Question

我有服务编写C＃和Visual Studio 2012以及NetFrameWork 4.5，这是OnStart-event的代码

安装服务后，我尝试启动它，它确实工作了错误是：

Windows无法在本地计算机上启动DPrintServer服务

错误1053：服务未响应启动或控制请求及时时尚。

    protected override void OnStart(string[] args)
    {




        string dirx = @"D:\tempsi";
        //Directory.SetCurrentDirectory(@"D:\tmp");
        Directory.SetCurrentDirectory(dirx);
        /*Ini.IniFile ini = new Ini.IniFile("PrintServer.ini");
        string smtp_server = ini.IniReadValue("smtp", "server");
        string from_email = ini.IniReadValue("from", "fromemail");
        string to_email = ini.IniReadValue("from", "toemail");
        string directory = ini.IniReadValue("settings", "dir");
        */
        string smtp_server = ConfigurationSettings.AppSettings.Get("server");
        string from_email = ConfigurationSettings.AppSettings.Get("fromemail");
        string to_email = ConfigurationSettings.AppSettings.Get("toemail");
        string ddirectory = ConfigurationSettings.AppSettings.Get("dir");
        string attach_dir = ConfigurationSettings.AppSettings.Get("attachdir");
        string filename = ConfigurationSettings.AppSettings.Get("filename");
        string subject = ConfigurationSettings.AppSettings.Get("subject");
        string mailcontent = ConfigurationSettings.AppSettings.Get("mailcontent");
        //string directory = ConfigurationSettings.AppSettings.Get("dir");
        /* while(true)
         {*/
        int counter = 1;
        while (true)
        {

            //counter++;
            if (File.Exists(attach_dir + filename))
            {


                ProcessStartInfo startInfo = new ProcessStartInfo();
                /*startInfo.FileName = "batti.bat";
                startInfo.CreateNoWindow = true;
                startInfo.UseShellExecute = false;
                //startInfo.Arguments = file;
                Process.Start(startInfo);
                startInfo.FileName = "batti2.bat";
                startInfo.CreateNoWindow = true;
                startInfo.UseShellExecute = false;
                Process.Start(startInfo);
                */
                startInfo.FileName = "blat.exe";
                startInfo.CreateNoWindow = true;
                startInfo.UseShellExecute = false;
                startInfo.Arguments = "-install " + smtp_server + " " + from_email;
                Process.Start(startInfo);
                startInfo.FileName = "blat.exe";
                startInfo.CreateNoWindow = true;
                startInfo.UseShellExecute = false;
                startInfo.Arguments = mailcontent + " -to " + to_email + " -subject " + subject + " -attach " + attach_dir + filename;
                var process = Process.Start(startInfo);
                process.WaitForExit();

                File.Delete(attach_dir + filename);
                //}
            }
            else
            {
                var stopwatch = Stopwatch.StartNew();
                stopwatch = Stopwatch.StartNew();
                Thread.Sleep(1000);
                stopwatch.Stop();


            }
        }
    }

Answer 1

你有一个无限循环。因此，就主机系统而言，您的应用程序永远不会响应。因此，它永远不会成功“启动”（OnStart永不返回），主机终止明显破坏的过程。

这意味着您希望您的服务每秒都做一些事情：

Thread.Sleep(1000);

您可以使用Timer对象完成此操作。您在OnStart中真正需要做的就是初始化该对象：

this.timer = new System.Timers.Timer();
this.timer.Interval = 1000;
this.timer.Elapsed += new System.Timers.ElapsedEventHandler(this.OnTimerTick);
this.timer.Start();

然后有一个tick事件的处理程序：

public void OnTimerTick(object sender, System.Timers.ElapsedEventArgs args)
{
    // perform your once-every-second logic here
}

这样可以更合适地使用系统而无需手动休眠线程并使用无限循环，正如您所看到的那样导致应用程序无限期地挂起。

启动jus编码的C＃服务。虽然（真） - 循环没有工作

1 个答案: