Calculate or Query Great Circle Distance Between Points of Latitude and Longitude Using The Haversine Formula (PHP, JavaScript, Java, Python, MySQL, MSSQL Examples)
This month I’ve been programming quite a bit in PHP and MySQL with respect to GIS. Snooping around the net, I actually had a hard time finding some of the Geographic calculations to find the distance between two locations so I wanted to share them here.
The simple way of calculating a distance between two points is using the Pythagorean formula to calculate the hypotenuse of a triangle (A² + B² = C²). This is known as the Euclidean distance.
That’s an interesting start but it doesn’t apply to Geography since the distance between lines of latitude and longitude are not equal distances apart. As you get closer to the equator, lines of latitude get further apart. If you use a simple triangulation equation, it may measure distance accurately in one location and wrong in the other, because of the curvature of the Earth.
Great Circle Distance
The routes traveled long distances around the Earth are known as the Great Circle Distance. That is… the shortest distance between two points on a sphere differs from the points on a flat map. Combine that with the fact that latitude and longitude lines aren’t equidistant… and you’ve got a difficult calculation.
Here’s a fantastic video explanation of how Great Circles work.
The Haversine Formula
The distance using the curvature of the Earth is incorporated in the Haversine formula, which uses trigonometry to allow for the earth’s curvature. When you’re finding the distance between 2 places on earth (as the crow flies), a straight line is actually an arc.
This is applicable in air flight – have you ever looked at the actual map of flights and noticed they are arched? That’s because flying in an arch between two points is shorter than directly to the location.
PHP: Calculate Distance Between 2 Points of Latitude and Longitude
Here’s the PHP formula for calculating the distance between two points (along with Mile vs. Kilometer conversion) rounded to two decimal places.
function getDistanceBetweenPointsNew($latitude1, $longitude1, $latitude2, $longitude2, $unit = 'miles') {
$theta = $longitude1 - $longitude2;
$distance = (sin(deg2rad($latitude1)) * sin(deg2rad($latitude2))) + (cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * cos(deg2rad($theta)));
$distance = acos($distance);
$distance = rad2deg($distance);
$distance = $distance * 60 * 1.1515;
switch($unit) {
case 'miles':
break;
case 'kilometers' :
$distance = $distance * 1.609344;
}
return (round($distance,2));
}The variables are:
- $Latitude1 – a variable for your first location’s latitude.
- $Longitude1 – a variable for your first location’s longitude
- $Latitude2 – a variable for your second location’s latitude.
- $Longitude2 – a variable for your second location’s longitude.
- $unit – the default being miles. This can be updated or passed as kilometers.
Java: Calculate Distance Between 2 Points of Latitude and Longitude
public static double getDistanceBetweenPointsNew(double latitude1, double longitude1, double latitude2, double longitude2, String unit) {
double theta = longitude1 - longitude2;
double distance = 60 * 1.1515 * (180/Math.PI) * Math.acos(
Math.sin(latitude1 * (Math.PI/180)) * Math.sin(latitude2 * (Math.PI/180)) +
Math.cos(latitude1 * (Math.PI/180)) * Math.cos(latitude2 * (Math.PI/180)) * Math.cos(theta * (Math.PI/180))
);
if (unit.equals("miles")) {
return Math.round(distance, 2);
} else if (unit.equals("kilometers")) {
return Math.round(distance * 1.609344, 2);
} else {
return 0;
}
}The variables are:
- latitude1 – a variable for your first location’s latitude.
- longitude1 – a variable for your first location’s longitude
- latitude2 – a variable for your second location’s latitude.
- longitude2 – a variable for your second location’s longitude.
- unit – the default being miles. This can be updated or passed as kilometers.
Javascript: Calculate Distance Between 2 Points of Latitude and Longitude
function getDistanceBetweenPoints(latitude1, longitude1, latitude2, longitude2, unit = 'miles') {
let theta = longitude1 - longitude2;
let distance = 60 * 1.1515 * (180/Math.PI) * Math.acos(
Math.sin(latitude1 * (Math.PI/180)) * Math.sin(latitude2 * (Math.PI/180)) +
Math.cos(latitude1 * (Math.PI/180)) * Math.cos(latitude2 * (Math.PI/180)) * Math.cos(theta * (Math.PI/180))
);
if (unit == 'miles') {
return Math.round(distance, 2);
} else if (unit == 'kilometers') {
return Math.round(distance * 1.609344, 2);
}
}The variables are:
- latitude1 – a variable for your first location’s latitude.
