OpenFileDialog 不显示

Question

我有这个简单的代码：

private void buttonOpen_Click(object sender, EventArgs e)
{
    if (openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        textBox2.Text = openFileDialog1.FileName;
    }
}

当我运行程序时，窗体不显示并退出调试模式。

在输出视图中写道：程序 '[4244] openfiledialog.vshost.exe: Managed (v4.0.30319)' 已退出，代码为 1073741855 (0x4000001f)。

我有 Visual Studio 2010 专业版。

编辑：form1.designer.cs

     private void InitializeComponent()
    {
        this.openFileDialog1 = new System.Windows.Forms.OpenFileDialog();
        this.buttonOpen = new System.Windows.Forms.Button();
        this.textBox1 = new System.Windows.Forms.TextBox();
        this.textBox2 = new System.Windows.Forms.TextBox();
        this.SuspendLayout();
        // 
        // openFileDialog1
        // 
        this.openFileDialog1.FileName = "openFileDialog1";
        // 
        // buttonOpen
        // 
        this.buttonOpen.Location = new System.Drawing.Point(13, 48);
        this.buttonOpen.Name = "buttonOpen";
        this.buttonOpen.Size = new System.Drawing.Size(75, 23);
        this.buttonOpen.TabIndex = 0;
        this.buttonOpen.Text = "open";
        this.buttonOpen.UseVisualStyleBackColor = true;
        this.buttonOpen.Click += new System.EventHandler(this.buttonOpen_Click);
        // 
        // textBox1
        // 
        this.textBox1.Location = new System.Drawing.Point(113, 50);
        this.textBox1.Name = "textBox1";
        this.textBox1.Size = new System.Drawing.Size(279, 20);
        this.textBox1.TabIndex = 1;
        // 
        // textBox2
        // 
        this.textBox2.Location = new System.Drawing.Point(13, 98);
        this.textBox2.Name = "textBox2";
        this.textBox2.Size = new System.Drawing.Size(385, 20);
        this.textBox2.TabIndex = 2;
        // 
        // Form1
        // 
        this.AutoScaleDimensions = new System.Drawing.SizeF(6F, 13F);
        this.AutoScaleMode = System.Windows.Forms.AutoScaleMode.Font;
        this.ClientSize = new System.Drawing.Size(445, 216);
        this.Controls.Add(this.textBox2);
        this.Controls.Add(this.textBox1);
        this.Controls.Add(this.buttonOpen);
        this.Name = "Form1";
        this.Text = "Form1";
        this.ResumeLayout(false);
        this.PerformLayout();

Answer 1

作为一般规则，我在调用 OpenFileDialog 的事件中初始化并使用它。我想不出在什么情况下我会希望它成为我的窗口的属性。我要做的第一件事是将其作为属性删除并在您的事件中初始化它。

private void buttonOpen_Click(object sender, EventArgs e)
{
    using (OpenFileDialog openFileDialog1 = new OpenFileDialog())
    {
        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            textBox2.Text = openFileDialog1.FileName;
        }
    }
}

您无需将 FileName 属性设置为任何内容，因为对话框会为您完成此操作。

我在你的错误代码中发现的唯一的东西是这个（程序和调试器退出而没有问题的指示）。在您当前的代码中我找不到任何会导致此问题的内容。如果您正在访问非托管代码，您可能需要启用非托管代码调试。