，并且未附加到订单或用户。

        $query = "SELECT comm_cart.id AS `cart_id`, (" . time() . " - comm_cart.update_date) AS `diff`, COUNT(comm_orders.cart_id) AS `c1`, COUNT(comm_users.cart_id) AS `c2` " .
             "FROM `comm_cart` " .
             "LEFT JOIN `comm_orders` ON comm_cart.id=comm_orders.cart_id " .
             "LEFT JOIN `comm_users` ON comm_cart.id=comm_users.cart_id " .
             "GROUP BY comm_cart.id ";
             "HAVING `diff`>86400 AND `c1`=0 AND `c2`=0";

该查询找到了太多的购物车：它还标记了c1> 0或c2> 0的购物车，我不知道为什么。有什么线索吗？

最佳答案

我怀疑您正在沿着两个不同的维度加入。简单的解决方法是使用distinct：

SELECT comm_cart.id AS `cart_id`, (" . time() . " - comm_cart.update_date) AS `diff`,
       COUNT(DISTINCT comm_orders.cart_id) AS `c1`, COUNT(DISTINCT comm_users.cart_id) AS `c2` " .

更好的解决方案是对这两个条件使用not exists：

FROM comm_carts cc
WHERE not exists (select 1 from comm_orders co where cc.id = co.cart_id )
      not exists (select 1 from comm_users cu where cc.id = cu.cart_id )

关于php - MySQL Join并排除count = 0的行，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/24409552/