LocationManager的服务 - 需要停止，如果在1分钟内未修复 [英] LocationManager in Service - need to stop if no fix in 1 minute

问题描述

我有定期的服务，我开始使用报警器每5分钟。 服务实现LocationListener的获得GPS定位，并将其保存到SQLite数据库。

I have regular service that I start using alarms every 5 minutes. Service Implements LocationListener to get GPS fix and save it into SqLite database.

我给服务1分钟，得到最好的解决成为可能。如果我得到的精度小于20之前 - 甚至更好

I give service 1 minute to get best fix possible. If I get accuracy <20 before that - even better.

所有这一切工作正常。我也有code，检查是否GPS禁用，因此我退出服务。

All that works fine. I also have code that checks if GPS disabled so I exit service.

我想要做的是：如果在1分钟内未修复 - 我要退出服务。目前，由于大多数的逻辑（校对时间等）是onLocationChanged - 我从来没有机会做，因为我从来没有得到位置

What I want to do is: If there is no fix within 1 minute - I want to exit service. Currently since most logic (time checks, etc) is in onLocationChanged - I never have chance to do so because I never get location.

有没有计时器之类的，我可以用的东西吗？我是新来的Java，例如将是很好。我认为是这样的：在开始创建计时器回调，并在这个回调函数我会清理和关闭。我不知道是否有将是可能的onLocationChanged运行产生任何影响？

Is there timer or something like that I can use? I'm new to Java, example would be nice. I see it like this: On start I create timer with callback and in this callback function I will cleanup and shutdown. I'm not sure if there going to be any implications with possible onLocationChanged running?

我还能当我得到onLocationChanged禁止定时器？

Can I possibly disable timer as soon as I get onLocationChanged?

推荐答案

使用 android.os.Handler ：

Handler timeoutHandler = new Handler();
timeoutHandler.postDelayed(new Runnable() {
    public void run() {
        stopSelf();
    }
}, 60 * 1000);

这篇关于LocationManager的服务 - 需要停止，如果在1分钟内未修复的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！