C＃-否定访问路径-openFileDialog

Question

我是开发新手，我制作了一个简单的应用程序来尝试打开文件，该应用程序返回一个错误：“对路径的访问无效”，但我是管理员，UAC被禁用，并且我是唯一的帐户，为什么它给了我这个错误吗？

这是我的代码：

private void button1_Click(object sender, EventArgs e)
    {
        Stream myStream = null;
        OpenFileDialog openFileDialog1 = new OpenFileDialog();

        openFileDialog1.InitialDirectory = "c:\\";
        openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
        openFileDialog1.FilterIndex = 2;
        openFileDialog1.RestoreDirectory = true;

        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            try
            {

                if ((myStream = openFileDialog1.OpenFile()) != null)
                {
                    using (myStream)
                    {
                        // Insert code to read the stream here.
                        string sourceCode = File.ReadAllText(Path.GetDirectoryName(openFileDialog1.FileName));
                        string colorizedSourceCode = new CodeColorizer().Colorize(sourceCode, Languages.Java);

                        textBox1.Text = colorizedSourceCode;
                    }
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
            }
        }
    }

Answer 1

在我看来，您正在尝试打开文件夹中的流。

尝试更换它

string sourceCode = File.ReadAllText(Path.GetDirectoryName(openFileDialog1.FileName))

有了这个：

string sourceCode = File.ReadAllText(openFileDialog1.FileName)

Answer 2

您打开流错误。

也许尝试这样的事情？

private void button1_Click(object sender, EventArgs e)
    {            
            Stream myStream = null;
            OpenFileDialog openFileDialog1 = new OpenFileDialog();

            openFileDialog1.InitialDirectory = "c:\\";
            openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
            openFileDialog1.FilterIndex = 2;
            openFileDialog1.RestoreDirectory = true;

            if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                try
                {
                    if ((myStream = openFileDialog1.OpenFile()) != null)
                    {
                        using (myStream)
                        {
                            string strfilename = openFileDialog1.FileName;
                            string filetext = File.ReadAllText(strfilename);
                            textBox1.Text = filetext;
                            button1.PerformClick();
                        }
                    }
                }
                catch (Exception ex)
                {
                    MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
                }
            }