- longitude1 – a variable for your first location’s longitude
- latitude2 – a variable for your second location’s latitude.
- longitude2 – a variable for your second location’s longitude.
- unit – the default being miles. This can be updated or passed as kilometers.
Python: Calculate Distance Between 2 Points of Latitude and Longitude
Anyways, here’s the Python formula for calculating the distance between two points (along with Mile vs. Kilometer conversion) rounded to two decimal places. Credit to my son, Bill Karr who is a Data Scientist for OpenINSIGHTS, for the code.
from numpy import sin, cos, arccos, pi, round
def rad2deg(radians):
degrees = radians * 180 / pi
return degrees
def deg2rad(degrees):
radians = degrees * pi / 180
return radians
def getDistanceBetweenPointsNew(latitude1, longitude1, latitude2, longitude2, unit = 'miles'):
theta = longitude1 - longitude2
distance = 60 * 1.1515 * rad2deg(
arccos(
(sin(deg2rad(latitude1)) * sin(deg2rad(latitude2))) +
(cos(deg2rad(latitude1)) * cos(deg2rad(latitude2)) * cos(deg2rad(theta)))
)
)
if unit == 'miles':
return round(distance, 2)
if unit == 'kilometers':
return round(distance * 1.609344, 2)The variables are:
- latitude1 – a variable for your first location’s latitude.
- longitude1 – a variable for your first location’s longitude
- latitude2 – a variable for your second location’s latitude.
- longitude2 – a variable for your second location’s longitude.
- unit – the default being miles. This can be updated or passed as kilometers.
MySQL: Retrieving All Records Within A Range By Calculating Distance In Miles Using Latitude and Longitude
It’s also possible to use SQL to calculate all records within a specific distance. In this example, I’m going to query MyTable in MySQL to find all the records that are less than or equal to variable $distance (in Miles) to my location at $latitude and $longitude:
The query for retrieving all of the records within a specific distance by calculating distance in miles between two points of latitude and longitude are:
$query = "SELECT *, (((acos(sin((".$latitude."*pi()/180)) * sin((`latitude`*pi()/180)) + cos((".$latitude."*pi()/180)) * cos((`latitude`*pi()/180)) * cos(((".$longitude."- `longitude`)*pi()/180)))) * 180/pi()) * 60 * 1.1515) as distance FROM `table` WHERE distance <= ".$distance."You’ll need to customize this:
- $longitude – this is a PHP variable where I’m passing the longitude of the point.
- $latitude – this is a PHP variable where I’m passing the longitude of the point.
- $distance – this is the distance that you would like to find all the records less or equal to.
- table – this is the table… you’ll want to replace that with your table name.
- latitude – this is the field of your latitude.
- longitude – this is the field of your longitude.
MySQL: Retrieving All Records Within A Range By Calculating Distance In Kilometers Using Latitude and Longitude
And here’s the SQL query using kilometers in MySQL:
$query = "SELECT *, (((acos(sin((".$latitude."*pi()/180)) * sin((`latitude`*pi()/180)) + cos((".$latitude."*pi()/180)) * cos((`latitude`*pi()/180)) * cos(((".$longitude."- `longitude`) * pi()/180)))) * 180/pi()) * 60 * 1.1515 * 1.609344) as distance FROM `table` WHERE distance <= ".$distance."You’ll need to customize this:
- $longitude – this is a PHP variable where I’m passing the longitude of the point.
- $latitude – this is a PHP variable where I’m passing the longitude of the point.
- $distance – this is the distance that you would like to find all the records less or equal to.
- table – this is the table… you’ll want to replace that with your table name.
- latitude – this is the field of your latitude.
- longitude – this is the field of your longitude.
I utilized this code in an enterprise mapping platform that we utilized for a retail store with over 1,000 locations across North America and it worked beautifully.
Microsoft SQL Server Geographic Distance: STDistance
If you’re utilizing Microsoft SQL Server, they offer their own function, STDistance for calculating the distance between two points using the Geography data type.
DECLARE @g geography;
DECLARE @h geography;
SET @g = geography::STGeomFromText('LINESTRING(-122.360 47.656, -122.343 47.656)', 4326);
SET @h = geography::STGeomFromText('POINT(-122.34900 47.65100)', 4326);
SELECT @g.STDistance(@h); Hat tip to Manash Sahoo, VP and Architect of Highbridge.
