c# – OpenFileDialog没有显示
作者:互联网
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;
}
}
当我运行程序窗体时不显示并退出调试模式.
In output view writes:The program ‘[4244] openfiledialog.vshost.exe: Managed (v4.0.30319)’ has exited with code 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();
解决方法:
作为一般规则,我在调用它的事件中初始化并使用我的OpenFileDialog.我想不出一种情况,我希望它成为我窗口的属性.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.
我在你的错误代码上找到的唯一一件事就是这个(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.
标签:c,visual-studio-2010,winforms,openfiledialog 来源: https://codeday.me/bug/20190708/1405296.html