Question

我正在尝试实施借记 - 贷方解决系统，但我很难根据集合来表达逻辑。

假设我有一个Orders表：

Id    OrderId     Amount     AdjustmentFlag
 1       1           10.00        0
 2       1           10.00        1
 3       1           10.00        2
 4       2           20.00        1
 5       2           20.00        2
 6       2           20.00        2
 7       3           30.00        1
 8       4           40.00        0
 9       4           40.00        0
10       4           40.00        1
11       5           50.00        0
12       5           50.00        1
13       5           60.00        2
14       5           60.00        1
15       5           60.00        2
16       5           70.00        1

我需要根据他们是否有匹配的“已取消”标记来挑选仍然有效的Id。

0 - Original Order
1 - Cancelled Order
2 - Adjusted Order

1与0或2匹配，优先于0。
如果1标志不匹配，则会被忽略。

鉴于以上例子：

Id 2将匹配Id 1，即Id 3。
Id 4将匹配Id 5或Id 6，但不能同时匹配。
Id 7将被忽略。
Id 10将匹配Id 8或Id 9，但不能同时匹配。
Id 12将与Id 11匹配。
Id 14将匹配Id 13或Id 15，但不能同时匹配。
Id 16将被忽略。

可能的结果是[1,2,4,5,7,8,10,11,12,13,14,16]（较低的Id有偏好）或[1,2,4,6,7 ，9,10,11,12,14,15,16]（更高的Id有偏好）。只要结果是确定性的，任何一个都可以工作。

创建脚本：

CREATE TABLE [Order]
(
     Id INT IDENTITY NOT NULL PRIMARY KEY
    ,OrderId INT NOT NULL
    ,Amount MONEY NOT NULL
    ,AdjustmentFlag TINYINT NOT NULL
);

INSERT INTO [Order](OrderId, Amount, AdjustmentFlag)
SELECT 1, 10.00, 0
UNION ALL
SELECT 1, 10.00, 1
UNION ALL
SELECT 1, 10.00, 2
UNION ALL
SELECT 2, 20.00, 1
UNION ALL
SELECT 2, 20.00, 2
UNION ALL
SELECT 2, 20.00, 2
UNION ALL
SELECT 3, 30.00, 1
UNION ALL
SELECT 4, 40.00, 0
UNION ALL
SELECT 4, 40.00, 0
UNION ALL
SELECT 4, 40.00, 1
UNION ALL
SELECT 5, 50.00, 0
UNION ALL
SELECT 5, 50.00, 1
UNION ALL
SELECT 5, 60.00, 2
UNION ALL
SELECT 5, 60.00, 1
UNION ALL
SELECT 5, 60.00, 2
UNION ALL
SELECT 5, 70.00, 1

这是我目前的部分解决方案：

WITH Orders AS
(
    SELECT
        Id,
        OrderId,
        Amount,
        AdjustmentFlag,
        EffectiveOrder = ROW_NUMBER() OVER (PARTITION BY OrderId, Amount ORDER BY AdjustmentFlag DESC),
        UnmatchedOrder = CASE WHEN EXISTS(SELECT 1 FROM [Order] uo WHERE uo.OrderId = o.OrderId GROUP BY uo.OrderId HAVING(COUNT(uo.OrderId) = 1)) THEN 1 ELSE 0 END,
        OriginalWithoutAdjustment = CASE WHEN EXISTS(SELECT 1 FROM [Order] uo WHERE uo.OrderId = o.OrderId AND uo.Amount = o.Amount GROUP BY uo.OrderId, uo.Amount HAVING (MAX(uo.AdjustmentFlag) = 1)) THEN 1 ELSE 0 END,
        AdjustmentWithoutOriginal = CASE WHEN EXISTS(SELECT 1 FROM [Order] uo WHERE uo.OrderId = o.OrderId AND uo.Amount = o.Amount GROUP BY uo.OrderId, uo.Amount HAVING (MIN(uo.AdjustmentFlag) = 1)) THEN 1 ELSE 0 END
    FROM [Order] o
)
,MatchedOrders AS
(
    SELECT
        Id
    FROM Orders
    WHERE
    -- Assume AdjustmentFlag = 2 and take everything else
    EffectiveOrder <> 1
    OR
    (
        -- Assume AdjustmentFlag = 2 and there is no Order with AdjustmentFlag = 0
        -- Take everything since the MIN AdjustmentFlag = 1
        AdjustmentWithoutOriginal = 1
        AND EffectiveOrder > 1
    )
    OR
    (
        -- Assume AdjustmentFlag = 1 and there are no other Orders, so ignore it
        AdjustmentFlag = 1
        AND UnmatchedOrder = 1
    )
    OR
    (
        -- We don't care about the orders if they don't have any Amount
        Amount = 0
        AND EffectiveOrder = 1
    )
    AND NOT
    (
        -- We have an Original without any other Orders
        EffectiveOrder = 1
        AND UnmatchedOrder = 1
        AND AdjustmentFlag = 0
    )
)
SELECT
    o.OrderId,
    o.AdjustmentFlag,
    o.Amount,
    o.EffectiveOrder,
    o.UnmatchedOrder,
    Excluded = CASE WHEN mo.Id IS NULL THEN 0 ELSE 1 END
FROM Orders o
LEFT OUTER JOIN MatchedOrders mo
ON o.Id = mo.Id
ORDER BY OrderId, Amount, AdjustmentFlag

结果：

Result

Answer 1

尝试：

with cte as
(select o.*, 
        case AdjustmentFlag when 1 then -1 else 1 end DrCr,
        row_number() over (partition by OrderId, Amount, case AdjustmentFlag when 1 then 1 end
                           order by AdjustmentFlag, Id) Rn
 from [Order] o)
select OrderId,
       max(case DrCr when 1 then Id end) DrId,
       sum(case DrCr when 1 then Amount else 0 end) DrAmount,
       max(case DrCr when 1 then AdjustmentFlag end) DrAdjustmentFlag,
       max(case DrCr when -1 then Id end) CrId,
       sum(case DrCr when -1 then Amount else 0 end) CrAmount,
       max(case DrCr when -1 then AdjustmentFlag end) CrAdjustmentFlag,
       sum(DrCr * Amount) BalanceAmount
from cte
group by OrderId, Amount, Rn
having sum(DrCr * Amount) >= 0 /* excludes unmatched cancelled orders */

- 如果您只想查看不匹配的原始/修订订单，请将having子句条件更改为> 0。

SQLFiddle here。

在SQL查询中实现借记信用解决系统

1 个答案: