Question

我有一张桌子，

<p>This one is with flex-grow</p>
<div class="flex">
  <div class="col1">1 col</div>
  <div class="col2 fg">2 col</div>
</div>

<p>This one is with width 100%</p>
<div class="flex">
  <div class="col1">1 col</div>
  <div class="col2 fw">2 col</div>
</div>

现在，我想获得在同一datetime（2014-08-29 08：44：34.122）上创建的Ticket_NUM，其最大时差小于60秒。谁能让我知道如何编写查询以获取此数据。我在桌子上使用了自我联接，但没有得到想要的Ticket_NUM（2，5，3）。

Answer 1

一种方法是使用LEAD和LAG

select * from(
    select 
       *, 
        myFlag = case 
                    when abs(datediff(second,[date],lag([Date]) over (order by [Date]))) < 60 then 1
                    when abs(datediff(second,[date],lead([Date]) over (order by [Date]))) < 60 then 1
                    else 0
                 end
    from yourtable)
where myFlag = 1
order by [Date]

您可以删除where myFlag = 1来查看标记了哪些。

或更丑陋的方法：

with cte as(
select *, RN = row_number() over (order by [Date])
from yourtable
),

staging as(
    select 
        c.ID
        ,c.Ticket_NUM
        ,c.[Date]
        ,c.Comments
        ,c2DT = min(c2.[Date])
        ,c3DT = max(c3.[Date])
    from 
        cte c
        left join cte c2 on c2.RN = c.RN + 1  --row above
        left join cte c3 on c3.RN = c.RN - 1  --row below
    group by
        c.ID
        ,c.Ticket_NUM
        ,c.[Date]
        ,c.Comments)

select 
    c.ID
    ,c.Ticket_NUM
    ,c.[Date]
    ,c.Comments
from stating c
where abs(datediff(second,c.[Date],c.c2DT)) < 60 or abs(datediff(second,c.[Date],c.c3DT)) < 60

SQL-获取在1分钟的时差中创建的一个表中的所有行

1 个答案: