根据键的支付日期比较（大于或小于）第一行值和第二行值，然后检查第三行是否大于第二行

Question

当前结果

-- Scenario1--对于ID 1和2，我想比较Diff Between (DD,Begin date of ID 2 , End Date OF ID 1) < = 1, 那么该值应该为真，否则显示不显示该值

Scenario2 对于 ID 10 和 11 -- ID 10 的开始日期和 ID 11 的开始日期相同，所以我想要根据付款日期进行比较，因为ID 11的付款日期> ID 10的付款日期。我希望显示的值更大

Id  Employeekey ReceiptNo   BeginDate   endDate     PaidDate    main    Supplier    RollNo
1   101         5505        3/28/2016   3/29/2016   4/29/2016   1       2001    655
2   101         5506        3/30/2016   4/1/2016    4/30/2016   1       2001    666
3   101         5507        4/5/2016    4/6/2016    4/30/2016   1       2001    155
4   101         5508        4/7/2016    4/10/2016   5/1/2016    1       2001    155
5   101         5509        4/11/2016   4/14/2016   5/5/2016    1       2001    155
6   101         5510        5/1/2016    5/3/2016    6/24/2016   1       2001    255
7   101         5511        5/1/2016    5/3/2016    6/30/2016   1       2001    265
8   102         5512        3/28/2017   3/29/2016   4/29/2017   1       2001    655
9   102         5513        3/28/2017   3/29/2016   4/29/2017   1       2001    655
10  102         5514        3/28/2017   3/29/2016   4/29/2017   1       2001    655
11  102         5515        3/28/2016   3/29/2016   5/29/2016   1       2001    655
12  102         5515        3/28/2016   3/29/2016   5/29/2016   1       2001    659

Answer 1

查看第二个场景的日期。 ID 10 和 ID 11 的 BeginDate 的日期不同。此外，PaidDate 不是 >，而是 <（查看两种情况下的年份）。话虽如此，我编辑了您的测试数据。这是您将如何执行此操作的方法。您可以更改它以满足您的需要，但这会让您继续前进。

declare @table table (id int, 
                     Employeekey int, 
                     RecieptNo int, 
                     BeginDate datetime, 
                     endDate datetime, 
                     PaidDate datetime, 
                     main int, 
                     Supplier int, 
                     RollNo int)
 insert into @table 
 values
(1,101,5505,'3/28/2016','3/29/2016','4/29/2016',1,2001,655),
(2,101,5506,'3/30/2016','4/1/2016','4/30/2016',1,2001,666),
(3,101,5507,'4/5/2016','4/6/2016','4/30/2016',1,2001,155),
(4,101,5508,'4/7/2016','4/10/2016','5/1/2016',1,2001,155),
(5,101,5509,'4/11/2016','4/14/2016','5/5/2016',1,2001,155),
(6,101,5510,'5/1/2016','5/3/2016','6/24/2016',1,2001,255),
(7,101,5511,'5/1/2016','5/3/2016','6/30/2016',1,2001,265),
(8,102,5512,'3/28/2017','3/29/2016','4/29/2017',1,2001,655),
(9,102,5513,'3/28/2017','3/29/2016','4/29/2017',1,2001,655),
(10,102,5514,'3/28/2016','3/29/2016','4/29/2016',1,2001,655),   --changed this to 2016 for being date and paid date
(11,102,5515,'3/28/2016','3/29/2016','5/29/2017',1,2001,655),   --changed this to 2017 for paid date
(12,102,5515,'3/28/2016','3/29/2016','5/29/2016',1,2001,659)



select
    *
    --scenario 1
    ,case 
        when datediff(day,lead(BeginDate) over (partition by EmployeeKey order by Id),endDate) <=1 then 'True' 
    end
    --scenario 2... id 10 and 11 have different years in your test but i fixed this in my test
    ,case 
        when lead(BeginDate) over (partition by EmployeeKey order by Id) = BeginDate
             and  lead(PaidDate) over (partition by EmployeeKey order by Id) > PaidDate then 'True' 
    end
from @table

根据键的支付日期比较（大于或小于）第一行值和第二行值，然后检查第三行是否大于第二行

Compare (greater than or less than) 1st row value with 2nd row value based on Paid date for the Key, and then check if 3rd row is greater than 2nd

sql-server

sql-server-2016