Entity Framework Code First Date field creation

Question

I am using Entity Framework Code First method to create my database table. The following code creates a DATETIME column in the database, but I want to create a DATE column.

[DataType(DataType.Date)]
[DisplayFormatAttribute(ApplyFormatInEditMode = true, DataFormatString = "{0:d}")]
public DateTime ReportDate { get; set; }

How can I create a column of type DATE, during table creation?

Answer 1

Try to use ColumnAttribute from System.ComponentModel.DataAnnotations (defined in EntityFramework.dll):

[Column(TypeName="Date")]
public DateTime ReportDate { get; set; }

Answer 2

I use following

    [DataType(DataType.Time)]
    public TimeSpan StartTime { get; set; }

    [DataType(DataType.Time)]
    public TimeSpan EndTime { get; set; }

    [DataType(DataType.Date)]
    [Column(TypeName = "Date")]
    public DateTime StartDate { get; set; }

    [DataType(DataType.Date)]
    [Column(TypeName = "Date")]
    public DateTime EndDate { get; set; }

With Entity Framework 6 & SQL Server Express 2012 - 11.0.2100.60 (X64). It works perfectly and generates time/date column types in sql server

Answer 3

the Best Way it using The

[DataType(DataType.Date)]
public DateTime ReportDate { get; set; }

but you must using the EntityFramework v 6.1.1

Answer 4

The EF6 version of David Roth's answer is as follows:

public class DataTypePropertyAttributeConvention 
    : PrimitivePropertyAttributeConfigurationConvention<DataTypeAttribute>
{
    public override void Apply(ConventionPrimitivePropertyConfiguration configuration, 
        DataTypeAttribute attribute)
    {
        if (attribute.DataType == DataType.Date)
        {
            configuration.HasColumnType("Date");
        }
    }
}

Register this as before:

protected override void OnModelCreating(DbModelBuilder modelBuilder)
{
     base.OnModelCreating(modelBuilder);

     modelBuilder.Conventions.Add(new DataTypePropertyAttributeConvention());
}

This has the same outcome as Tyler Durden's approach, except that it's using an EF base class for the job.

Answer 5

If you prefer not to decorate your classes with attributes, you can set this up in the DbContext's OnModelCreating like this:

public class DatabaseContext: DbContext
{
    // DbSet's

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        base.OnModelCreating(modelBuilder);

        // magic starts
        modelBuilder.Entity<YourEntity>()
                    .Property(e => e.ReportDate)
                    .HasColumnType("date");
        // magic ends

        // ... other bindings
    }
}

Answer 6

I found this works in EF6 nicely.

I created a convention for specifying my data types. This convention changes the default DateTime data type in the database creation from datetime to datetime2. It then applies a more specific rule to any properties that I have decorated with the DataType(DataType.Date) attribute.

public class DateConvention : Convention
{
    public DateConvention()
    {
        this.Properties<DateTime>()
            .Configure(c => c.HasColumnType("datetime2").HasPrecision(3));

        this.Properties<DateTime>()
            .Where(x => x.GetCustomAttributes(false).OfType<DataTypeAttribute>()
            .Any(a => a.DataType == DataType.Date))
            .Configure(c => c.HasColumnType("date"));
    }
}

Then register then convention in your context:

protected override void OnModelCreating(DbModelBuilder modelBuilder)
{
    modelBuilder.Conventions.Remove<PluralizingTableNameConvention>();
    modelBuilder.Conventions.Add(new DateConvention());
    // Additional configuration....
}

Add the attribute to any DateTime properties that you wish to be date only:

public class Participant : EntityBase
{
    public int ID { get; set; }

    [Required]
    [Display(Name = "Given Name")]
    public string GivenName { get; set; }

    [Required]
    [Display(Name = "Surname")]
    public string Surname { get; set; }

    [DataType(DataType.Date)]
    [Display(Name = "Date of Birth")]
    public DateTime DateOfBirth { get; set; }
}