SQL 计算一列中两个日期之间的天数

Question

我正在尝试计算同一列中两个日期之间的天数！当 ID 相同时。

下面是一个示例...

ID       Date    
a       01/12/2024    
a       01/13/2024    
b       01/12/2024    
b       01/20/2024    
b       01/25/2024    
c       01/12/2024    
c       01/12/2024    
   

计数天数应出现在新列中。

ID       Date        Days 
a       01/12/2024    0   
a       01/13/2024    1   
b       01/12/2024    0   
b       01/20/2024    8   
b       01/25/2024    5   
c       01/12/2024    0   
c       01/12/2024    0   

非常感谢任何帮助（y）。

Answer 1

您可以使用窗口函数

LAG()

来检索先前的行，并使用

DATEDIFF

函数来计算两个日期之间的天数差异：

SELECT *, DATEDIFF(dt, LAG(dt, 1, dt) OVER (PARTITION BY ID ORDER BY dt)) as Days 
FROM mytable;

演示在这里

Answer 2

这个例子有帮助吗？

create table #temp(

    id varchar(10) null,
    theDate datetime null
    )
    go

    insert into #temp (id, theDate) values ('a','20240112')
    insert into #temp (id, theDate) values ('a','20240113')
    insert into #temp (id, theDate) values ('b','20240112')
    insert into #temp (id, theDate) values ('b','20240120')
    insert into #temp (id, theDate) values ('b','20240125')
    insert into #temp (id, theDate) values ('c','20240112')
    insert into #temp (id, theDate) values ('c','20240112')


select d.id, d.theDate, case when  case when DATEDIFF(day,LAG(d.theDate) over (order by d.id, d.theDate),d.theDate) < 0 then 0 else  DATEDIFF(day,LAG(d.theDate) over (order by d.id, d.theDate),d.theDate) end is null then '' else case when DATEDIFF(day,LAG(d.theDate) over (order by d.id, d.theDate),d.theDate) < 0 then 0 else  DATEDIFF(day,LAG(d.theDate) over (order by d.id, d.theDate),d.theDate) end end Days
from #temp d


id  theDate                 Days
a   2024-01-12 00:00:00.000 0
a   2024-01-13 00:00:00.000 1
b   2024-01-12 00:00:00.000 0
b   2024-01-20 00:00:00.000 8
b   2024-01-25 00:00:00.000 5
c   2024-01-12 00:00:00.000 0
c   2024-01-12 00:00:00.000 0