I want to sort a NSMutableArray, where each row is a NSMutableDictionary, with my GPS position from CoreLocation framework.
This is an example of my array of POI
arrayCampi = (
{
cap = 28100;
"cell_phone" = "";
championship = "IBL 1D";
citta = Novara;
division = "";
email = "";
fax = 0321457933;
indirizzo = "Via Patti, 14";
latitude = "45.437174";
league = "";
longitude = "8.596029";
name = "Comunale M. Provini";
naz = Italy;
prov = NO;
reg = Piemonte;
sport = B;
surname = "Elettra Energia Novara 2000";
telefono = 03211816389;
webaddress = "http://www.novarabaseball.it/";
})
I need to sort this array with my location (lat and long) with field 'latitude' and 'longitude' of each row in ascending mode (first row is POI nearest to me).
I have tried this solution without success:
+ (NSMutableArray *)sortBallparkList:(NSMutableArray *)arrayCampi location:(CLLocation *)myLocation {
if ([arrayCampi count] == 0) {
return arrayCampi;
}
if (myLocation.coordinate.latitude == 0.00 &&
myLocation.coordinate.longitude == 0.00) {
return arrayCampi;
}
NSMutableArray *sortedArray = [NSMutableArray arrayWithArray:arrayCampi];
BOOL finito = FALSE;
NSDictionary *riga1, *riga2;
while (!finito) {
for (int i = 0; i < [sortedArray count] - 1; i++) {
finito = TRUE;
riga1 = [sortedArray objectAtIndex: i];
riga2 = [sortedArray objectAtIndex: i+1];
CLLocationDistance distanceA = [myLocation distanceFromLocation:
[[CLLocation alloc]initWithLatitude:[[riga1 valueForKey:@"latitude"] doubleValue]
longitude:[[riga1 valueForKey:@"longitude"] doubleValue]]];
CLLocationDistance distanceB = [myLocation distanceFromLocation:
[[CLLocation alloc]initWithLatitude:[[riga2 valueForKey:@"latitude"] doubleValue]
longitude:[[riga2 valueForKey:@"longitude"] doubleValue]]];
if (distanceA > distanceB) {
[riga1 retain];
[riga2 retain];
[sortedArray replaceObjectAtIndex:i+1 withObject:riga2];
[sortedArray replaceObjectAtIndex:i withObject:riga1];
[riga1 release];
[riga2 release];
finito = FALSE;
}
}
}
return sortedArray;
}
Can anyone help me, also with other solution?
Alex.