Question

如何将数据插入3个相关表（SQL Server）

例如，我有表Customer＆lt; - Customer_Address - ＆gt;地址将数据插入Customer和Address后，如何将Customer和Address中的ID插入Customer_Address？（加入表）

谢谢！

Answer 1

使用SCOPE_IDENTITY，@@ IDENTITY可以从任何范围返回值：

DECLARE @CustomerId INT
DECLARE @AddressId INT

BEGIN TRANSACTION

  INSERT INTO CUSTOMER (blah, blah) values (blah, blah)
  SET @CustomerId = SCOPE_IDENTITY

  INSERT INTO ADDRESS (blah, blah) values (blah, blah)
  SET @AddressId = SCOPE_IDENTITY 

  INSERT INTO CUSTOMERADDRESS (CustomerId,AddressId) values (@CustomerId,@AddressId)

COMMIT TRANSACTION

Answer 2

如果要插入多行，可以使用输出子句：

declare @insertedAddresses table (OriginalID int, AddressID int);
declare @insertedCustomers table (OriginalID int, CustomerID int);

insert into dbo.Addresses (AddressData)
output source.OriginalID, inserted.AddressID into @insertedAddresses
select AddressData from source;

insert into dbo.Customers (CustomerData)
output source.OriginalID, inserted.CustomerID into @insertedCustomers
select CustomerData from source;

insert into dbo.Customer_Address (AddressID, CustomerID)
select a.AddressID, c.CustomerID
from @insertedAddresses a inner join @insertedCustomers c on c.OriginalID=a.OriginalID;

Answer 3

如果Customer和Address表中的ID是Identity列，则可以将新ID存储到变量中。

DECLARE @CustomerID int

SELECT @CustomerID = @@IDENTITY FROM TABLE CUSTOMER

类似的语法可用于Address表。然后在INSERT语句中，您可以执行此操作：

INSERT INTO Customer_Address (CustomerID, AddressID)
VALUES (@CustomerID, @AddressID)

Answer 4

使用交易，并记住身份。在sql server中 -

declare @CustomerId int
declare @AddressId int

begin tran
    insert into Customer (blah, blah) values (blah, blah)
    set @CustomerId = @@IDENTITY    --assuming there are no triggers
    insert into [Address] (blah, blah) values (blah, blah)
    set @AddressId = @@IDENTITY     --once again, no triggers to mess up the @@IDENTITY
    insert into CustomerAddress(CustomerId,AddressId) values (@CustomerId,@AddressId)
commit

Answer 5

这是一种方式。

Declare CustomerID int, AddressID int

insert into Customer (list, of, fields) values (list, of, values)
select @CustomerID=scope_Identity()
insert into Address (list, of, fields) values (list, of, values)
select @AddressID =scope_Identity()

insert into Customer_Address (CustomerID, AddressID) values (@CustomerID, @AddressID)

Answer 6

如果是应用程序，为什么不使用Linq，你可以在几分钟内以更加压缩的方式完成这样的事情，并且更加大胆地使用业务逻辑模型。

如果不是，我强烈建议使用事务和SCOPE_IDENTITY版本，因为如果您处于负载较重的系统中，则其他操作可以执行插入并更改标识值，因此您将以不一致的数据结束。

将数据插入3个相关表中

6 个答案: