从 LocationServices 返回的位置年龄

Question

我需要确保我使用的位置是新鲜的：

有没有办法找出 LocationServices.FusedLocationApi.getLastLocation 返回的定位结果的年代？

如果不是：如果我向 LocationServices.FusedLocationApi.requestLocationUpdates（使用 setNumUpdates(1) 和 setMaxWaitTime(0)）注册一个位置侦听器，那么如果位置与LocationServices.FusedLocationApi.getLastLocation?

Answer 1

是的，有一个非常简单的方法。您可以像这样调用 getTime() 来获取位置修复的时间：

Location currentLocation = LocationServices.FusedLocationApi.getLastLocation(apiClient);
long locationAge = System.currentTimeMillis() - currentLocation.getTime();

if (locationAge <= 60 * 1000) { // not older than 60 seconds
    // do something with the location
}

文档建议不要使用 System.currentTimeMillis() 进行时间比较，但我从未遇到过这种方法的任何缺陷。但是，您应该考虑阅读（简短的）文档：

https://developer.android.com/reference/android/location/Location.html#getTime()

Answer 2

为了扩展 Illiminat 的回答，从 API 17 开始，添加了 getElapsedRealtimeNanos() 方法。从方法的文档...

Return the time of this fix, in elapsed real-time since system boot.

This value can be reliably compared to SystemClock.elapsedRealtimeNanos(), to calculate the age of a fix and to compare Location fixes. This is reliable because elapsed real-time is guaranteed monotonic for each system boot and continues to increment even when the system is in deep sleep (unlike getTime().

https://developer.android.com/reference/android/location/Location.html#getElapsedRealtimeNanos()

因此，以下应该是最精确的计算方法

Location currentLocation = LocationServices.FusedLocationApi.getLastLocation(apiClient);
long locationAge = SystemClock.elapsedRealtimeNanos() - currentLocation.getElapsedRealtimeNanos();
long ageLimitNanoSec = 60_000_000_000; // 60 seconds in nano seconds
if (locationAge <= ageLimitNanoSec) { 
    // do something with the location
}

从 LocationServices 返回的位置年龄

Age of location returned from LocationServices

android

location-services