Question

我有一个带有子查询的查询。两者都返回用户ID列表。在内部查询中，我给出了在给定时间范围内在特定位置使用现金的所有客户。

第二个，选择在特定日期之后订购的第一个用户ID，并且不使用现金作为付款方式。

select distinct c.user_id
from (
    select distinct o.user_id
    from `order` as o
    inner join payments as p on p.id = o.id
    where o.orderplaced_ts > "2016-01-01 00:00:00"
    and o.store_id = "12"
    and p.payment_method = "Cash"
) as c
inner join `order` as o on c.user_id = o.user_id
inner join `payments` as p on o.id = p.id
where o.orderplaced_ts > "2016-03-13 00:00:00"
and o.store_id = "12"
and p.payment_method != "Cash"

如何在2016-03-13以后没有订单的用户之间进行交叉。

供参考：

select distinct o.user_id
from `order` as o
inner join payments as p on p.id = o.id
where o.orderplaced_ts > "2016-01-01 00:00:00"
and o.store_id = "12"
and p.payment_method = "Cash"

返回236个唯一ID - 调用此集A

两个查询组合返回160唯一ID - 调用此集B

集合A中的哪些ID在集合B中是not

Answer 1

我更愿意从你这边得到一些方法。

但是既然我没有，这是我的尝试：

select distinct c.user_id
from (
    select distinct o.user_id
    from `order` as o
    inner join payments as p on p.id = o.id
    where o.orderplaced_ts > "2016-01-01 00:00:00"
    and o.store_id = "12"
    and p.payment_method = "Cash"
) as c
LEFT JOIN `order` as o 
ON c.user_id = o.user_id
   AND o.orderplaced_ts > "2016-03-13 00:00:00"
   AND o.store_id = "12"
LEFT JOIN `payments` as p 
ON o.id = p.id
   AND p.payment_method != "Cash"
WHERE p.id IS NULL

Answer 2

在两个查询之间使用not in子句

select distinct o.user_id
from `order` as o
inner join payments as p on p.id = o.id
where o.orderplaced_ts > "2016-01-01 00:00:00"
and o.store_id = "12"
and p.payment_method = "Cash"
and o.user_id not in (

    select distinct c.user_id
    from (
        select distinct o.user_id
        from `order` as o
     inner join payments as p on p.id = o.id
        where o.orderplaced_ts > "2016-01-01 00:00:00"
        and o.store_id = "12"
        and p.payment_method = "Cash"
    ) as c
    inner join `order` as o on c.user_id = o.user_id
    inner join `payments` as p on o.id = p.id
    where o.orderplaced_ts > "2016-03-13 00:00:00"
    and o.store_id = "12"

)

MySQL：在两个查询之间获得不同的，不同的值

2 个答案: