安卓：可能code当设备进入睡眠状态，并从睡眠中唤醒启动服务 [英] Android: possible code to start a service when device goes to sleep and wakes from sleep

问题描述

我要当设备进入，当它醒来睡觉，也是我的服务自动启动。我该怎么做就无需在后台持续运行的服务，检查设备的状态？

I want my service to start automatically when the device goes to sleep and also when it wakes up. How do i do it without making the service run continuously in background to check the state of device?

例如，我可以我的服务添加到上手起来的系统默认的服务列表中，当设备被唤醒？如果是的话，该怎么办呢？

For example, can i add my service to the list of system default services that get started up when a device wakes up?if so, how to do it?

推荐答案

如果你正在谈论的电话事件，则广播接收机是解决方案

If you are talking about phone events then broadcast receivers is the solution

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