如何在兩個單獨的類之間進行通信？

Question

我創建了單獨的類稱爲連接，我已經回到了SqlConnection這樣的：

class connection 
{ 
    public SqlConnection Functions() 
    { 
     string connetionString = null; 
     SqlConnection cnn; 
     Boolean flag = false; 
     connetionString = "Data Source=(LocalDB)\\v11.0;AttachDbFilename=|DataDirectory|\\Database1.mdf;Integrated Security=True"; 
     cnn = new SqlConnection(connetionString); 

     return cnn; 
    } 

} 

而且我想使用另一種單獨的類返回的連接，但它似乎錯誤，我已經編碼如下

using System; 
using System.Collections.Generic; 
using System.ComponentModel; 
using System.Data; 
using System.Drawing; 
using System.Linq; 
using System.Text; 
using System.Threading.Tasks; 
using System.Windows.Forms; 
using System.Data.SqlClient; 
using WindowsFormsApplication1.deepak; 


namespace WindowsFormsApplication1 
{ 
    public partial class Form1 : Form 
    { 
     public Form1() 
     { 
      InitializeComponent(); 
     } 

     private void Form1_Load(object sender, EventArgs e) 
     { 

     } 

     private void button1_Click(object sender, EventArgs e) 
     { 

      SqlConnection cnn ; 

      connection c = new connection(); 
      cnn = new connection.Function(); 
      try 
      { 
       cnn.Open(); 
       MessageBox.Show ("Connection Open ! "); 
       cnn.Close(); 
      } 
      catch (Exception ex) 
      { 
       MessageBox.Show("Can not open connection ! "); 
      } 
     } 
    } 
} 

Answer 1

我覺得你找cnn = c.Functions();，而不是cnn = new connection.Function();
我也建議使用standard C# naming conventions，並在你的類名的開頭使用大寫字母和給你的方法吶我會描述它的目的，如GetConnection()。

Answer 2

你connection類（我更喜歡Connection代替connection），應該從外部訪問，所以我把它修正至public：

public class Connection 
{ 
    public SqlConnection Function() 
    { 
     var connetionString = /*your connection string*/ 
     return new SqlConnection(connetionString); 
    } 
} 

現在，你可以用它如下：

var conn = new Connection().Function(); //<--- you forgot the parentheses 
try 
{ 
    conn.Open(); 
    MessageBox.Show ("Connection Open ! "); 
    conn.Close(); 
} 
catch (Exception ex) 
{ 
    MessageBox.Show("Can not open connection ! "); 
}