c# - 从 Windows 窗体应用程序向 EXE 传递多个参数

Question

我有一个 app.exe 应用程序要求输入 input path 字符串，一旦我输入，它会询问 output path 字符串...现在当我输入时，app .exe 执行一些操作

我需要从我的窗口窗体应用程序传递这些路径 我看到了很多这样的问题，但无法实现我的要求，因为我从未使用过流程和 Stream Reader 或 Writer 请提供任何帮助...将感谢示例..谢谢..

        string input = @"C:\Documents and Settings\pankaj\Desktop\My File\greetingsfreinds.ppt";
        string output = @"C:\Documents and Settings\pankaj\Desktop\test";
        Process process = new Process();
        process.StartInfo.FileName = @"C:\Program Files\Wondershare\MyApp\app.exe";
        process.StartInfo.UseShellExecute = false;
        process.StartInfo.CreateNoWindow = false;
        process.StartInfo.RedirectStandardInput = true;
        process.Start();
        process.WaitForExit(3000);
        process.Close();

---

好的，我试过了 但它给出了一些异常(exception) StandardOut 尚未重定向或进程尚未启动... 我的代码是

        string input = @"C:\Documents and Settings\pankaj\Desktop\My File\greetingsfreinds.ppt";
        string output = @"C:\Documents and Settings\pankaj\Desktop\test";
        Process process = new Process();
        process.StartInfo.FileName = @"C:\Program Files\Wondershare\MyApp\app.exe";
        process.StartInfo.UseShellExecute = false;
        process.StartInfo.CreateNoWindow = false;
        process.StartInfo.RedirectStandardInput = true;
        process.StartInfo.Arguments = input + ";" + output;
        process.Start();
        string Strout = process.StandardOutput.ReadToEnd();
        process.WaitForExit();
        process.Close();

Answer 1

您可以使用 ProcessStartInfo.Arguments为此。

    Process process = new Process()
    process.StartInfo.FileName = @"C:\Program Files\Wondershare\MyApp\app.exe";
    process.StartInfo.UseShellExecute = false;
    ....
    process.Arguments = input + " " + output;