设置Windows对话框输入

Question

I'm trying to write a C# code that handles file uploading automaticly. The follow I need to implement is to choose a file from an open file dialog:

• The Dialog I need to control

I managed to find the window using users32.dll FindWindow() method. But i have no idea how to set the input if the dialog and approve the chosen file (choose a file & press OK).

My Code so far:

[DllImport("user32.dll")]
public static extern int FindWindow(string lpClassName, string lpWindowName);

public const int WM_SYSCOMMAND = 0x0112;
public const int SC_CLOSE = 0xF060;
public const int WM_SETTEXT = 0x000C;

private void ChooseFile()
{
    // retrieve the handler of the window  
    int iHandle = FindWindow("#32770", "File Upload");
    if (iHandle > 0)
    {
        //Choose File
        //Press OK
    }
}

Any help will be much appreciated.

Answer 1

您有C＃类OpenFileDialog（http://www.dotnetperls.com/openfiledialog），对于user32.dll来说不是必需的。

Answer 2

What you are trying to do do is a bit odd. You are calling Win32 function, but all you need is to use OpenFileDialog class, which is proper .NET way here ( MSDN OpenFileDialog )

OpenFileDialog dlg = new OpenFileDialog();
DialogResult res = dlg.ShowDialog();
if (res == DialogResult.OK)
{
    string filePath = dlg.FileName;
    // do your upload logic here
}

Answer 3

After allot of research i found the solution.

I used Windodows.Form.SendKeys class which simulates the keyboard and send the string to the focused window.

here is the code:

 SendKeys.SendWait(fileInfo.FullName);
 Thread.Sleep(2000);
 SendKeys.SendWait("{ENTER}");
 Thread.Sleep(5000);