Thank you very much for sharing. This was an easy copy and paste job and works great. You’ve saved me alot of time.
FYI for anyone porting to C:
double deg2rad(double deg) { return deg*(3.14159265358979323846/180.0); }
Very nice piece of posting – worked very nice – I only had to change the name of the table holding the lat-long. It works pretty fast to.. I have a reasonably small number of lat-longs (< 400) but I think this would scale nicely. Nice site too – i have just added it to my del.icio.us account and will check back regularly.
Thanks very much Peter and Kerry! If you like working on GIS projects, I’d recommend:
Thank you very much… 😀
I searched the whole day for distance calculations and found the harversine algorithm, thanks to you for giving the example on how to put it in an sql statement. Thanks and greets, Daniel
Glad to help out, rails friend!
Now I’m in search of an ‘in Polygon’ PHP function that will take an array of sequenced latitude and longitude coordinates and figure out if another point is within or outside of the polygon.
I found the equation to figure out if a point in a polygon!
i think your SQL needs a having statement.
instead of WHERE distance <= $distance you might need to
use HAVING distance <= $distance
otherwise thanks for saving me a bunch of time and energy.
Hi David,
If you’re doing any type of GROUP BY statement, you will need HAVING. I am not doing that in the example above.
Doug
As of MySQL 5.x, you can’t use an alias on a WHERE clause see http://dev.mysql.com/doc/refman/5.0/en/problems-with-alias.html
Use HAVING instead of WHERE in the above querys
Thank you very much. You have done great job That’s the thing what i actually want. Thanks alot.
Thanks so much for sharing this code. It saved me a lot of development time. Also, thanks to your readers for pointing out that a HAVING statement is necessary for MySQL 5.x. Very helpful.
I’m blessed to have readers much smarter than I am!
🙂
The above formula is saving me a lot of time. Thank you very much.
I also have to switch between the NMEA format and Degrees. I found a formula at this URL at the bottom of the page. http://www.errorforum.com/knowledge-base/16273-converting-nmea-sentence-latitude-longitude-decimal-degrees.html
Does anyone know how to verify this?
Thank you!
Harry
Hello,
Another question. Is there a formula for NMEA strings like the one below ?
1342.7500,N,10052.2287,E
$GPRMC,032731.000,A,1342.7500,N,10052.2287,E,0.40,106.01,101106,,*0B
Thanks,
Harry
I also found that WHERE did not work for me. Changed it to HAVING and everything works perfect. At first i didnt read the comments and rewrote it using a nested select. Both will work just fine.
Thank you very mouch for the script written in mysql, just had to make few minor adjustments (HAVING) 🙂
Gret job
Incredibly helpful, thank you very much! I was having some problems with the new “HAVING”, rather than “WHERE”, but once I read the comments here (after about half an hour of grinding my teeth in frustration =P), I got it working nicely. Thank you ^_^
thanks a lot works great
Keep in mind that a select statement like that will be very computationally intense and therefore slow. If you have a lot of those queries, it can bog things down quite quickly.
A much less intense approach is to run a first (crude) select using a SQUARE area defined by a calculated distance i.e. “select * from tablename where latitude between lat1 and lat2 and longitude between lon1 and lon2”. lat1 = targetlatitude – latdiff, lat2 = targetlatitude + latdiff, similar with lon. latdiff ~= distance / 111 (for km), or distance/69 for miles since 1 degree of latitude is ~ 111 km (slight variation since earth is slightly oval, but sufficient for this purpose). londiff = distance / (abs(cos(deg2rad(latitude))*111)) — or 69 for miles (you can actually take a slightly larger square in order to account for variations). Then take the result of that and feed it into the radial select. Just don’t forget to account for out-of-bounds coordinates – i.e. the range of acceptable longitude is -180 to +180 and the range of acceptable latitude is -90 to +90 — in case your latdiff or londiff runs outside this range. Note that in most cases this may not be applicable since it only affects calculations over a line through the pacific ocean from pole to pole, though it does intersect part of chukotka and part of alaska.
