How to configure AutoFixture to use an enum value as seed when creating many of a certain type?

StackOverflow https://stackoverflow.com/questions/17111765

  •  31-05-2022
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Question

I have the following types:

public enum Status
{
    Online,
    Offline
}

public class User
{
    private readonly Status _status;
    public User(Status status) { _status = status; }
    public Status Status {get {return _status; }}
    public string Name {get;set;}
}

Now, when executing fixture.CreateMany<User> I want AutoFixture to return two Users, one per status. All other properties - like Name - should be filled with anonymous data.

Question:
How to configure AutoFixture to do this?


I tried the following this:

  1. Register collection that news up the User object:

    fixture.Register(
        () => Enum.GetValues(typeof(Status)).Cast<Status>().Select(s => 
            new User(s)));
    

    The problem with this approach is that AutoFixture doesn't fill the other properties like Name

  2. Customize User to use a factory and register a collection that uses fixture.Create:

        f.Customize<User>(c => c.FromFactory((Status s) => new User(s)));
        f.Register(() =>
            Enum.GetValues(typeof(Status))
                .Cast<Status>()
                .Select(s => (User)f.Create(new SeededRequest(typeof(User), s),
                                            new SpecimenContext(f))));
    

    That didn't work either. The seed isn't being used.

Was it helpful?

Solution

You could do this:

var users = new Fixture().Create<Generator<User>>();

var onlineUser = users.Where(u => u.Status == Status.Online).First();
var offlineUser = users.Where(u => u.Status == Status.Offline).First();

If you're using AutoFixture.Xunit, the declarative equivalent is:

[Theory, AutoData]
public void CreateOneOfEachDeclaratively(Generator<User> users)
{
    var onlineUser = users.Where(u => u.Status == Status.Online).First();
    var offlineUser = users.Where(u => u.Status == Status.Offline).First();

    // Use onlineUser and offlineUser here...
}

OTHER TIPS

You may declare and use a customization, e.g. StatusGenerator:

var fixture = new Fixture();
fixture.RepeatCount = 2;
fixture.Customizations.Add(new StatusGenerator());

var result = fixture.CreateMany<User>();

A hypothetical implementation of the StatusGenerator could be the following:

internal class StatusGenerator : ISpecimenBuilder
{
    private readonly Status[] values;
    private int i;

    internal StatusGenerator()
    {
        this.values =
            Enum.GetValues(typeof(Status)).Cast<Status>().ToArray();
    }

    public object Create(object request, ISpecimenContext context)
    {
        var pi = request as ParameterInfo;
        if (pi == null || !pi.ParameterType.IsEnum)
            return new NoSpecimen(request);

        return this.values[i == this.values.Length - 1 ? i = 0 : ++i];
    }
}

Based on Mark's answer, this is what I am using now:

fixture.Customize<User>(c => c.Without(x => x.Status));
fixture.Customize<IEnumerable<User>>(
    c =>
    c.FromFactory(
        () => Enum.GetValues(typeof(Status)).Cast<Status>()
                  .Select(s => users.First(u => u.Status == s))));

fixture.Create<IEnumerable<User>>(); // returns two Users

I know it is already answered and the Generator was a very interesting finding. I think there is a much simpler approach for this problem.

        var numberOfEnumValues = Enum.GetValues(typeof(Status)).Length;
        var users = fixture.CreateMany<User>(numberOfEnumValues);

In case the constructor is more complicated, with multiple Status values, or the model has property setters of Status type. Then you generally have a problem, and the generator might blow as well.

Say that:

    public class SuperUser : User
    {
        public SuperUser(Status status, Status shownStatus): base(status)
        {
        }
    }

Then this will never be evaluated:

    var users = fixture.Create<Generator<SuperUser>>();
    var offlineUser = users.Where(u => u.Status == Status.Offline).First();
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