打开记事本并添加文本不起作用 [英] Open Notepad and add Text not working

问题描述

我刚刚从互联网上看到了此片段，但对我来说却无效。
应该打开一个新的记事本应用程序，并在其中添加 asdf。

I just saw this snippet from the internet but it doesn't work from me. It's suppose to open a new notepad application and add "asdf" into it.

代码是否有错误？

[DllImport("User32.dll")]     
        public static extern int SendMessage(IntPtr hWnd, int uMsg, int wParam, [MarshalAs(UnmanagedType.LPStr)] string lParam);

void Test()
{
         const int WM_SETTEXT = 0x000C;

    ProcessStartInfo startInfo = new ProcessStartInfo("notepad.exe");
    startInfo.UseShellExecute = false;
    Process notepad = System.Diagnostics.Process.Start(startInfo);
    SendMessage(notepad.MainWindowHandle, WM_SETTEXT, 0, "asdf");
}

推荐答案

要确保该过程是准备接受输入，请调用 notepad.WaitForInputIdle（）。而且重要的是使用刚刚创建的进程的 MainWindowHandle 而不是任何记事本进程。

To make sure that process is ready to accept input, call notepad.WaitForInputIdle(). And it is important to use MainWindowHandle of the process that was just created, not the any notepad process.

[DllImport("user32.dll")]
public static extern IntPtr FindWindowEx(IntPtr hwndParent, IntPtr hwndChildAfter, string lpszClass, string lpszWindow);

[DllImport("User32.dll")]
public static extern int SendMessage(IntPtr hWnd, int uMsg, int wParam, string lParam);

static void ExportToNotepad(string text)
{
    ProcessStartInfo startInfo = new ProcessStartInfo("notepad");
    startInfo.UseShellExecute = false;

    Process notepad = Process.Start(startInfo);
    notepad.WaitForInputIdle();

    IntPtr child = FindWindowEx(notepad.MainWindowHandle, new IntPtr(0), null, null);
    SendMessage(child, 0x000c, 0, text);
}

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