c# - Must declare the table variable @table

Question

I'm a beginner in C# and SQL, i have this SQL insert statement that i want to perform. It asks for the table name among the other variables that i want to insert.

But when i run this console app i get this error :

Must declare the table variable @table

This is a part of the code :

StreamReader my_reader =  getFile(args);
string CS = formCS();
try
{
    using (SqlConnection con = new SqlConnection(CS))
    {
        SqlCommand com = new SqlCommand("insert into @table (time, date, pin) values (@time, @date, @pin)", con);                    
        con.Open();
        Console.WriteLine("Enter table name:");
        Console.Write(">> ");
        string tblname = Console.ReadLine();
        com.Parameters.AddWithValue("@table", tblname);

        string line = "";
        int count = 0;
        while ((line = my_reader.ReadLine()) != null)
        {
            Dictionary<string, string> result = extractData(line);                        
            com.Parameters.AddWithValue("@time", result["regTime"]);
            com.Parameters.AddWithValue("@date", result["regDate"]);
            com.Parameters.AddWithValue("@pin", result["regPin"]);
            count += com.ExecuteNonQuery();
            com.Parameters.Clear();                        

        }
        Console.WriteLine("Recoreds added : {0}", count.ToString());
        Console.WriteLine("Press Enter to exit.");
    }
    Console.ReadLine();
}
catch (SqlException ex)
{
    Console.WriteLine(ex.Message);
}
catch (Exception ex)
{
    Console.WriteLine(ex.Message);                
}

Answer 1

You can't do this. You can't pass the table name as a parameter the way you did:

SqlCommand com = new SqlCommand("insert into @table ...");
...
com.Parameters.AddWithValue("@table", tblname);

You can do this instead:

Console.WriteLine("Enter table name:");
Console.Write(">> ");
string tblname = Console.ReadLine();

string sql = String.Format("insert into {0} (time, date, pin) values ... ", tblname);

SqlCommand com = new SqlCommand(sql, con);                    

...