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PHP and MySQL: Calculating Distance

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.

Flight Map EuropeIf you remember ‘the old days’ of calculating a distance between two points, it was simply the hypotenuse of a triangle (A² + B² = C²).

That’s an interesting start but it doesn’t apply with Geography since the distance between lines of latitude and longitude are not an equal distance apart. As you get closer to the equator, lines of latitude get further apart. If you use some kind of simple triangulation equation, it may measure distance accurately in one location and terribly wrong in the other, because of the curvature of the Earth.

That brings up the Haversine formula, which uses trigonometry to allow for the curvature of the earth. When you’re finding the distance between 2 places on earth (as the crow flies), a straight line is really 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 it’s shorter to fly in an arch between two points sometimes than directly to the location.

Anyways, 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 = '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));

It’s also possible to use MySQL to do a calculation to find all records within a specific distance. In this example, I’m going to query MyTable to find all the records that are less than or equal to variable $distance (in Miles) to my location at $latitude and $longitude:
$qry = "SELECT *,(((acos(sin((".$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) as distance
FROM `MyTable`
WHERE distance >= ".$distance."

For Kilometers:
$qry = "SELECT *,(((acos(sin((".$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."

I utilized similar computations in the Wild Birds Unlimited mapping platform that we built.

About Douglas Karr

Douglas Karr is the founder of The Marketing Technology Blog. Doug is the CMO of CircuPress and CEO of DK New Media, an agency specializing in assisting marketing technology companies with their inbound marketing - leveraging social media, blogging, search engine optimization, pay per click and public relations. Their clients include Angie's List, GoDaddy, Mindjet and many more. Douglas is also the author of Corporate Blogging for Dummies.

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  1. 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); }

  2. 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.

  3. Thank you very much… ūüėÄ

  4. 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

  5. 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.

  6. 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

  7. Thank you very much. You have done great job That’s the thing what i actually want. Thanks alot.

  8. 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.

  9. 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!

  10. Hello,

    Another question. Is there a formula for NMEA strings like the one below ?




  11. 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.

  12. Thank you very mouch for the script written in mysql, just had to make few minor adjustments (HAVING) :)
    Gret job

  13. 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 ^_^

  14. thanks a lot works great

  15. 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.

  16. 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…

  17. 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.

  18. Hi Douglas,
    thank you so much for this article – you just saved me alot of time.
    take care,
    nimrod @Israel

  19. 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.

  20. Thanks a lot works good

  21. Thank you much for this formula. It shaved some time on a store location project that was eating at me.

  22. Thanks a bundle. This little line of code saved me some considerable time in a store location project!

  23. #1054 – Unknown column ‘distance’ in ‘where clause’

  24. 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

  25. instead of WHERE clause use :
    HAVING distance < 50

  26. 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!

  27. 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!

  28. Thanks a Lot!

  29. Thanks a Lot!

  30. I don’t think the code is showing up anymore. Maybe it’s firefox?

  31. Hi. Thanks a lot. This works like a charm.

  32. Thanks a lot Douglas. This is working perfect.

  33. 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 ?

  34. Beautiful! This has helped me immensely!

  35. Great stuff Douglas.  Have you tried getting the intersection point given the Long/Lat/Bearing of two points?

  36. 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!

  37. I keep getting Errormessage: Unknown column ‘Distance’ in ‘where clause’¬† on the MySQL Query.

  38. Thank you for this great article! Just tested the code on my DB and worked great! 

  39. 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.

  40. 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

  41. 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

  42. Thank you for this article. Working fine with my code. :)

  43. 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.

  44. here while using with function we are getting one type of distance..while using query its coming other type of distance

  45. I wont calculate distance between two state

  46. Muchas gracias por tan hermoso codigo…

  47. Thisd i good cosinus functions. I dont know math, but thanks!

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