Question

我坚持如何在t-sql中解决这个问题。我一直在想通过c＃或某种递归调用来做这件事。下面的测试用例给出了这个想法，找出了重新开启索赔票的次数。

--assumptions
-- A claim is considered reopened only and only when is opened after
-- its previous state is closed. 
-- an open claim cannot have a closed date


--TEST CASE 1

DECLARE @ClaimHistory TABLE (UserId INT, ClaimdId INT, RefClaimStatus VARCHAR(10), DateOpened DATETIME NOT NULL, DateClosed DATETIME NULL)
DELETE @ClaimHistory
INSERT INTO @ClaimHistory
VALUES
(1,1,'open','1/1/2015',null), 
(1,1,'closed','1/2/2015','1/2/2015'), 
(1,1,'open','1/3/2015',null)

-- First determine the range of dates 
DECLARE @BeginDate DATETIME = '1/1/2014';
DECLARE @EndDate DATETIME = '1/4/2015';

SELECT UserId, ClaimdId, RefClaimStatus, DateOpened, DateClosed 
FROM @ClaimHistory
WHERE DateOpened >= @BeginDate   AND  DateOpened <= @EndDate


--Answer the claim was reopened 1 time for user id 1
--TEST Case 2
DELETE @ClaimHistory
INSERT INTO @ClaimHistory
VALUES
(1,1,'open','1/1/2015',null), 
(1,1,'closed','1/2/2015','1/2/2015'), 
(1,1,'open','1/3/2015',null), -- in a sense a claim was transferred to to userid2 . 
(2,1,'open','1/17/2015',null),
(2,1,'closed','1/17/2015','1/19/2015'),
(2,1,'open','1/18/2015',null)

--Answer the claim was 
--1.) reopened 1 time for user 1
--2.) reopened 1 time for user 2


--Test Case 3
DELETE @ClaimHistory
INSERT INTO @ClaimHistory
VALUES
(1,1,'open','1/1/2015',null), 
(1,1,'closed','1/2/2015','1/2/2015'), 
(1,1,'open','1/3/2015',null), 
(1,1,'closed','1/3/2015','1/5/2015'),  -- this means that the last time it was opened was 1/3/2015 and finally closed 1/5/2015
(1,1,'open','1/6/2015',null), -- This guy  is horrible at his job... opens the claim again
(2,1,'open','1/17/2015',null),--boss decides user 1 not good transfers to to userid2 . 
(2,1,'closed','1/17/2015','1/19/2015')

--Answer user id 1 had claimed reopened twice
--user id 2 had his claim was never reopened

Answer 1

用户是否可以打开票证（由同一用户或其他用户打开）？

如果没有，或者如果这种行为也被视为“重新开放”，那么这样的话：

select UserId, ClaimdId, count(*) - 1 as reopen_count
from @ClaimHistory
where RefClaimStatus = 'open'
group by UserId, ClaimdId
having count(*) > 1

更新：

下面的脚本适用于案例1和案例3，但不适用于案例2.似乎案例2不正确，票证由同一用户打开两次，在1/17，然后在1/18，然后关闭1/19。如果最后两行具有正确的日期，则脚本也适用于案例2

SELECT ch1.UserId, ch1.ClaimdId, ch1.RefClaimStatus, ch1.DateOpened, ch1.DateClosed 
FROM @ClaimHistory ch1
WHERE ch1.RefClaimStatus = 'Open'
    and ch1.ClaimdId in 
    (SELECT Q.ClaimdId
    FROM
        (SELECT TOP 1 ch2.ClaimdId, ch2.RefClaimStatus
        FROM @ClaimHistory ch2 
        WHERE ch2.ClaimdId = ch1.ClaimdId AND ch1.DateOpened > ISNULL(ch2.DateClosed, ch2.DateOpened)
        ORDER BY ISNULL(ch2.DateClosed, ch2.DateOpened) DESC) Q
    WHERE Q.RefClaimStatus = 'Closed')

T-SQL审计表确定何时重新打开索赔票证

1 个答案: