Great - thank you. I'll remember to read the documentation next
time :)
I've pasted my code below in case anyone wants to borrow it: it keeps
polling for up-to-date and accurate GPS data, up to a maximum of 10
seconds. If by then the GPS data is old or not accurate enough, it
just returns false.
The code feels a bit dubious, but it seems to do the job.
Anna
long locationTime = location.getTime();
long currentTime = System.currentTimeMillis();
timeDifference = (currentTime - locationTime) / 1000;
float accuracy = location.getAccuracy();
int count = 0;
// Wait for accurate GPS data, up to a maximum of 10
seconds before
// throwing an error
while (((timeDifference > 10) || (accuracy > 20.0)) &&
(count <
20)) {
location =
locationmanager.getLastKnownLocation("gps");
locationTime = location.getTime();
currentTime = System.currentTimeMillis();
timeDifference = (currentTime - locationTime) /
1000;
accuracy = location.getAccuracy();
Log.d(LOG_TAG, "getting up to date GPS data,
time diff = "
+ timeDifference + " & accuracy
= " + accuracy
+ " & count = " + count);
try {
Thread.currentThread();
Thread.sleep(500);
} catch (InterruptedException ie) {
}
count++;
}
// No accurate GPS data? Exit here and warn the user
if ((timeDifference > 10) || (accuracy > 20.0)) {
return false;
}
On 27 Jan, 03:38, gjs <[email protected]> wrote:
> Hi,
>
> Subtract the Location.getTime() value from the current time to get the
> age of the last fix.
>
> Seehttp://code.google.com/android/reference/android/location/Location.html
>
> Regards
>
> On Jan 23, 10:07 am, Anna PS <[email protected]> wrote:
>
> > Hi
>
> > When you get GPS location using getLastKnownLocation, is there a way
> > to check how old the data is, i.e. when the location was last
> > updated?
>
> > I'm noticing that sometimes my app is giving me an out-of-date
> > location (usually because the sky is not visible when the app starts)
> > - it'd be good to warn the user about this.
>
> > It's possible to check the age of the data on the iPhone I believe, is
> > it possible in Android?
>
> > thanks!
> > Anna
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