MySQL Join并排除count = 0的行

Question

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？

Answer 1

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` " .

更好的解决方案是针对这两个条件使用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 )

Answer 2

您甚至不需要在可能起作用的奇迹所在的情况下进行分组，当然，我建议使用Gordon关于不存在的建议，但是如果您希望进行最小的更改，那就是。

SELECT
    comm_cart.id AS `cart_id`,
    (UNIX_TIMESTAMP() - comm_cart.update_date) AS `diff`
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
WHERE
  comm_orders.cart_id IS NULL
  AND
  comm_users.cart_id IS NULL

哦，我已经使用UNIX_TIMESTAMP()代替了PHP时间函数，效果相同，但是这避免了将PHP和SQL混合使用。

Answer 3

如果您只想获取c1 = 0和c2 = 0的数据，则需要编写一个where条件，而不是使用group by之前，

$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 " . 
         " where c1=0 and c2 =0 and diff >86400 GROUP BY comm_cart.id;