Display a specific result using foreach in Dictionary C# - c#-4.0

Question

Problem that I'm unable to solve is that when I search a contact, if found than I need that contact to be the only output but it's displaying whole record. What should I do?
here is my code
namespace Telephone_Directory
{
class Program
{
public bool info()
{
Console.WriteLine("ENTER THE NAME TO BE SEARCHED");
Dictionary<string, uint> contact = new Dictionary<string, uint>();
contact.Add("usman", 03453648729);
contact.Add("Iqtiqa", 03159825052);
contact.Add("Aamir", 03343315412);
contact.Add("Ghous", 03323142783);
var items = from pair in contact
orderby pair.Value ascending
select pair;
string chk = Console.ReadLine();
if (contact.ContainsKey(chk))
{
foreach (KeyValuePair<string, uint> pair in contact)
{
Console.WriteLine("{0}: {1}", pair.Key, pair.Value);
} return true;
}
else
return false;
}
static void Main(string[] args)
{
Program ob =new Program();
bool a=ob.info();
if (a == true)
{
Console.WriteLine("Your contact is found successfully");
}
else
Console.WriteLine("Not found");
Console.ReadLine();
}
}
}

Answer 1

You are checking if the Contact is available but you are iterating over the whole collection as an output. So you can completely remove the first query (var item = ...) because it does not have any effect because you are not using it anywhere.
I would propose accessing everything directly by the Dictionary Key like this
uint srchNumber;
contact.TryGetValue(chk, out srchNumber);
if (srchNumber != null)
Console.WriteLine("{0}: {1}", chk, srchNumber);
this will get you what you want.
But I would suggest to change the implementation because a Dictionary may be not optimal if you will have multiple items with the same name because it only allows you to have every key once. Instead I would suggest to use something like SortedList> or List<...>. Then you may query all possible results with a LINQ-Statement like
var results = from c in contact select c where c.key == chk
this will get you all Key-Value-Pairs belonging to a certain name.
But it is a matter of what you are trying to achieve.