如何在C＃中将查询保存到字典中？

Question

I am using Visual Studio 2012 and C#. I have a problem: I want to collect all results of a SQL query into dictionary.

This is my code:

Dictionary<int,List<string>> dic = new Dictionary<int, List<string>>();

string query = "select request_number, service_category ,service_type from enugro.enugro_service_requests_info;";

MySqlConnection connec = new MySqlConnection(strcon);
connec.Open();

MySqlCommand cmd = new MySqlCommand(query,connec);
MySqlDataReader reader = cmd.ExecuteReader();

while(reader.Read())
{
    dic.Add((int)reader["request_number"], {               reader["service_category"], reader["service_type"] });
}

// display result
foreach(var disp in dic)
{
    MessageBox.Show(disp.Key.ToString()+"= "+disp.Value.ToString());
}

As you can see my SQL query returns 3 columns to be retrieved and to store into dictionary. Could you help me?

Answer 1

Since the type of the dictionary values is List<string> , then you need to create a new list for each row and add it to the dictionary like this:

dic
  .Add(
      (int)reader["request_number"],
      new List<string>
      {
          reader["service_category"].ToString(),
          reader["service_type"].ToString()
      });

Consider creating a class to hold the information that you want from each row and use it as the type for the dictionary values. Such class would look something like this:

public class MyValue
{
    public string ServiceCategory {get;set;}
    public string ServiceType {get;set;}
}

Then you can have your dictionary be of type Dictionary<int, MyValue> which will allow you to do this:

dic
  .Add(
      (int)reader["request_number"],
      new MyValue
      {
          ServiceCategory = reader["service_category"].ToString(),
          ServiceType = reader["service_type"].ToString()
      });

Answer 2

Why would you do that? You can create DTO (data transfer object) with properties according to column names in database, and bind query result to it. Then you can return list of DTO objects. It would be better solution, then complicating with dictionary.