What we accomplish by this is a significant reduction in the number of points against which you make this calculation. If you have a million global points in the database distributed roughly evenly and you want to search within 100 km, then your first (fast) search is of an area 10000 sq km and will probably yield about 20 results (based on even distribution over a surface area of about 500M sq km), which means that you run the complex distance calculation 20 times for this query instead of a million times.
Minor mistake in the example… that would be for within 50 km (not 100) since we are looking at the “radius” of our… square.
Fantastic advice! I actually worked with a developer who wrote a function that pulled the inside square and then a recursive function that made ‘squares’ around the perimeter to include and exclude the remaining points. The result was an incredibly fast result – he could evaluate millions of points in microseconds.
My approach above is definitely ‘crude’ but capable. Thanks again!
Doug,
I have been trying to use mysql and php to evaluate whether a lat long point is within a polygon. Do you know if your developer friend published any examples on how to accomplish this task. Or do you know any good examples. Thanks in advance.
Hi everyone this is my test SQL statement:
SELECT DISTINCT area_id, (
(
(
acos( sin( ( 13.65 * pi( ) /180 ) ) * sin( (
`lat_dec` * pi( ) /180 ) ) + cos( ( 13.65 * pi( ) /180 ) ) * cos( (
`lat_dec` * pi( ) /180 )
) * cos( (
( 51.02 - `lon_dec` ) * pi( ) /180 )
)
)
) *180 / pi( )
) *60 * 1.1515 * 1.609344
) AS distance
FROM `post_codes` WHERE distance <= 50
and Mysql is telling me that distance, doesn’t exist as a column, i can use order by, i can do it without WHERE, and it works, but not with it…
Replace “WHERE distance” with “HAVING distance”.
Works like a charm, thanks, Douglas!
This is great, however it is just as the birds fly. It would be great to try and incorporate the google maps API to this somehow (maybe using roads etc.) Just to give an idea using a different form of transportation. I still have yet to make a simulated annealing function in PHP that would be able to offer an efficient solution to the traveling salesman problem. But I think that I may be able to reuse some of your code to do so.
Hi Douglas,
thank you so much for this article – you just saved me alot of time.
take care,
nimrod @Israel
Good article! I found a lot of articles describing how to compute distance between two points but I was really looking for the SQL snippet.
Thanks a lot works good
Thank you much for this formula. It shaved some time on a store location project that was eating at me.
Thanks a bundle. This little line of code saved me some considerable time in a store location project!
#1054 – Unknown column ‘distance’ in ‘where clause’
approve
Same here! What’s the problem :-/? how to solve the “distance” – Column problem? Help us, please!! 🙂
Try using HAVING instead of WHERE
2 days of research to finally find this page that solves my problem. Looks like i better bust out my WolframAlpha and brush up on my maths. The change from WHERE to HAVING has my script in working order. THANK YOU
instead of WHERE clause use :
HAVING distance < 50
Thanks Georgi. I kept getting column ‘distance’ not found. Once I change the WHERE to HAVING it worked like a charm!
I wish this was the first page i’d found on this. After trying many different commands this was the only one to work properly, and with minimal changes needed to fit my own database.
Thanks a lot!
I wish this was the first page i’d found on this. After trying many different commands this was the only one to work properly, and with minimal changes needed to fit my own database.
Thanks a lot!
Thanks a Lot!
Thanks a Lot!
I don’t think the code is showing up anymore. Maybe it’s firefox?
I just tested both in Firefox and Chrome and it’s appearing. Try again?
Hi. Thanks a lot. This works like a charm.
Thanks a lot Douglas. This is working perfect.
I know this formula works, but I can’t see where the radius of the earth is taken into account. Can anyone enlighten me, please ?
Tim, for a full explanation of the Haversine formula (that’s not code), check out Wikipedia’s article: http://en.wikipedia.org/wiki/Haversine_formula
Beautiful! This has helped me immensely!
Great stuff Douglas. Have you tried getting the intersection point given the Long/Lat/Bearing of two points?
Haven’t done that yet, Khanh!
Thank you Douglas, the SQL Query is exactly what I needed, and I thought I’d have to write it myself. You’ve saved me from possibly hours of latitude longitude learning curve!
I keep getting Errormessage: Unknown column ‘Distance’ in ‘where clause’ on the MySQL Query.
Peter, please read through the other comments. It appears that some folks had to use a different syntax for WHERE/HAVING.
Thank you for this great article! Just tested the code on my DB and worked great!
