C# C语言中的OpenFileDialog#

如何从打开的文件对话框中获得结果（即文件名及其位置）

我的代码：

private void selectFileButton_Click( object sender, EventArgs e ) {
    var selectedFile = selectFileDialog.ShowDialog();
    //label name = fileName
    fileName.Text = //the result from selectedFileDialog
}

---

OpenFileDialog类具有该类的FileName属性

通常，您希望确保用户没有取消该对话框：

using (var selectFileDialog = new OpenFileDialog()) {
  if (selectFileDialog.ShowDialog() == DialogResult.OK) {
    fileName.Text = selectFileDialog.FileName;
  }
}

---

您可以尝试使用此代码

if(selectFileDialog.ShowDialog() == DialogResult.OK)
{
        var result = selectFileDialog.FileName; 

        if((myStream = selectFileDialog.OpenFile())!= null)
        {
            // Insert code to read the stream here.
            ..........
            myStream.Close();
        }
    }

---

出现了什么问题？我不知道如何从对话框中获取所选文件的名称和位置您是否查看了

FileDialog

对象的属性？

if(selectFileDialog.ShowDialog() == DialogResult.OK)
{
        var result = selectFileDialog.FileName; 

        if((myStream = selectFileDialog.OpenFile())!= null)
        {
            // Insert code to read the stream here.
            ..........
            myStream.Close();
        }
    }

private void selectFileButton_Click( object sender, EventArgs e ) 
{
    Stream fileStream = null;
    //Update - remove parenthesis
    if (selectFileDialog.ShowDialog() == DialogResult.OK && (fileStream = selectFileDialog.OpenFile()) != null)
    {
        string fileName = selectFileDialog.FileName;
        using (fileStream)
        {
           // TODO
        }
    }
}