linq conditional query

Question

What would be the best practice for setting a status depending on several other "columns" retrieved in a linq query.

  var result = (from q in query
                select new Item
                         {
                           ApprovedDate = q.ApprovedDate,
                           CreatedDate = q.CreatedDate,
                           DeclinedDate = q.DeclinedDate,
                           Status = 0
                         });

I'd like to set the status to either 0, 1, 2.

(ApprovedDate == null and DeclinedDate == null) --> 0
(ApprovedDate != null and DeclinedDate == null) --> 1
(DeclinedDate != null) --> 3

So perhaps something like:

  var result = (from q in query
                select new Item
                         {
                           ApprovedDate = q.ApprovedDate,
                           CreatedDate = q.CreatedDate,
                           DeclinedDate = q.DeclinedDate,
                           Status = (q.CreatedDate == null && q.DeclinedDate == null) ? 0 : (q.ApprovedDate != null && q.DeclinedDate == null) ? 1 : 2
                         });

I might add even more status combinations, so should I try and do this in the linq select query, in my repository object.. Or later on in the controller where I would do a .ToList() and then foreach the list to set the correct status code?

Having even more than 3 statuscodes, the linq query gets "hard" to read.

Sergey Berezovskiy · Accepted Answer · 2012-11-12 20:39:45Z

2

What about moving status calculation to Item class? If status property depends on other properties value, then it's definitely calculated property:

var result = from q in query
             select new Item
                         {
                           ApprovedDate = q.ApprovedDate,
                           CreatedDate = q.CreatedDate,
                           DeclinedDate = q.DeclinedDate
                         });

And

public class Item
{
  // other properties

  public int Status
  {
      get
      {
          if (ApprovedDate == null and DeclinedDate == null)
              return 0;
          if (ApprovedDate != null and DeclinedDate == null)
              return 1;
          if (DeclinedDate != null)
              return 3;
          // etc
      }
  }
}

Actually I think it's best option, because in this case status calculation logic will be close to required data. If (for some reason) you can't use this approach, then move setting statuses to local items collection:

var items = result.ToList().ForEach(i => i.Status = CalculateStatus(i));

edited Nov 12, 2012 at 20:39

answered Nov 12, 2012 at 20:33

Sergey Berezovskiy

237k44 gold badges441 silver badges468 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Kman Over a year ago

Thank you. I agree 100% with you. This is of course the most obvious and best solution. Thanks again..!! Now why didn't that cross my mind 5 minutes ago :)

Jorge · Accepted Answer · 2012-11-12 20:33:45Z

Maybe wrapped all in a function An do a linq like this

var result = (from q in query sele q).AsEnumerable()
                                       .Select( x => new Item()
                                       {
                                           ApprovedDate = x.ApprovedDate,
                                           CreatedDate = x.CreatedDate,
                                           DeclinedDate = x.DeclinedDate,
                                           Status = MyStatusFunction(x.CreatedDate,q.DeclinedDate)
                                       });


public int MyStatusFunction(DateTime ApprovedDate , Datetime DeclinedDate)
{
    if (ApprovedDate == null and DeclinedDate == null) return 0;
    else if(ApprovedDate != null and DeclinedDate == null) return 1;
    else if (DeclinedDate != null) return 3;
}

Collectives™ on Stack Overflow

linq conditional query

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related