从 cmd 读取输出得到错误

Reading Output from cmd getting errors

我一直在查找 "how to get standard output from cmd" 并找到了一些教程,但是 none,是的 NONE 似乎有效,我正在尝试阅读 [=19] 的所有输出=] 对我来说。

这里是全部代码,向下滚动查看错误位置

    public static string C(string arguments, bool b)
    {
        System.Diagnostics.Process process = new 
        System.Diagnostics.Process();
        System.Diagnostics.ProcessStartInfo startInfo = new 
        System.Diagnostics.ProcessStartInfo();
        startInfo.WindowStyle = 
        System.Diagnostics.ProcessWindowStyle.Hidden;
        process.StartInfo.RedirectStandardOutput = true;
        process.StartInfo.RedirectStandardInput = true;
        process.StartInfo.UseShellExecute = false;
        startInfo.FileName = "cmd.exe";
        startInfo.Arguments = arguments;
        process.StartInfo = startInfo;
        process.Start();

        string res = "";
        if (b)
        {
            StringBuilder q = new StringBuilder();
            while (!process.HasExited)
            {
                q.Append(process.StandardOutput.ReadToEnd());
            }
            string r = q.ToString();
            res = r;
        }
        if(res == "" || res == null)
        {
            return "NO-RESULT";
        }
        return res;

    }

哪里出现错误(System.InvalidOperationException:“StandardOut 尚未重定向或进程尚未启动。”)

   string res = "";
   StringBuilder q = new StringBuilder();
   while (!process.HasExited)
   {
       q.Append(process.StandardOutput.ReadToEnd()); // Right here
   }
   string r = q.ToString();
   res = r;

您正在创建一个名为 startInfoProcessStartInfo,然后在 process.StartInfo 上设置一些属性,然后将 startInfo 分配给 process.StartInfo 基本上恢复您的设置以前。

您应该在 startInfo 上设置 RedirectStandardOutputRedirectStandardInputUseShellExecute

System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo.RedirectStandardOutput = true;
startInfo.RedirectStandardInput = true;
startInfo.UseShellExecute = false;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = arguments;
process.StartInfo = startInfo;
process.Start();