使用 IP 地址在远程计算机上打开 EXE

本文关键字：EXE 计算机 IP 地址 使用 | 更新日期: 2023-09-27 17:56:33

我想在IP为10.64.4.38的远程计算机上打开一个程序通过基于C#Windows的程序。

我要打开的文件位于 上。@"C：''Program Files''RealVNC''VNC Viewer''vncviewer.exe"

我正在使用以下代码，但它不起作用。 请帮帮我

ConnectionOptions options = new ConnectionOptions();
options.Impersonation = ImpersonationLevel.Impersonate; 
options.Authentication = AuthenticationLevel.Default; 
options.Username = userName; 
options.Password = password; 
options.Authority = null; 
options.EnablePrivileges = true;
ManagementScope scope = new ManagementScope(path, options); 
scope.Connect();
// Create the process 
using (ManagementClass process = new ManagementClass("Win32_Process")) 
{
    ConnectionOptions conn = new ConnectionOptions();
    conn.Username = "support";
    conn.Password = "password";
    process.Scope = scope; 
    process.InvokeMethod("Create", commandLine); 
    ManagementScope ms = new ManagementScope(@"''10.64.4.38'root'cimv2", conn);
    Process g = Process.Start(@"C:''Program Files'RealVNC'VNC Viewer'vncviewer.exe");

使用 IP 地址在远程计算机上打开 EXE

这是我写的一个程序的摘录，它做了你想做的事情。 您使用了进程 g = .。这将创建一个本地进程。这不是你想要的。

 ConnectionOptions connOptions = new ConnectionOptions();
  connOptions.Impersonation = ImpersonationLevel.Impersonate;
  connOptions.EnablePrivileges = true;
  ManagementScope manScope = new ManagementScope(String.Format(@"''{0}'ROOT'CIMV2", remoteMachine),
 connOptions);
  manScope.Connect();
  ObjectGetOptions objectGetOptions = new ObjectGetOptions();
  ManagementPath managementPath = new ManagementPath("Win32_Process");
  ManagementClass processClass = new ManagementClass(manScope, managementPath, objectGetOptions);
  ManagementBaseObject inParams = processClass.GetMethodParameters("Create");
  Console.WriteLine("{0}>Remoteexec: {1}",remoteMachine, sBatFile);
  inParams["CommandLine"] = sBatFile;
  inParams["CurrentDirectory"] = !nocopy?@"c:'656":@"c:'"; // this is part of my current process so this needs to be the remote folder you wish to work from, so c:'656 is just a default
  ManagementBaseObject outParams = processClass.InvokeMethod("Create", inParams, null);
  UInt32 rv = (UInt32) outParams["returnValue"];
  uint ProcessId = (uint) outParams["processId"];
  String result = "";
  switch (rv)
  {
case 0:
 result = "ok";
 break;
case 2:
 result = "Access denied";
 break;
case 3:
 result = "Insufficient privilage";
 break;
case 8:
 result = "Unknown failure";
 break;
case 9:
 result = "Path not found";
 break;
case 21:
 result = "Unknown failure";
 break;
default:
 result = "Mega bad";
 break;
  }
  Console.WriteLine("{0}>Creation of the process returned: {1}",remoteMachine,result);

这还将告诉您远程进程失败的原因。 如果它无法启动

你为什么要用这条线？

Process g = Process.Start(@"C:''Program Files'RealVNC'VNC Viewer'vncviewer.exe");

所有这些都将尝试在本地启动该过程。

而是像这样初始化MagemementClass（其中服务器是启用WMI的计算机的路径）： 

process = new ManagementClass(@"''server'root'cimv2:Win32_Process");

然后调用该方法：

object[] theProcessToRun = { @"C:''Program Files'RealVNC'VNC Viewer'vncviewer.exe" };
process.InvokeMethod("Create", theProcessToRun);
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