C#启动另外一个程序

          string exe_Path = "C:\\";//程序路径
                System.Diagnostics.ProcessStartInfo info = new System.Diagnostics.ProcessStartInfo();
                info.FileName = "xxx.exe";
                info.WorkingDirectory = exe_Path;
                System.Diagnostics.Process proc;                
                try
                {
                    proc = System.Diagnostics.Process.Start(info);
                }
                catch (Exception ex)
                {
                    MessageBox.Show(ex.Message);
                }

 

posted @ 2013-11-27 14:07  EleMMent  阅读(896)  评论(0编辑  收藏  举报