Linq - Distinct list of object fill in int array

Question

I am trying to do, get randomly one of "list of objects" from all lists. I am getting NullReferenceException I also tried List couldn't make it work.

List<BL.Test.Test> Tests = BL.Test.GET.TestGroup(CategoryId);

// NullReferenceException on the line below:
int[] Groups = Tests.Select(d => d.TestGroupId).Distinct().ToArray(); 

Session["TestGroup"] = Tests.Select(t => t.TestGroupId = Groups[rnd.Next(Groups.Length)]);

Tests not null

Can you please provide us the "Test" class and the return value of the "TestGroup" method? — Kevin Brechbühl
– Kevin Brechbühl, Commented Dec 25, 2013 at 16:39
Please check what Tests.Select(d => d.TestGroupId) returns. — user176134
– user176134, Commented Dec 25, 2013 at 16:56
The property get accessor for TestGroupId throws an exception. This is an example why it is confusing when get accessors throw; they should not. — Jeppe Stig Nielsen
– Jeppe Stig Nielsen, Commented Dec 25, 2013 at 16:58

astef · Accepted Answer · 2013-12-25 16:57:47Z

2

Obviously, BL.Test.GET.TestGroup is the method which returns null.

That's the most probable explanation for a NullReferenceException in a second line of your example.

And if Select, Distinct and ToArray are extension methods declared in System.Linq, this reason is the only possible, so check your method.

UPD.

Sorry guys, I am wrong.

TestGroupId member of BL.Test.Test class is missed.

UPD-2

This is a good example of community debugging question. As I know it is not appreciated here

edited Dec 25, 2013 at 16:57

answered Dec 25, 2013 at 16:44

astef

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1 Comment

astef Over a year ago

How did you checked that?

Ankush Madankar · Accepted Answer · 2013-12-26 06:50:47Z

0

Since TestGroupId would be null hence null.Distinct() throws NullArgumentReference exception. Change ur code with following code:

List<BL.Test.Test> Tests = BL.Test.GET.TestGroup(CategoryId);

int[] Groups = Tests.Where(d=>d.TestGroupId.HasValue).Select(d => d.TestGroupId).Distinct().ToArray(); 

Session["TestGroup"] = Tests.Select(t => t.TestGroupId = Groups[rnd.Next(Groups.Length)]);

Make use of HasValue to find if TestGroupId has some value.

answered Dec 26, 2013 at 6:50

Ankush Madankar

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