由网友(獣潮:ORgASM.)分享简介：ProcessStartInfo procStartInfo = new ProcessStartInfo("cmd", "/c " +"processNeedToRun"){RedirectStandardError = true,RedirectStandardOutput = true,UseShellExec...

ProcessStartInfo procStartInfo = new ProcessStartInfo("cmd", "/c " +"processNeedToRun")
{
    RedirectStandardError = true,
    RedirectStandardOutput = true,
    UseShellExecute = false,
    CreateNoWindow = true,
    Verb ="runas"
};

我用上面的code通过cmd，在C＃中运行一个进程。

I use the above code to run a process through cmd in C#.

不过，问题是：

在processNeedToRun运行时需要的参数。 即使我设置动词=运行方式时，Windows 7仍然提示了迅速提升对话框。

是否有可能满足所有要求？

Is it possible to meet all the requirements?

推荐答案

如果是这样的启动进程 processNeedToRun 不升高，再有就是没有办法避免在提升对话框。这样做将是一个安全漏洞。所以，你只是将不得不忍受的提升提示。

If the process that's launching processNeedToRun is not elevated, then there is no way to avoid the elevation dialog. Doing so would be a security hole. So you're just going to have to live with the elevation prompt.

添加参数 processNeedToRun 是没有问题的，但。您可以将它们只是添加到您传递给的ProcessStartInfo 参数：

Adding arguments to processNeedToRun is no problem, though. You can just add them to the arguments you pass to ProcessStartInfo:

var procStartInfo = new ProcessStartInfo("cmd", "/c processNeedToRun arg1 arg2");