Reusing OpenFileDialog

I have 2 textboxes and 2 button [...] next to each textbox. Is it possible to use one OpenFileDialog and pass the FilePath to the respective textbox, based on which button is clicked? i.e...if I click buttton one and laod the dialog, when 开发者_开发百科I click open on the dialog, it passes the fileName to the first textbox.

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Whenever you think "there's common functionality!" you should consider a method to implement it. It could look like this:

    private void openFile(TextBox box) {
        if (openFileDialog1.ShowDialog(this) == DialogResult.OK) {
            box.Text = openFileDialog1.FileName;
            box.Focus();
        }
        else {
            box.Text = "";
        }
    }

    private void button1_Click(object sender, EventArgs e) {
        openFile(textBox1);
    }

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There are several ways to do this. One is to have a Dictionary<Button, TextBox> that holds the link between a button and its related textbox, and use that in the click event for the button (both buttons can be hooked up to the same event handler):

public partial class TheForm : Form
{
    private Dictionary<Button, TextBox> _buttonToTextBox = new Dictionary<Button, TextBox>();
    public Form1()
    {
        InitializeComponent();
        _buttonToTextBox.Add(button1, textBox1);
        _buttonToTextBox.Add(button2, textBox2);
    }

    private void Button_Click(object sender, EventArgs e)
    {
        OpenFileDialog ofd = new OpenFileDialog();
        if (ofd.ShowDialog() == DialogResult.OK)
        {
            _buttonToTextBox[sender as Button].Text = ofd.FileName;
        }
    }
}

Of course, the above code should be decorated with null-checks, nice encapsulation of behavior and so on, but you get the idea.

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This worked for me (and it's simpler than the other posts, but either of them would work as well)

private void button1_Click(object sender, EventArgs e)
{
    openFileDialog1.ShowDialog();
    textBox1.Text = openFileDialog1.FileName;
}

private void button2_Click(object sender, EventArgs e)
{
    openFileDialog1.ShowDialog();
    textBox2.Text = openFileDialog1.FileName;
}

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Yes it is, basically you need to keep a reference to the button that was clicked, and then a mapping of a textbox to each button:

public class MyClass
{
  public Button ClickedButtonState { get; set; }
  public Dictionary<Button, TextBox> ButtonMapping { get; set; }

  public MyClass
  {
    // setup textbox/button mapping.
  } 

   void button1_click(object sender, MouseEventArgs e)
   {
     ClickedButtonState = (Button)sender;
     openDialog();
   }

   void openDialog()
   {
     TextBox current = buttonMapping[ClickedButtonState];
     // Open dialog here with current button and textbox context.
   }
}