openfiledialogde 的showdialog方法

Question

private openFileDialog=new OpenFileDialog();
OpenFileDialogForm1() 
                { 
                        InitializeComponent(); 
                        this.openFile.Click   +=   new   EventHandler(openFile_Click); 
                } 
                private   void   openFile_Click(object   sender,System.EventArgs   e) 
                { 
                        openFileDialog.ShowDialog(); 
                } 

Answer 1

  public   Form1() 
  { 
     InitializeComponent(); 
     this.openFile.Click   +=   new   EventHandler(openFile_Click);
     OpenFileDialog mODlg ;
   } 
   private   void   openFile_Click(object   sender,System.EventArgs   e) 
   { 
       this.mODlg = new OpenFileDialog();
       DialogResult dResult = this.mODlg.ShowDialog();
       if (dResult == DialogResult.OK)
           {
             //Todo somthing     
                
           }
   } 

Answer 2

先实例化，再ShowDialog()OpenFileDialog dlgFile = new OpenFileDialog();
dlgFile.ShowDialog();如果要得到返回值，可以这样
OpenFileDialog dlgFile = new OpenFileDialog();
if (dlgFile.ShowDialog() == DialogResult.OK)
{
    MessageBox.Show(dlgFile.FileName);
}