计算从一种状态到另一种状态需要多少天：SQL

Question

请尽情欣赏我们数据库的当前结构。

我们的 DBA 目前要离开两周，我的 SQL 知识非常有限，我喜欢留在 UI 和中间层。

我们要弄清楚的是如何执行以下操作，我们需要编写一个查询来计算所有佣金从“已验证”过渡到“已支付”的平均时间（以天为单位）经销商，目前状态为

1. 已创建
2. 已验证
3. 拒绝
4. 等待付款
5. 付费
6. 已退款

我认为这个查询需要直接针对佣金历史表？

由于我对 SQL 的了解有限，我不确定如何编写这样的查询...

任何帮助都会很棒。

Answer 1

这是一种实现您所追求的方法，尽管它可能不是最有效的。在我看来，它更像是您希望运行的一次性查询，而不是您要频繁运行以影响数据库性能的查询。

测试台设置：

CREATE TABLE Commission
(
    CommissionId INT,
    DealerId INT
)

CREATE TABLE CommissionHistory
(
    CommissionId INT,
    ActionDate DATETIME,
    NewPaymentStatusId INT
)

插入虚拟数据 - 1 个经销商的 5 个佣金：

INSERT INTO dbo.Commission
        ( CommissionId ,
          DealerId
        )
VALUES ( 1 , 1 ),
       ( 2 , 1 ),
       ( 3 , 1 ),
       ( 4 , 1 ),
       ( 5 , 1 ),

INSERT INTO dbo.CommissionHistory
        ( CommissionId ,
          ActionDate ,
          NewPaymentStatusId
        )
VALUES ( 1 , GETDATE() -25, 1 ),
       ( 1 , GETDATE() -21, 2 ),
       ( 1 , GETDATE() -18, 3 ),
       ( 1 , GETDATE() -16, 4 ),
       ( 1 , GETDATE() -5, 5 ),
       ( 2 , GETDATE() -10, 1 ),
       ( 2 , GETDATE() -9, 2 ),
       ( 2 , GETDATE() -8, 3 ),
       ( 2 , GETDATE() -7, 4 ),
       ( 2 , GETDATE() -6, 5 ),
       ( 3 , GETDATE() -10, 1 ),
       ( 3 , GETDATE() -8, 2 ),
       ( 3 , GETDATE() -6, 3 ),
       ( 3 , GETDATE() -4, 4 ),
       ( 3 , GETDATE() -2, 5 ),
       ( 3 , GETDATE() -25, 6 ),
       ( 4 , GETDATE() -10, 1 ),
       ( 4 , GETDATE() -7, 2 ),
       ( 4 , GETDATE() -6, 3 ),
       ( 4 , GETDATE() -4, 4 ),
       ( 4 , GETDATE() -1, 5 ),
       ( 5 , GETDATE() -1, 1 ),
       ( 5 , GETDATE() -1, 2 )

因此，对于虚拟数据，佣金 1、2 和 4 被归类为有效记录，因为它们的状态为 2 和 5。3 被排除在外，因为它已退还，5 被排除在外，因为它没有被支付。

为了生成平均值，我编写了以下查询：

-- set the required dealer id
DECLARE @DealerId INT = 1

-- return all CommissionId's in to a temp table that have statuses 2 and 5, but not 6
SELECT DISTINCT CommissionId
INTO #DealerCommissions
FROM dbo.CommissionHistory t1      
WHERE CommissionId IN (SELECT CommissionId 
                       FROM dbo.Commission 
                       WHERE    DealerId = @DealerId)
AND NOT EXISTS (SELECT CommissionId 
                FROM dbo.CommissionHistory t2 
                WHERE t2.NewPaymentStatusId = 6 AND t2.CommissionId = t1.CommissionId)
AND EXISTS (SELECT CommissionId 
            FROM dbo.CommissionHistory t2 
            WHERE t2.NewPaymentStatusId = 2 AND t2.CommissionId = t1.CommissionId)
AND EXISTS (SELECT CommissionId 
            FROM dbo.CommissionHistory t2 
            WHERE t2.NewPaymentStatusId = 5 AND t2.CommissionId = t1.CommissionId)

-- use the temp table to return average difference between the MIN & MAX date
;WITH cte AS (
    SELECT CommissionId FROM #DealerCommissions
)
SELECT  AVG(CAST(DaysToCompletion AS DECIMAL(10,8)))
FROM    (
         SELECT DATEDIFF(DAY, MIN(ch.ActionDate), MAX(ch.ActionDate)) DaysToCompletion
         FROM cte
         INNER JOIN dbo.CommissionHistory ch ON ch.CommissionId = cte.CommissionId
         GROUP BY ch.CommissionId
) AS averageDays

-- remove temp table
DROP TABLE #DealerCommissions

Answer 2

对于历史表中的每个佣金，您可以获得最大验证日期和最小支付日期，假设支付日期总是晚于验证日期。然后您可以加入佣金表，按经销商 ID 分组，以获取平均持续时间（以天为单位）。

with comm as(
select 
    commissionid,
    max(case NewPamentStatus when 'Verified' then ActionDate else null end) as verified_date,  
    min(case NewPamentStatus when 'Paid' then ActionDate else null end) as paid_date
        --using max or min just incase that same status will be recorded more than one time.
from 
    CommissionHistory
group by 
    commistionid
)
select 
    c.DealerId,
    avg(datediff(day,comm.verified_date,comm.paid_date))
from
    comm
inner join
    commission c
on  c.commissionid = comm.commissionid
where 
    datediff(day,comm.verified_date,comm.paid_date)>0
    -- to get rid off the commissions with paid date before the verified date or in same day
group by
    c.DealerId