Question

我有两张桌子：

delivery包含uid，dtime，candy列，用于记录哪些用户何时获赠糖果
lookup列uid和ltime，用于记录用户口袋的检查时间

我需要知道lookup的结果，即结果表应该有uid，ltime，candy列，告诉我在用户的口袋（假设用户在给予新糖果时会吃旧糖果。）

每次查询前都有几次交付。我只需要最新的一个。

，例如，

select l.uid, l.ltime, 
       d.candy  /* ... for max(d.dtime):
          IOW, I want to sort (d.dtime, d.candy) 
          by the first field in decreasing order, 
          then take the second field in the first element */
from delivery d
join lookup l 
on d.uid = l.uid
and d.dtime <= l.ltime
group by l.uid, l.ltime

那么，我怎么知道查找找到了什么？

Answer 1

我建议outer apply：

select l.*, d.candy
from lookup l outer apply
     (select top 1 d.*
      from delivery d
      where d.uid = l.uid and d.dtime <= l.ltime
      order by d.dtime desc
     ) d;

这回答了你的问题。但是，自上次查找以来，用户不会所有糖果吗？或者，我们是否假设当用户获得另一个用户时，用户会吃掉糖果？也许口袋里只有一个糖果。

Answer 2

使用Top with with Ties获取最新的交付并加入Backup Table

Select * from lookup
    Inner Join (
        Select Top 1  with Ties uid,dtime
          From delivery 
          Order by row_number() over (partition by uid order by dtime desc)) as Delivery
          on lookup.uid = Delivery.uid  and lookup.ltime >= delivery.dtime

如何加入最新更新？

2 个答案: