Question

    private void browsebtn_Click(object sender, EventArgs e)
    {
        OpenFileDialog fdlg = new OpenFileDialog();
        fdlg.Title = "Select Song";
        fdlg.InitialDirectory = @"c:\";
        fdlg.Filter = "All files (*.*)|*.*|All files (*.*)|*.*";
        fdlg.FilterIndex = 2;
        fdlg.RestoreDirectory = true;
        if (fdlg.ShowDialog() == DialogResult.OK)
        {
            filedialoglbl.Text = fdlg.FileName;
        }
    }

    private void runbtn_Click(object sender, EventArgs e)
    {
        var Path = fdlg.FileName;
        var pi = new System.Diagnostics.ProcessStartInfo(Path)
        {
            Arguments = Path.GetFileName(Path),
            UseShellExecute = true,
            WorkingDirectory = Path.GetDirectoryName(Path),
            FileName = "C:\\relax.exe",
            Verb = "OPEN"
        };
        System.Diagnostics.Process.Start(pi);
    }
}

如您所见，我正在尝试选择一个文件并使用该文件运行另一个程序，但我不想使用Windows默认值。

我的错误在

var Path = fdlg.FileName;

问题是fdlg在同一上下文中不是，我需要使用之前选择的目录，有关如何执行此操作的任何想法？

根据我尝试过的答案：

    private OpenFileDialog fdlg = new OpenFileDialog();
    private void browsebtn_Click(object sender, EventArgs e)
    {
        fdlg.Title = "Select Song";
        fdlg.InitialDirectory = @"c:\";
        fdlg.Filter = "All files (*.*)|*.*|All files (*.*)|*.*";
        fdlg.FilterIndex = 2;
        fdlg.RestoreDirectory = true;
        if (fdlg.ShowDialog() == DialogResult.OK)
        {
            filedialoglbl.Text = fdlg.FileName;
        }
    }

    private void runbtn_Click(object sender, EventArgs e)
    {
        var Path = fdlg.Text;
        var pi = new System.Diagnostics.ProcessStartInfo(Path)
        {
            Arguments = Path.GetFileName(Path),
            UseShellExecute = true,
            WorkingDirectory = Path.GetDirectoryName(Path),
            FileName = "C:\\relax.exe",
            Verb = "OPEN"
        };
        System.Diagnostics.Process.Start(pi);
    }

，我收到一个错误消息，说“ OpenFileDialog”没有针对“ Text”的定义或扩展方法，并且在类中包含以下行：

private OpenFileDialog fdlg = new OpenFileDialog();

我收到一个新错误，说“ GetFileName”和“ GetDirectoryName”在“字符串”中没有定义或扩展方法。

编辑

我重命名了目录的路径，但存在相同的错误

        private void runbtn_Click(object sender, EventArgs e)
    {
        var directory =filedialoglbl.Text;
        var pi = new System.Diagnostics.ProcessStartInfo(directory)
        {
            Arguments = directory.GetFileName(directory),
            UseShellExecute = true,
            WorkingDirectory = directory.GetDirectoryName(directory),
            FileName = "C:\\relax.exe",
            Verb = "OPEN"
        };
        System.Diagnostics.Process.Start(pi);
    }

Answer 1

您仅在FileName中使用runbtn_Click方法中的对话框。您将其保存到名为filedialoglbl.Text的标签中。使用它。

private void browsebtn_Click(object sender, EventArgs e)
{
    OpenFileDialog fdlg = new OpenFileDialog();
    fdlg.Title = "Select Song";
    fdlg.InitialDirectory = @"c:\";
    fdlg.Filter = "All files (*.*)|*.*|All files (*.*)|*.*";
    fdlg.FilterIndex = 2;
    fdlg.RestoreDirectory = true;
    if (fdlg.ShowDialog() == DialogResult.OK)
    {
        filedialoglbl.Text = fdlg.FileName;
    }
}

private void runbtn_Click(object sender, EventArgs e)
{
    var path = filedialoglbl.Text;
    var pi = new System.Diagnostics.ProcessStartInfo(path)
    {
        Arguments = Path.GetFileName(path),
        UseShellExecute = true,
        WorkingDirectory = Path.GetDirectoryName(path),
        FileName = "C:\\relax.exe",
        Verb = "OPEN"
    };
    System.Diagnostics.Process.Start(pi);
}

修改

您不能将Path大写使用，因为您正在破坏Path类，并且不能使用它们的方法。

Answer 2

将方法中的OpenFileDialog fdlg声明为类成员（字段），例如，private OpenFileDialog fdlg = new OpenFileDialog();

args.gn

如何在不同的上下文中使用变量？

2 个答案: