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;
    }
}

当我运行程序表单时，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 Professional。

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，我在调用它的事件中初始化并使用我的OpenFileDialog。我无法想到一种情况，我希望它成为我窗户的财产。我要做的第一件事就是将其删除为属性并在事件中初始化它。

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

您不需要将FileName属性设置为任何内容，因为对话框将为您执行此操作。

我在您的错误代码中找到的唯一内容就是这个（Program and debugger quit without indication of problem）。在您当前的代码中，我找不到任何可能导致此问题的内容。如果要访问非托管代码，则可能需要启用非托管代码调试。

OpenFileDialog没有显示

1 个答案: