How to fix the datetime2 out-of-range conversion error using DbContext and SetInitializer?

You have to ensure that Start is greater than or equal to SqlDateTime.MinValue (January 1, 1753) - by default Start equals DateTime.MinValue (January 1, 0001).

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Simple. On your code first, set the type of DateTime to DateTime?. So you can work with nullable DateTime type in database. Entity example:

public class Alarme
    {
        [Key]
        [DatabaseGenerated(DatabaseGeneratedOption.Identity)]
        public int Id { get; set; }

        public DateTime? DataDisparado { get; set; }//.This allow you to work with nullable datetime in database.
        public DateTime? DataResolvido { get; set; }//.This allow you to work with nullable datetime in database.
        public long Latencia { get; set; }

        public bool Resolvido { get; set; }

        public int SensorId { get; set; }
        [ForeignKey("SensorId")]
        public virtual Sensor Sensor { get; set; }
    }

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In some cases, DateTime.MinValue (or equivalenly, default(DateTime)) is used to indicate an unknown value.

This simple extension method can help handle such situations:

public static class DbDateHelper
{
    /// <summary>
    /// Replaces any date before 01.01.1753 with a Nullable of 
    /// DateTime with a value of null.
    /// </summary>
    /// <param name="date">Date to check</param>
    /// <returns>Input date if valid in the DB, or Null if date is 
    /// too early to be DB compatible.</returns>
    public static DateTime? ToNullIfTooEarlyForDb(this DateTime date)
    {
        return (date >= (DateTime) SqlDateTime.MinValue) ? date : (DateTime?)null;
    }
}

Usage:

 DateTime? dateToPassOnToDb = tooEarlyDate.ToNullIfTooEarlyForDb();

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You can make the field nullable, if that suits your specific modeling concerns. A null date won't be coerced to a date that isn't within the range of the SQL DateTime type the way a default value would. Another option is to explicitly map to a different type, perhaps with,

.HasColumnType("datetime2")

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Even though this question is quite old and there are great answers already, I thought I should put one more which explains 3 different approaches to solve this problem.

1st Approach

Explicitly map DateTime property public virtual DateTime Start { get; set; } to datetime2 in corresponding column in the table. Because by default EF will map it to datetime.

This can be done by fluent API or data annotation.

Fluent API In DbContext class overide OnModelCreating and configure property Start (for explanation reasons it's a property of EntityClass class). protected override void OnModelCreating(DbModelBuilder modelBuilder) { //Configure only one property modelBuilder.Entity() .Property(e => e.Start) .HasColumnType("datetime2"); //or configure all DateTime Preperties globally(EF 6 and Above) modelBuilder.Properties() .Configure(c => c.HasColumnType("datetime2")); } Data annotation [Column(TypeName="datetime2")] public virtual DateTime Start { get; set; }

2nd Approach

Initialize Start to a default value in EntityClass constructor.This is good as if for some reason the value of Start is not set before saving the entity into the database start will always have a default value. Make sure default value is greater than or equal to SqlDateTime.MinValue ( from January 1, 1753 to December 31, 9999)

public class EntityClass
{
    public EntityClass()
    {
        Start= DateTime.Now;
    }
    public DateTime Start{ get; set; }
}

3rd Approach

Make Start to be of type nullable DateTime -note ? after DateTime-

public virtual DateTime? Start { get; set; }

For more explanation read this post

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If your DateTime properties are nullable in the database then be sure to use DateTime? for the associated object properties or EF will pass in DateTime.MinValue for unassigned values which is outside of the range of what the SQL datetime type can handle.

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My solution was to switch all datetime columns to datetime2, and use datetime2 for any new columns. In other words make EF use datetime2 by default. Add this to the OnModelCreating method on your context:

modelBuilder.Properties<DateTime>().Configure(c => c.HasColumnType("datetime2"));

That will get all the DateTime and DateTime? properties on all your entities.

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initialize the Start property in the constructor

Start = DateTime.Now;

This worked for me when I was trying to add few new fields to the ASP .Net Identity Framework's Users table (AspNetUsers) using Code First. I updated the Class - ApplicationUser in IdentityModels.cs and I added a field lastLogin of type DateTime.

public class ApplicationUser : IdentityUser
    {
        public ApplicationUser()
        {
            CreatedOn = DateTime.Now;
            LastPassUpdate = DateTime.Now;
            LastLogin = DateTime.Now;
        }
        public String FirstName { get; set; }
        public String MiddleName { get; set; }
        public String LastName { get; set; }
        public String EmailId { get; set; }
        public String ContactNo { get; set; }
        public String HintQuestion { get; set; }
        public String HintAnswer { get; set; }
        public Boolean IsUserActive { get; set; }

        //Auditing Fields
        public DateTime CreatedOn { get; set; }
        public DateTime LastPassUpdate { get; set; }
        public DateTime LastLogin { get; set; }
    }

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Based on user @andygjp's answer, its better if you override the base Db.SaveChanges() method and add a function to override any date which does not fall between SqlDateTime.MinValue and SqlDateTime.MaxValue.

Here is the sample code

public class MyDb : DbContext
{
    public override int SaveChanges()
    {
        UpdateDates();
        return base.SaveChanges();
    }

    private void UpdateDates()
    {
        foreach (var change in ChangeTracker.Entries().Where(x => (x.State == EntityState.Added || x.State == EntityState.Modified)))
        {
            var values = change.CurrentValues;
            foreach (var name in values.PropertyNames)
            {
                var value = values[name];
                if (value is DateTime)
                {
                    var date = (DateTime)value;
                    if (date < SqlDateTime.MinValue.Value)
                    {
                        values[name] = SqlDateTime.MinValue.Value;
                    }
                    else if (date > SqlDateTime.MaxValue.Value)
                    {
                        values[name] = SqlDateTime.MaxValue.Value;
                    }
                }
            }
        }
    }
}

Taken from the user @sky-dev's comment on https://stackoverflow.com/a/11297294/9158120

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I had the same issue and in my case I was setting the date to new DateTime() instead of DateTime.Now

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In my case this happened when I used entity and the sql table has default value of datetime == getdate(). so what I did to set a value to this field.

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I'm using Database First and when this error happened to me my solution was to force ProviderManifestToken="2005" in edmx file (making the models compatible with SQL Server 2005). Don't know if something similar is possible for Code First.

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One line fixes this:

modelBuilder.Properties<DateTime>().Configure(c => c.HasColumnType("datetime2"));

So, in my code, I added:

protected override void OnModelCreating(DbModelBuilder modelBuilder)
{
    modelBuilder.Properties<DateTime>().Configure(c => c.HasColumnType("datetime2"));
}

Adding that one line to the DBContext subclass override void OnModelCreating section should work.

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In my case, after some refactoring in EF6, my tests were failing with the same error message as the original poster but my solution had nothing to do with the DateTime fields.

I was just missing a required field when creating the entity. Once I added the missing field, the error went away. My entity does have two DateTime? fields but they weren't the problem.

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In case anyone is as dopey as me, double check the year of your date. I was converting a date from a text file in YYMMDD format so was creating a date with a year of 0020, not 2020. Obvious error but I spent more time looking at it but not seeing it than I should have!

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