什么是需要超过一天的卸货，我有订单日期和卸货日期

Question

我开始了我的职业生涯中的数据分析，我必须在日常工作中使用SQL语句。我正在学习，但还需要提供一些快速解答。所以我想我会在这个小组中提出一些问题。

我需要帮助来编写SQL查询来获取需要超过一天或两天（基于rquirement）从该位置释放的订单。

活动类型列代表1,2,3,4

1-Order placed 
2-Order discharged 

日期记录在列日的相应行中

现在我想呼吁所有的订单，要花了超过一定的天数'n'

这是表格的例子，我的表格如何。

活动表

|Order Nr| activity|date| 
|  1 |  1 | date1| order placed 
|  1 |  3 | date2| order approved 
|  1 |  4 | date3| order packed 
|  1 |  2 | date4| order discharged 

Answer 1

你为了得到拆放同

select * from activities where activity = 1 

，你可能已经猜到如何获得订单排放:-)

所以将二者结合起来，只保留行与过高的区别：

select p.order_nr, p.date as placed, d.date as discharged 
from (select * from activities where activity = 1) p 
join (select * from activities where activity = 2) d 
    on d.order_nr = p.order_nr and datediff(d.date, p.date) > 1; 

你可以得到相同的，每个订单的集合：

select 
    order_nr, 
    any_value(case when activity = 2 then date end) as placed, 
    any_value(case when activity = 1 then date end) as discharged 
from activities 
group by order_nr 
having datediff(any_value(case when activity = 2 then date end), 
       any_value(case when activity = 1 then date end)) > 1; 

如果您要包括未结订单，你会做几乎相同的。对于没有解除记录的订单，可能会在今天输入，因此昨天发出的订单可能仍然正常，而之前发出的订单已经过久。所以如果没有dicharged记录，我们想假装有一个date = today。

查询＃1：

select p.order_nr, p.date as placed, d.date as discharged 
from (select * from activities where activity = 1) p 
left join (select * from activities where activity = 2) d 
    on d.order_nr = p.order_nr and datediff(coalesce(d.date, curdate()), p.date) > 1; 

查询＃2：

select 
    order_nr, 
    any_value(case when activity = 2 then date end) as placed, 
    any_value(case when activity = 1 then date end) as discharged 
from activities 
group by order_nr 
having datediff(any_value(case when activity = 2 then date end), 
       coalesce(any_value(case when activity = 1 then date end), curdate())) > 1; 

Answer 2

Not exists是一个方法：

select a.* 
from activities a 
where a.activity = 'placed' and 
     not exists (select 1 
        from activities a2 
        where a2.activity = 'discharged' and 
         a2.ordernum = a.ordernum and 
         a2.date >= a.date and 
         a2.date <= a.date + interval 1 day 
       );