首先转到所需的表名

步骤2 ->右击表名

选择设计

单击列名

步骤5)进入列属性，然后将No设置为Identity Specification

[注意:插入到显式值后，如果你想，你可以恢复到标识规范为true，然后它会再次生成值]

如果你使用SQL server管理studio你可以使用下面的方法

步骤1) 步骤2)

This occurs when you have a (Primary key) column that is not set to Is Identity to true in SQL and you don't pass explicit value thereof during insert. It will take the first row, then you wont be able to insert the second row, the error will pop up. This can be corrected by adding this line of code [DatabaseGenerated(DatabaseGeneratedOption.Identity)] in your PrimaryKey column and make sure its set to a data type int. If the column is the primary key and is set to IsIDentity to true in SQL there is no need for this line of code [DatabaseGenerated(DatabaseGeneratedOption.Identity)] this also occurs when u have a column that is not the primary key, in SQL that is set to Is Identity to true, and in your EF you did not add this line of code [DatabaseGenerated(DatabaseGeneratedOption.Identity)]

请注意，如果您用;结束每一行，SET IDENTITY_INSERT mytable ON命令将不适用于以下行。

即。 像这样的查询

SET IDENTITY_INSERT mytable ON;
INSERT INTO mytable (VoucherID, name) VALUES (1, 'Cole');

给出错误 当IDENTITY_INSERT设置为OFF时，无法为表'mytable'中的身份列插入显式值。

但是像这样的查询是可以的:

SET IDENTITY_INSERT mytable ON
INSERT INTO mytable (VoucherID, name) VALUES (1, 'Cole')
SET IDENTITY_INSERT mytable OFF;

SET IDENTITY_INSERT命令似乎只适用于事务，而;将表示事务的结束。

如果您使用Oracle SQL Developer进行连接，请记住添加/sqldev:stmt/

/sqldev:stmt/ set identity_insert TABLE on

每个人都评论SQL，但是在EntityFramework中发生了什么?我看了整篇文章，没人能解出EF。所以几天后，我找到了一个解决方案: 在EF Core的上下文中创建模型有这样一条指令:modelBuilder。实体<客户端>(实体=>{实体。属性(e => e.d id).ValueGeneratedNever();

这也会产生错误，解决方案:你必须通过ValueGeneratedOnAdd()改变它的工作!

要非常小心地将IDENTITY_INSERT设置为ON。这是一种糟糕的实践，除非数据库处于维护模式并设置为单用户。这不仅会影响您的插入，还会影响试图访问表的其他人的插入。

为什么要在单位字段中输入一个值?