我想要一個聲明,如果兩個日期(MaturityDate,PaymentDate)之間的差異大於1個月,將返回1。 到期日是客戶應付款的到期日。子查詢返回的值超過1。這是不允許的
我已經試過:
CASE WHEN DATEDIFF(MONTH,
(SELECT Date FROM dim.RepaymentSchedule rs JOIN dim.Calendar cal ON cal.DateID = rs.MaturityDateID),
(SELECT Date FROM dim.RepaymentSchedule rs JOIN dim.Calendar cal ON cal.DateID = rs.PaymentDateID))
BETWEEN 1 AND 2 THEN 1 ELSE 0 END AS OneMonthDelay
這是模擬到
CASE WHEN DATEDIFF(MONTH, 2017-06-30, 2017-08-01) BETWEEN 1 AND 2 THEN 1 ELSE 0 END AS OneMonthDelay
,可惜的返回
子查詢返回多個值。當 子查詢遵循=,!=,<,< =,>,> =或當子查詢用作 表達式時,這是不允許的。
任何幫助,將不勝感激。
編輯:整個查詢
SELECT
pt.ProductType
,sr.SalesRegionName
,ca.CreditAdvisorID
,cal.CalendarYearMonth
,CASE WHEN DATEDIFF(MONTH,
(SELECT Date FROM dim.RepaymentSchedule rs JOIN dim.Calendar cal ON cal.DateID = rs.MaturityDateID),
(SELECT Date FROM dim.RepaymentSchedule rs JOIN dim.Calendar cal ON cal.DateID = rs.PaymentDateID))
BETWEEN 1 AND 2 THEN 1 ELSE 0 END AS OneMonthDelay -- >1 <2 not =>1 =<2
,CASE WHEN DATEDIFF(MONTH,
(SELECT Date FROM dim.RepaymentSchedule rs JOIN dim.Calendar cal ON cal.DateID = rs.MaturityDateID),
(SELECT Date FROM dim.RepaymentSchedule rs JOIN dim.Calendar cal ON cal.DateID = rs.PaymentDateID))
BETWEEN 2 AND 3 THEN 1 ELSE 0 END AS TwoMonthsDelay
,RANK() OVER (PARTITION BY c.ApplicationID ORDER BY rs.RepaymentNumber, rs.Amount) AS RankID
INTO #Frauds
FROM
dim.Contract c
JOIN dim.Application a ON c.ApplicationID = a.ApplicationID
JOIN dim.Calendar cal ON a.ApplicationDateID = cal.DateId
JOIN dim.CreditAdvisor ca ON a.OriginalCreditAdvisorID = ca.CreditAdvisorId
JOIN dim.SalesRegion sr ON ca.SalesRegionID = sr.SalesRegionId
JOIN dim.ProductType pt ON a.ProductTypeID = pt.ProductTypeID
JOIN dim.RepaymentSchedule rs ON c.ContractID = rs.ContractID
WHERE
((cal.CalendarYear >= @YearId) AND (rs.MaturityDateID < @DateId)) -- Since given year to this date
AND ((rs.PaymentDateID = 19000101) OR (rs.PaymentDateID > rs.MaturityDateID))
AND rs.Amount <> 0
錯誤非常明顯。你能發佈你的整個查詢嗎? – HoneyBadger
使用相關的子查詢或連接。 – jarlh
@HoneyBadger爲整個查詢編輯。 –