Returning a Dictionary<string, string> from a linq query

Question

I have a table with 2 columns defined as varchar(50): Column1 and Column2. I want to return a dictionary of <string, string> where each row is in the dictionary and where Column1 is the key and Column2 is the value. This is what I have:

public Dictionary<string, string> LoadAllTheDataFromDB()
{
   using (MyDC TheDC = new MyDC())
   {
       return (from c in TheTable
               select new Dictionary<string, string>()
               {
                     //stuck here

               }).FirstOrDefault();

   }
}

How do I make it that the dictionary is filled?

First Select those two columns (new {c.Column1, c.Column2}), then materialize (ToArray), then call ToDictionary. — Evk
– Evk, Commented Nov 19, 2016 at 22:03

CodingYoshi · Accepted Answer · 2018-10-22 19:35:48Z

Try this:

var dict = TheTable.Select( t => new { t.Col1, t.Col2} )
               .ToDictionary( t => t.Col1, t => t);

Remember in select lambda you will perform projection and create some anonymous object. Then in ToDictionary you will pass two parameters: First Parameter is a lambda to specify the key; in code above we are choosing Col1 to be the key. Second parameter is a lambda to specify the value; in code above we are choosing the object itself to be the value.

If you want the value to be an anonymous type, change the 2nd lambda like this:

ToDictionary( t => t.Col1, t => new { t.Col2 });

If you want the value to be a type you have defined, change the 2nd lambda like this:

ToDictionary( t => t.Col1, t => new YourType { Prop1 = t.Col2 });

Please, do not forget to mark the answer you perceive as correct.

Alexei Levenkov · Accepted Answer · 2016-11-19 22:04:26Z

3

Since you just need value of one first row why not to do that first:

 var row = TheTable.FirstOrDefault();

And than just construct that dictionary if you got the result:

 return row == null ? null : 
       new Dictionary<string,string>{ {row.Column1, row.Column2 } };

answered Nov 19, 2016 at 22:04

Alexei Levenkov

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