Douglas, thank you for this amazing code. Been cracking my head on how to do this on my GPS community portal. You’ve saved me hours.
Great to hear, Ash!
thanks for posting this helpful article,
but for some reason I’d like to ask
how to get the distance between coords inside mysql db and coords inserted to php by user?
for more clearly describe:
1.user have to insert [id] for selecting specified data from db and user itself’s coords
2.the php file get the target data (coords) using [id] and then calculate distance between user and target point
or can just simply get distance from the code below?
$qry = “SELECT *,(((acos(sin((“.$latitude.”*pi()/180)) * sin((`Latitude`*pi()/180))+cos((“.$latitude.”*pi()/180)) * cos((`Latitude`*pi()/180)) * cos(((“.$longitude.”- `Longitude`)*pi()/180))))*180/pi())*60*1.1515*1.609344) as distance FROM `MyTable` WHERE distance >= “.$distance.” >>>>can I “take out” the distance from here?
thanks again,
Timmy S
never mind, I’ve figure out how the “function” work in php
$dis=getDistanceBetweenPointsNew($userLati, $userLongi, $lati, $longi, $unit = ‘Km’)
thanks a lot!!
ok, everything I have tried is not working. I mean, what I have works, but the distances are way off.
Could anybody possibly see what is wrong with this code?
if(isset($_POST[‘submitted’])){ $z = $_POST[‘zipcode’]; $r = $_POST[‘radius’]; echo “Results for “.$z; $sql = mysql_query(“SELECT DISTINCT m.zipcode, m.MktName,m.LocAddSt,m.LocAddCity,m.LocAddState,m.x1,m.y1,m.verified,z1.lat,z2.lon,z1.city,z1.state FROM mrk m, zip z1, zip z2 WHERE m.zipcode = z1.zipcode AND z2.zipcode = $z AND (3963 * acos( truncate( sin( z2.lat / 57.2958 ) * sin( m.y1 / 57.2958 ) + cos( z2.lat / 57.2958 ) * cos( m.y1 / 57.2958 ) * cos( m.x1 / 57.2958 – z2.lon / 57.2958 ) , 8 ) ) ) <= $r ") or die(mysql_error()); while($row = mysql_fetch_array( $sql )) { $store1 = $row['MktName']."”; $store = $row[‘LocAddSt’].””; $store .= $row[‘LocAddCity’].”, “.$row[‘LocAddState’].” “.$row[‘zipcode’]; $latitude1 = $row[‘lat’]; $longitude1 = $row[‘lon’]; $latitude2 = $row[‘y1’]; $longitude2 = $row[‘x1’]; $city = $row[‘city’]; $state = $row[‘state’]; $dis = getnew($latitude1, $longitude1, $latitude2, $longitude2, $unit = ‘Mi’); // $dis = distance($lat1, $lon1, $lat2, $lon2); $verified = $row[‘verified’]; if($verified == ‘1’){ echo “”; echo “”.$store.””; echo $dis . ” mile(s) away”; echo “”; } else { echo “”.$store.””; echo $dis . ” mile(s) away”; echo “”; } }}
my functions.php code
function getnew($latitude1, $longitude1, $latitude2, $longitude2, $unit = ‘Mi’) { $theta = $longitude1 – $longitude2; $distance = (sin(deg2rad($latitude1)) * sin(deg2rad($latitude2))) + (cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * cos(deg2rad($theta))); $distance = acos($distance); $distance = rad2deg($distance); $distance = $distance * 60 * 1.1515; switch($unit) { case ‘Mi’: break; case ‘Km’ : $distance = $distance * 1.609344; } return (round($distance,2)); }
Thank you in advance
Thank you for this article. Working fine with my code. 🙂
Hey Douglas, great article. I found your explanation of the geographical concepts and the code really interesting. My only suggestion would be to space and indent the code for display (like Stackoverflow, for example). I understand that you want to conserve space, but conventional code spacing / indentation would make it a lot easier for me, as a programmer, to read and dissect. Anyhow, that’s a small thing. Keep up the great work.
Thanks! I’ve modified the post a little… but the equations take up so much room and are so long that I’m not sure that it helps too much.
Thank you SO much.
here while using with function we are getting one type of distance..while using query its coming other type of distance
I wont calculate distance between two state
Muchas gracias por tan hermoso codigo…
Thisd i good cosinus functions. I dont know math, but thanks!
