Ordering with SQLITE by nearest latitude & longitude coordinates

Question

I must obtain a SQLite SQL Sentence for ordering by nearest latitude and longitude coordinates, given a initial location.

THis is a example sentence for my table in the sqlite database:

SELECT id, name, lat, lng FROM items

EXAMPLE RESULT: 1, Museu, 41375310.0, 2175970.0

I must achieve this with SQLite, and with that table. I can't use another techniques because this is for a existen SQlite database that i can't change.

Exists a way to achieve this with Android and SQlite? I checked a lot of stackoverflow posts and i didn't find the way to achieve that

Thanks

Answer 1

SELECT * AS distance FROM items ORDER BY ABS(location_lat - lat) + ABS(location_lng - lng) ASC

This should roughly sort the items on distance in MySQL, and should work in SQLite.
If you need to sort them preciser, you could try using the Pythagorean theorem (a^2 + b^2 = c^2) to get the exact distance.

Answer 2

The answer from Darkwater is nearly correct. To be correct, you need to use the square of the differences. As the square function is not available on SqLite, you need to multiply the differences by themselves. No need to calculate the square roots.

SELECT * AS distance FROM items ORDER BY ((location_lat-lat)*(location_lat-lat)) + ((location_lng - lng)*(location_lng - lng)) ASC

Answer 3

The solution proposed by Chris:

SELECT * AS distance FROM items ORDER BY ((location_lat-lat)*(location_lat-lat)) + ((location_lng - lng)*(location_lng - lng)) ASC

is correct when we are close to the equator. For it to work correctly in other latitudes, I propose:

SELECT * AS distance FROM items ORDER BY ((location_lat-lat)*(location_lat-lat)) + ((location_lng - lng)*(location_lng - lng)* cos_lat_2 ) ASC

where we must precompute:

cos_lat_2 = cos(location_lat * PI / 180) ^ 2

THE PROBLEM:

If we are in the equator and we move one degree in longitude (east or west), we do so on a circumference of 40,000 km, representing a distance of 40.000 / 360. If we move one degree in latitude (north or south), we do so on a circle that crosses both poles, which also involves a distance of 40.000 / 360 (considering that the earth is a sphere).

But, if we are in the south of England, with a latitude of 50°, and we move one degree in longitude (east or west) we do it on the 50th parallel, which has a smaller circumference than the equator. The distance is perimeter_parallel_50 / 360. Calculating this perimeter is simple: perimeter_parallel_50 = cos (50) * 2 * PI * EARHT_RADIUS = 0.64 * 40,000 km. This reduction in distance is not seen if we move one degree south or north. The circumference by which we move still has a perimeter of 40,000 Km.

SOLUTION:

Since location_lat is a value known beforehand, we can pre-calculate the value of cos(location_lat) so that it can be used as a scaling factor so the displacements in longitude and latitude are equivalent. Moreover, we pre square it to avoid having to multiply it twice.

NOTE:

This is still an approximation, and it will give wrong results when moving big distances, especially near the poles and when crossing the 180th meridian.

Answer 4

If you're able to load the records; convert them to Locations and then use the distanceTo function I'd recommend that but...

You can approximate the distance between two points using plain SQL and the various approaches are laid out clearly here . The further apart the points you are comparing become then your values can become increasingly incorrect if you use a simple calculation

If you're calculating these things yourself and your locations can be anywhere then you might need to be aware of values wrapping around if you're comparing locations around the international date line.

Answer 5

If you know, that 1 degree of latitude is about 111111 meters and 1 degree of longitude is 111111*cos(latitude) meters, then you can easily get all places inside a specific square.

SELECT * FROM items WHERE latitude BETWEEN %f AND %f AND longitude BETWEEN %f AND %f

This query is very fast, even with millions of rows. But don't forget to create an index for latitude and longitude:

CREATE INDEX position ON items (latitude, longitude)

I use this in Objective-C to obtain all places of interest that are within 3km around the current location:

double latDist = 1.0 / 111111.0 * 3.0;
double lonDist = 1.0 / ABS(111111.0*cos(location.coordinate.latitude)) * 3.0;

FMResultSet *results = [database executeQueryWithFormat:@"SELECT * FROM items WHERE latitude BETWEEN %f AND %f AND longitude BETWEEN %f AND %f", location.coordinate.latitude - latDist, location.coordinate.latitude + latDist, location.coordinate.longitude - lonDist, location.coordinate.longitude + lonDist];

You can now calculate the exact distance and sort your results...