Question

How do i return matching entities in a random order?
Just to be clear this is Entity Framework stuff and LINQ to Entities.

(air code)

IEnumerable<MyEntity> results = from en in context.MyEntity                                 where en.type == myTypeVar                                 orderby ?????                                 select en; 

Thanks

Edit:
I tried adding this to the context:

public Guid Random() {     return new Guid(); } 

And using this query:

IEnumerable<MyEntity> results = from en in context.MyEntity                                 where en.type == myTypeVar                                 orderby context.Random()                                 select en; 

But i got this error:

System.NotSupportedException: LINQ to Entities does not recognize the method 'System.Guid Random()' method, and this method cannot be translated into a store expression.. 

Edit (Current code):

IEnumerable<MyEntity> results = (from en in context.MyEntity                                  where en.type == myTypeVar                                  orderby context.Random()                                  select en).AsEnumerable(); 

Answer 1

The simple solution would be creating an array (or a List<T>) and than randomize its indexes.

EDIT:

static IEnumerable<T> Randomize<T>(this IEnumerable<T> source) {   var array = source.ToArray();   // randomize indexes (several approaches are possible)   return array; } 

EDIT: Personally, I find the answer of Jon Skeet is more elegant:

var results = from ... in ... where ... orderby Guid.NewGuid() select ... 

And sure, you can take a random number generator instead of Guid.NewGuid().