我写了一个非常简单的应用程序,它向LocationProvider提供一个位置并将其打印到System.out。这在模拟器中很好用。但是,当我在我的黑莓手机上运行它时,getLocation调用似乎无限期地挂起。我在一个单独的线程中运行代码,只需获取提供程序并请求位置。我用空标准(它应该给我的默认值是正确的?)以及一个应该提供助手然后自治的标准。我已经在下面包含了我的代码。当我在我的设备上运行此它挂在调用getLocation.Here是我的代码below..plzz告诉我可能是做错了......Blackberry - LocationProvider.getLocation()挂在
public void getLocation() {
Thread t = new Thread(new Runnable() {
private double lat;
private double lon;
public void run() {
Criteria cr = new Criteria();
cr.setHorizontalAccuracy(Criteria.NO_REQUIREMENT);
cr.setVerticalAccuracy(Criteria.NO_REQUIREMENT);
cr.setCostAllowed(false);
cr.setPreferredPowerConsumption(Criteria.NO_REQUIREMENT);
cr.setPreferredResponseTime(1000);
LocationProvider lp = null;
try {
lp = LocationProvider.getInstance(cr);
} catch (LocationException e) {
// System.out.println("*****************Exception" + e);
}
Coordinates c = null;
if (lp == null) {
UiApplication.getUiApplication().invokeLater(
new Runnable() {
public void run() {
Dialog.alert("GPS not supported!");
return;
}
});
} else {
// System.out.println("OK");
switch (lp.getState()) {
case LocationProvider.AVAILABLE:
// System.out.println("Provider is AVAILABLE");
Location l = null;
try {
l = lp.getLocation(-1);
} catch (LocationException e) {
// System.out.println("***Location Exception caught "
// + e);
} catch (InterruptedException e) {
// System.out.println("***Interrupted Exception aught:"
// + e);
} catch (Exception e) {
// System.out.println("***Exception caught :"
// ;+ e);
}
if (l != null && l.isValid()) {
c = l.getQualifiedCoordinates();
}
if (c != null) {
lat = c.getLatitude();
lon = c.getLongitude();
System.out.println(lat + " - " + lon);
}
}
}
}
});
t.start();
}