Great Job… 🙂 (y)
it seems faster (mysql 5.9) to use twice the formula in the select and where:
$formula = “(((acos(sin((“.$latitude.”*pi()/180)) * sin((`Latitude`*pi()/180))+cos((“.$latitude.”*pi()/180)) * cos((`Latitude`*pi()/180)) * cos(((“.$longitude.”- `Longitude`)*pi()/180))))*180/pi())*60*1.1515*1.609344)”;
$sql = ‘SELECT *, ‘.$formula.’ as distance FROM table WHERE ‘..$formula.’ <= '.$distance;
thanks…
not working if
“WHERE distance”
working if
“HAVING distance”
Thanks a lot for shear this article.it’s very helpful.
PHP was at first created as a simple scripting platform called “Personal Home Page”. Nowadays PHP (the short for Hypertext Preprocessor) is an alternative of the Microsoft’s Active Server Pages (ASP) technology.
PHP is an open source server-side language which is used for creating dynamic web pages. It can be embedded into HTML. PHP is usually used in conjunction with a MySQL database on Linux/UNIX web servers. It is probably the most popular scripting language.
I found above solution not working properly.
I need to change to :
$qqq = “SELECT *,(((acos(sin((“.$latitude.”*pi()/180)) * sin((`latt`*pi()/180))+cos((” . $latitude . “*pi()/180)) * cos((`latt`*pi()/180)) * cos(((” . $longitude . “- `longt`)*pi()/180))))*180/pi())*60*1.1515) as distance FROM `register` “;
Thanks Kupendra!
thank you sir wroking perfectly.. but i have one question if i want to output without decimal point then what can i do..?
Thanks in advance.
Hello, please i will really need your help on this.
I made a get request to my web-server http://localhost:8000/users/findusers/53.47792/-2.23389/20/
53.47792 = $latitude
-2.23389 = $longitude
and 20 = the distance i want to retrieve
However using you formula, it retrieves all rows in my db
$results = DB::select( DB::raw(“SELECT *, (((acos(sin((“.$latitude.”*pi()/180)) * sin((lat*pi()/180))+cos((“.$latitude.”*pi()/180)) * cos((lat*pi()/180)) * cos(((“.$longitude.”- lng)*pi()/180))))*180/pi())*60*1.1515*1.609344) as distance FROM markers HAVING distance >= “.$distance ));
[{“id”:1,”name”:”Frankie Johnnie & Luigo Too”,”address”:”939 W El Camino Real, Mountain View, CA”,”lat”:37.386337280273,”lng”:-122.08582305908,”distance”:16079.294719663},{“id”:2,”name”:”Amici’s East Coast Pizzeria”,”address”:”790 Castro St, Mountain View, CA”,”lat”:37.387138366699,”lng”:-122.08323669434,”distance”:16079.175940152},{“id”:3,”name”:”Kapp’s Pizza Bar & Grill”,”address”:”191 Castro St, Mountain View, CA”,”lat”:37.393886566162,”lng”:-122.07891845703,”distance”:16078.381373826},{“id”:4,”name”:”Round Table Pizza: Mountain View”,”address”:”570 N Shoreline Blvd, Mountain View, CA”,”lat”:37.402652740479,”lng”:-122.07935333252,”distance”:16077.420540582},{“id”:5,”name”:”Tony & Alba’s Pizza & Pasta”,”address”:”619 Escuela Ave, Mountain View, CA”,”lat”:37.394012451172,”lng”:-122.09552764893,”distance”:16078.563225154},{“id”:6,”name”:”Oregano’s Wood-Fired Pizza”,”address”:”4546 El Camino Real, Los Altos, CA”,”lat”:37.401725769043,”lng”:-122.11464691162,”distance”:16077.937560795},{“id”:7,”name”:”The bars and grills”,”address”:”24 Whiteley Street, Manchester”,”lat”:53.485118865967,”lng”:-2.1828699111938,”distance”:8038.7620112314}]
I want to retrieve just rows with 20 miles but it brings all rows. Please what am i doing wrong
I’m looking for a similar query but stepped up a bit – in short this is to group all coordinates within 2 miles of each coordinate and then count how many coordinates in each group and output only one group that has the most coordinates – even if you have more than one group among groups that have the most number of coordinates – simply output the random group from the groups with same largest number –
Thank you very much for sharing.