C# winforms如何在不同的事件处理程序中访问同一对象

Question

我正在尝试创建一个 WinForm 来读取选定的文件并将其显示到文本框中。我的问题是让其他事件处理程序访问我的对象。我基本上希望在单击选择后将文件名显示在文本框中，但我希望在单击打开按钮后将文件内容显示在单独的文本框中。当我将对象放入选择按钮内时，会显示文件名，但当我尝试将其放入打开按钮时，我无法再次访问该内容。你可以看到我注释掉的所有我尝试过的东西

public partial class xmlForm : Form
{
    OpenFileDialog openFileDialog1 = new OpenFileDialog();

    public xmlForm()
    {
        InitializeComponent();
    }

    public void btnSelect_Click(object sender, System.EventArgs e)
    {
        // Displays an OpenFileDialog so the user can select a Cursor.
        // OpenFileDialog openFileDialog1 = new OpenFileDialog();
        openFileDialog1.Filter = "XML files|*.xml";
        openFileDialog1.Title = "Select a XML File";

        // Show the Dialog.  
        // If the user clicked OK in the dialog and  
        // a .xml file was selected, open it.  
        if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
        {
            displayBox.Text = openFileDialog1.FileName;
            var onlyFileName = System.IO.Path.GetFileName(openFileDialog1.FileName);
            displayBox.Text = onlyFileName;

            /* Button btn = sender as Button;
             if (btn != null)
             {
                 if (btn == opnButton)
                 {
                     string s = System.IO.File.ReadAllText(openFileDialog1.FileName);
                     fileBox.Text = s;
                 } 
             }*/
            /* if (opnButtonWasClicked)
             {
                 string s = System.IO.File.ReadAllText(openFileDialog1.FileName);
                 fileBox.Text = s;
                 opnButtonWasClicked = false;
             } */
        }

        /*string s = System.IO.File.ReadAllText(openFileDialog1.FileName);
        fileBox.Text = s; */
    }

    public void opnButton_Click(object sender, EventArgs e)
    {
        string s = System.IO.File.ReadAllText(openFileDialog1.FileName);
        fileBox.Text = s;

        /*if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
        {
            Button btn = sender as Button;
            if (btn != null)
            {
                if (btn == opnButton)
                {
                    string s = System.IO.File.ReadAllText(openFileDialog1.FileName);
                    fileBox.Text = s;
                }
                else { }
            }
        }*/
    }
}

Answer 1

由于您已经在 btnSelect_Click 处理程序中打开了 FileDialog；因此，在关闭对话框之前，您无法打开已打开的对话框。因此，为了再次打开该对话框，您必须先将其关闭。然后您可以使用以下语句:

string s=System.IO.File.ReadAllText(openFileDialog1.FileName);
fileBox.Text = s;

但就您而言，为了达到目的，您不需要在关闭后再次打开对话框。因此，只需将 displayBox.Text 作为 opnButton_Click 处理程序内 ReadAllText 方法的参数传递，因为文本字段已包含文件名。

string System.IO.File.ReadAllText(displayBox.Text);
fileBox.Text = s;

谢谢