SqlException:当IDENTITY_INSERT设置为OFF时,无法在表'AspNetUsers'中为identity列插入显式值

时间:2018-05-15 05:57:25

标签: c# asp.net asp.net-mvc asp.net-identity

我已使用Microsoft.AspNet.Identity的enable-migration将aspnetuser表列数据类型字符串更改为long。

使用以下迁移脚本成功创建表:

CreateTable(
"dbo.AspNetUsers",
c => new
{
Id = c.Long(nullable:false,identity:true),
Email = c.String(maxLength: 256),
EmailConfirmed = c.Boolean(nullable: false),
PasswordHash = c.String(),
SecurityStamp = c.String(),
PhoneNumber = c.String(),
PhoneNumberConfirmed = c.Boolean(nullable: false),
TwoFactorEnabled = c.Boolean(nullable: false),
LockoutEndDateUtc = c.DateTime(),
LockoutEnabled = c.Boolean(nullable: false),
AccessFailedCount = c.Int(nullable: false),
UserName = c.String(nullable: false, maxLength: 256),
FullName = c.String(nullable: false, maxLength: 50),
CreatedBy = c.Long(nullable: true),
CreatedDate = c.DateTime(),
ModifiedDate = c.DateTime(nullable: false, defaultValue: null, defaultValueSql: null),
ModifiedBy = c.Long(nullable: true),
IsDeleted = c.Boolean(nullable: false, defaultValue: false, defaultValueSql: "0")
})
.PrimaryKey(t => t.Id)
.Index(t => t.UserName, unique: true, name: "UserNameIndex");

但是当我尝试使用以下代码创建用户时获取错误为“SqlException:当IDENTITY_INSERT设置为OFF时,无法在表'AspNetUsers'中为标识列插入显式值”。

// first we create Admin rool
var role = new Microsoft.AspNet.Identity.EntityFramework.IdentityRole();
role.Name = SuperAdminRole;
roleManager.Create(role);

//Here we create a Admin super user who will maintain the website               

var user = new ApplicationUser();
user.UserName = "Admin";
user.Email = "xyz@gmail.com";
user.FullName = "xyz";
user.IsDeleted = false;
user.ModifiedBy = null;
user.ModifiedDate = null;
user.CreatedBy = string.Empty;
user.CreatedDate = DateTime.Now;

string userPWD = "admin_123";
var chkUser = UserManager.Create(user, userPWD);

如果有人为此提供解决方案,请分享。

感谢您的快速回复!

1 个答案:

答案 0 :(得分:0)

我相信IDENTITY_INSERT是自动增量功能,它设置为OFF。因此,验证字段'Id'是sql数据库中的标识列,并在代码第一实体框架的'Id'列中添加以下属性

[DatabaseGeneratedAttribute(DatabaseGeneratedOption.Identity)]
public int Id { get; set; }
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