I try to get random record from database:
personToCall = db.Persons.Skip(toSkip).Take(1).First();
but I get exception which tells me:
{"The method 'Skip' is only supported for sorted input in LINQ to Entities. The method 'OrderBy' must be called before the method 'Skip'."}
Can I do it without OrderBy? Sorting data structure (O(nlogn)) to pick random element(which should be constant) doesn't look wise.
EDIT: I use Entity Framework 6.1.1.
You can have something like :
personToCall = db.Persons.OrderBy(r => Guid.NewGuid()).Skip(toSkip).Take(1).First();
You should use FirstOrDefault to be mode defensive.
Here the dark lord teaching the force to yoda! what is the world coming to!
First you need to get the random number from 1 to max record, see this
Random rand = new Random();
int toSkip = rand.Next(0, db.Persons.Count());
db.Persons.Skip(toSkip).Take(1).First();
with order by you can use the Guid.NewGuid()
db.Persons.OrderBy(x=>x.Guid.NewGuid()).Skip(toSkip).Take(1).FirstOrDefault();
Count on the second line of the example code.There is no way to do this without an ordering clause.
personToCall = db.Persons.OrderBy(r => Random.Next()).First();
That could be slow depending on the size of your Persons table, so if you wanted to make this fast, you'd have to add a column to Person, or join it to a dictionary of random numbers and Person keys, then order by that. But that is rarely a wise solution.
Better to ask a higher level question about the overall task at hand.
To avoid the OrderBy, dump to a List and random pick against the Index:
VB
With New List(Of Persons)
.AddRange(db.Persons)
PersonToCall = .Item(New Random().Next(0, .Count - 1))
End With
C#
var people = new List<Persons>();
people.AddRange(db.Persons);
personToCall = people.Item(new Random().Next(0, people.Count - 1));
Identityinteger primary keys?public class Person { public int Id { get; set; } etc....}and Entity Framework created table basing on this code.