Question

我想查找并删除满足依赖关系的行，但前提是所有兄弟节点都满足该依赖关系。孩子也可以有多个父母，这使得这种毛茸茸但是这里有完整的解释：

我正在开发一个遗留系统，其数据结构非常糟糕，我无从事任何选择。以下是问题的简化，但我认为这是解释问题的最佳方式。

关于我正在做的事情：

在数据库中，我有货运记录A - G

我想删除所有已完成的货件。出货A - E已完成，但不是F或G.

系统中还有几个发票。

规则：

如果已完成的货件是包含未的其他货件的发票的一部分，则无法将其删除
如果已完成的货件是包含其他无法删除的货件的发票的一部分，则无法将其删除

问题：
- 发票I1包含货件A，B和C.由于这些都已完成，我可以全部删除
- 发票I2包含货件A，F和G.作为F＆amp; G 不已完成，我无法删除

...然后我们需要检查任何其他发票B＆amp; C是......的一部分。

...等等到无限......

...对于我的生活，我想不出一个简单，快速的SQL语句或算法，而不是通过强力矿石递归 - 直到结束

感激地收到任何其他建议！

Answer 1

我想我有个主意。假设您从描述中得到类似这三个表的内容（下面的SQL Server语法）：

CREATE TABLE Invoice
(
   Id INT PRIMARY KEY
)

CREATE TABLE Shipment
(
   Id INT PRIMARY KEY,
   Completed BIT
)

CREATE TABLE Invoice_Shipment
(
   InvoiceId INT FOREIGN KEY REFERENCES Invoice(Id),
   ShipmentId INT FOREIGN KEY REFERENCES Shipment(Id)
)

然后你可以：

删除＆＃34;不必要的＆＃34;来自Invoice_Shipment表
的连接
从Invoice_Shipment删除 WHERE InvoiceId不在（从Invoice_Shipment中选择InvoiceId 在哪里ShipmentId IN （SELECT Id FROM Shipment WHERE Completed = 0））
删除＆＃34;未使用＆＃34;来自Shipment表
的货件
从货件中删除 WHERE完成= 1并且不在（从Invoice_Shipment中选择InvoiceId）

重新编写此内容以满足您的需求，并且不要忘记包装交易。

Answer 2

我尝试过这样的事情 - 这是导致我头疼的递归：

TRUNC Shipments_To_Delete

-- Grab all the 'Complete' Shipments
INSERT INTO Shipments_To_Delete
  ( SELECT * FROM Shipment WHERE Complete )

-- Make sure we iterate at least once
Shipments_To_Remove = True

WHILE Shipments_To_Remove

  -- Test if there's anything left to do..
  Shipments_To_Remove = 
    (
      -- Count if there are any Shipments marked for deletion..
      --   ..which can't be because they're part of an Invoice..
      --   ..which contains *other* Shipments that *can't* be deleted
      --
      SELECT COUNT(1) FROM Shipments_To_Delete WHERE Shipment_ID IN

        -- Find all the Shipments on Invoices that have Shipments that can't be deleted
        ( SELECT Shipment_ID FROM Invoice_Shipments WHERE Invoice_ID IN 

          -- Find all the Invoices that have Shipments which can't be deleted
          ( SELECT Invoice_ID FROM Invoice_Shipments WHERE Shipment_ID NOT IN
            ( SELECT Shipment_ID FROM Shipments_To_Delete )
          )

        )
     ) > 0

  IF Shipments_To_Remove THEN

    -- Reuse the query above to make it a DELETE
    --
    DELETE FROM Shipments_To_Delete WHERE Shipment_ID IN
      ( SELECT Shipment_ID FROM Invoice_Shipments WHERE Invoice_ID IN 
        ( SELECT Invoice_ID FROM Invoice_Shipments WHERE Shipment_ID NOT IN
          ( SELECT Shipment_ID FROM Shipments_To_Delete )
        )
      )

  ENDIF

-- If we've changed anything, loop around again and check whether there's anything left to do
DO

解决多个多次排除的算法

2 个答案: