如何在processStartInfo中传递多个参数？

Question

我想cmd从c#代码中运行一些命令.我跟着一些博客和教程得到了答案,但我有点困惑,即我应该如何传递多个参数？

我使用以下代码:

System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Normal;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = 
...

startInfo.Arguments以下命令行代码的值是多少？

• makecert -sk server -sky exchange -pe -n CN=localhost -ir LocalMachine -is Root -ic MyCA.cer -sr LocalMachine -ss My MyAdHocTestCert.cer

• netsh http add sslcert ipport=127.0.0.1:8085 certhash=0000000000003ed9cd0c315bbb6dc1c08da5e6 appid={00112233-4455-6677-8899-AABBCCDDEEFF} clientcertnegotiation=enable

Answer 1

它纯粹是一个字符串:

startInfo.Arguments = "-sk server -sky exchange -pe -n CN=localhost -ir LocalMachine -is Root -ic MyCA.cer -sr LocalMachine -ss My MyAdHocTestCert.cer"

当然,当参数包含空格时,您必须使用\"\"来转义它们,如:

"... -ss \"My MyAdHocTestCert.cer\""

有关此信息,请参阅MSDN.

Answer 2

请记住包含 System.Diagnostics

ProcessStartInfo startInfo = new ProcessStartInfo("myfile.exe");        // exe file
startInfo.WorkingDirectory = @"C:\..\MyFile\bin\Debug\netcoreapp3.1\"; // exe folder

//here you add your arguments
startInfo.ArgumentList.Add("arg0");       // First argument          
startInfo.ArgumentList.Add("arg2");       // second argument
startInfo.ArgumentList.Add("arg3");       // third argument
Process.Start(startInfo);