Question

我在CMD中创建了一个运行命令。

var process = Process.Start("CMD.exe", "/c apktool d app.apk");
process.WaitForExit();

如何在不显示实际CMD窗口的情况下运行此命令？

Answer 1

您可以将WindowsStyle-Property用于indicate whether the process is started in a window that is maximized, minimized, normal (neither maximized nor minimized), or not visible

process.StartInfo.WindowStyle = ProcessWindowStyle.Hidden

<强>来源：物业：MSDN Enumartion：MSDN

并将代码更改为此，因为您在初始化对象时启动了该过程，因此将无法识别属性（在启动过程后设置的属性）。

Process proc = new Process();
proc.StartInfo.FileName = "CMD.exe";
proc.StartInfo.Arguments = "/c apktool d app.apk";
proc.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
proc.Start();
proc.WaitForExit();

Answer 2

正如各种评论和答案中所指出的，您的计划存在一些问题。我试着在这里解决所有问题。

ProcessStartInfo psi = new ProcessStartInfo();
psi.FileName = "apktool";

//join the arguments with a space, this allows you to set "app.apk" to a variable
psi.Arguments = String.Join(" ", "d", "app.apk");

//leave it to the application, not the OS to launch the file
psi.UseShellExecute = false;

//choose to not create a window
psi.CreateNoWindow = true;

//set the window's style to 'hidden'
psi.WindowStyle = ProcessWindowStyle.Hidden;

var proc = new Process();
proc.StartInfo = psi;
proc.Start();
proc.WaitForExit();

主要问题：

在没有必要时使用cmd /c
启动应用程序而不设置隐藏它的属性

Answer 3

试试这个：

     proc.StartInfo.CreateNoWindow = true;
     proc.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
     proc.WaitForExit();

Answer 4

ProcessStartInfo startInfo = new ProcessStartInfo();
startInfo.CreateNoWindow = true;
startInfo.UseShellExecute = false;
startInfo.FileName = "dcm2jpg.exe";
startInfo.WindowStyle = ProcessWindowStyle.Hidden;
startInfo.Arguments = "-f j -o \"" + ex1 + "\" -z 1.0 -s y " + ex2;

运行CMD命令而不显示它？

4 个答案: