Entity Framework 核心 2.1:无法在实体类型 'BarCodeDevice' 上为 属性 'Model' 找到指定的字段 'Model'

Entity Framework Core 2.1 : The specified field 'Model' could not be found for property 'Model' on entity type 'BarCodeDevice'

我需要使用 Code First 方法通过 Entity Framework Core 2.1 生成数据库,但出现此错误:

The specified field 'Model' could not be found for property 'Model' on entity type 'BarCodeDevice'.

这是我曾经这样做过的 classes

public class BarCodeDevice
    {
        public int SerialNumber { get; set; }
        public string Model { get; set; }
        public virtual ICollection<ClientBarCodeDevice> ClientBarCodeDeviceList { get; set; }
    }

和配置class

public class BarCodeDeviceConfiguration : IEntityTypeConfiguration<BarCodeDevice>
    {
        public void Configure(EntityTypeBuilder<BarCodeDevice> builder)
        {
            builder.HasKey(x => x.SerialNumber);
            builder.Property(t => t.Model)
              .IsRequired()
              .HasField("Model");
        }
    }

和 DbContext Class

public class SegregationDbContext : DbContext, IDisposable
    {
        public SegregationDbContext(DbContextOptions<SegregationDbContext> options) : base(options)
        { }


        protected override void OnModelCreating(ModelBuilder modelBuilder)
        {                        
            modelBuilder.ApplyConfiguration(new BarCodeDeviceConfiguration());            
        }

        public DbSet<BarCodeDevice> BarCodeDevices { get; set; }
    }

最后是配置

public void ConfigureServices(IServiceCollection services)
        {
            services.AddDbContext<SegregationDbContext>(options => options.UseSqlServer(Configuration.GetConnectionString("Default")));
            services.AddMvc();
        }

问题是这个流畅的配置行:

.HasField("Model")

HasField用于指定backing field for the property being configured, when the backing field name does not conform to the conventions.

但是你 Model 属性 是自动 属性 并且没有名为 Model 的支持字段,因此例外。

所以要么删除该行,例如

builder.Property(t => t.Model)
    .IsRequired();

或者如果您想强制使用名称未知的支持字段(自动属性就是这种情况),请改用 UsePropertyAccessMode 方法,例如

builder.Property(t => t.Model)
    .IsRequired()
    .UsePropertyAccessMode(PropertyAccessMode.Field);