I have a HTML list of about 500 items and a "filter" box above it. I started by using jQuery to filter the list when I typed a letter (timing code added later): $('#filter').keyup( function() { var jqStart = (new Date).getTime(); var search = $(this).val().toLowerCase(); var $list = $('ul.ablist > li'); $list.each( function() { if ( $(this).text().toLowerCase().indexOf(search) === -1 ) $(this).hide(); else $(this).show(); } ); console.log('Time: ' + ((new Date).getTime() - jqStart)); } ); However, there was a couple of seconds delay after typing each letter (particularly the first letter). So I thought it may be slightly quicker if I used plain Javascript (I read recently that jQuery's each function is particularly slow). Here's my JS equivalent: document.getElementById('filter').addEventListener( 'keyup', function () { var jsStart = (new Date).getTime()...
we can find city temp by using this find_Location Class. in which firstly find the city location by use by longitude and latitude in android? And after that use the method SendToUrl to find the temperature by use GoogleXML "https://www.google.com/ig/api?weather="
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{
Log.d("Find Location", "in find_location");
this.con=con;
String location_context = Context.LOCATION_SERVICE;
locationManager = (LocationManager)con.getSystemService(location_context);
List<String> providers = locationManager.getProviders(true);
for (String provider : providers)
{
locationManager.requestLocationUpdates(provider, 1000, 0,new LocationListener()
{
public void onLocationChanged(Location location) {}
public void onProviderDisabled(String provider){}
public void onProviderEnabled(String provider){}
public void onStatusChanged(String provider, int status,Bundle extras){}
});
Location location = locationManager.getLastKnownLocation(provider);
if (location != null)
{
latitude = location.getLatitude();
longitude = location.getLongitude();
addr=ConvertPointToLocation(latitude,longitude);
String temp_c=SendToUrl(addr);
}
}
}
public String ConvertPointToLocation(double pointlat,double pointlog) {
String address = "";
Geocoder geoCoder = new Geocoder(con,
Locale.getDefault());
try {
List<Address> addresses = geoCoder.getFromLocation(pointlat,pointlog, 1);
if (addresses.size() > 0) {
for (int index = 0; index < addresses.get(0)
.getMaxAddressLineIndex(); index++)
address += addresses.get(0).getAddressLine(index) + " ";
}
}
catch (IOException e) {
e.printStackTrace();
}
return address;
}
private String SendToUrl(String string) {
// TODO Auto-generated method stub
try{
string=string.replace(" ","%20");
String queryString="https://www.google.com/ig/api?weather="+string;
url= new URL(queryString);
conec= url.openConnection();
stream = conec.getInputStream();
xpp = XmlPullParserFactory.newInstance().newPullParser();
xpp.setInput(stream, null);
int eventType = xpp.getEventType();
while(eventType != XmlPullParser.END_DOCUMENT)
{
if(eventType == XmlPullParser.START_TAG)
{String elementName = xpp.getName();
if("temp_c".equals(elementName))
{
int acount=xpp.getAttributeCount();
for(int x=0;x<acount;x++)
{
xpp.getAttributeValue(x);
String str=xpp.getAttributeValue(x);
return str;
}
}
}
eventType = xpp.next();
}
}
My answer is not totally perfect but Yahoo Weather API may help you. This worked perfectly for me.
ReplyDeleteIn this api you have to select one of the City and then you will get weather updates. However you have to go through the source for extracting your portion.
Hope this helps.
There appears to be a full code example on anddev.org google weather api description which should include everything you require.
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