Calculating distances from latitude and longitude (PHP)

14 Responses to Calculating distances from latitude and longitude (PHP)

1. mandy says:

   May 27, 2003 at 2:50 pm

   thanks for the code. and the redesign!
2. Shashank says:

   Jun 01, 2003 at 12:55 am

   Thanks Mandy, just redesigned this thing. Will get back the comments back up too as I get the time.
3. Zak says:

   Jan 12, 2004 at 1:24 pm

   FYI: This tool is useless for places that don’t fall on the same side of the prime meridian and/or equator. You should work the cardnal directions into your formula.
4. sniptools says:

   Jan 13, 2004 at 12:09 am

   Thanks Zak, useful bit of info. My code above is simply a PHP version of something I came across in C. Will read up on “cardinal directions”

   -Shanx.
5. william says:

   Mar 23, 2004 at 7:34 am

   thank you kindly for the code! i’m having a hell of a time making it work though. do you have the full page-code (php included) posted? i used the php from the text field, but somewhere between the variables being sent, and the calculation i get a “parse error” on the last line, which can’t be accurate since the last line is .

   thank you,
   -Confused
6. sniptools says:

   Mar 24, 2004 at 1:03 am

   Hi William,

   Pls send me your code through email. When PHP throws a last line error, it is usually because a semicolon is missing from somewhere in your code, or perhaps a closing bracket (the “}” character i.e.) for some loop in your code etc.

   HTH,
   Shanx
7. wyldie says:

   Apr 12, 2004 at 7:25 am

   To pull distances from one UK postcode to another use the following SQL dump:-
   http://www.good-stuff.co.uk/useful/pcodedb.sql.gz

   Now, presuming you’re using MySQL, use the following converted code to have MySQL do all the math on your behalf.

   SELECT
   a.Town,
   a.Postcode,
   b.lat_dec AS lat1, b.long_dec AS long1,
   c.lat_dec AS lat2, c.long_dec AS long2,
   round(degrees(acos(sin(radians(b.lat_dec))
   * sin(radians(c.lat_dec))
   + cos(radians(b.lat_dec))
   * cos(radians(c.lat_dec))
   * cos(radians(b.long_dec – c.long_dec))))*69, 2) AS distance
   FROM
   postcode_names a, postcode_data b, postcode_data c
   WHERE
   c.Postcode = ‘NR32′
   AND
   a.Postcode = b.Postcode
   AND
   round(degrees(acos(sin(radians(b.lat_dec))
   * sin(radians(c.lat_dec))
   + cos(radians(b.lat_dec))
   * cos(radians(c.lat_dec))
   * cos(radians(b.long_dec – c.long_dec))))*69, 2)
8. Stew says:

   Apr 20, 2004 at 7:24 pm

   Thats very nice but when I ran the query I got every postcode returned.

   I found adding a ‘HAVING’ clause to the query did the trick – viz…

   SELECT
   a.Town,
   a.Postcode,
   b.lat_dec AS lat1, b.long_dec AS long1,
   c.lat_dec AS lat2, c.long_dec AS long2,
   round(degrees(acos(sin(radians(b.lat_dec))
   * sin(radians(c.lat_dec))
   + cos(radians(b.lat_dec))
   * cos(radians(c.lat_dec))
   * cos(radians(b.long_dec – c.long_dec))))*69, 2) AS distance
   FROM
   postcode_names a, postcode_data b, postcode_data c
   WHERE
   c.Postcode = ‘NR32′
   AND
   a.Postcode = b.Postcode
   AND
   round(degrees(acos(sin(radians(b.lat_dec))
   * sin(radians(c.lat_dec))
   + cos(radians(b.lat_dec))
   * cos(radians(c.lat_dec))
   * cos(radians(b.long_dec – c.long_dec))))*69, 2)
   HAVING (distance
9. win says:

   Jul 05, 2004 at 12:57 pm

   How do you modify above script, if I want to display distance from 1 city to another. Not all.

   Having clause is missing from previous post.
   I’m having same thing happening. Shows all results, and if I use limit 1 it shows first record info.
10. Tony Fox says:

    Jan 19, 2005 at 7:26 pm

    Looks like the end of the script has been missed off. Try This:-

    SELECT a.Town,a.Postcode,b.lat_dec AS lat1,b.long_dec AS long1,c.lat_dec AS lat2,c.long_dec AS long2,round(degrees(acos(sin(radians(b.lat_dec))* sin(radians(c.lat_dec))+cos(radians(b.lat_dec))* cos(radians(c.lat_dec))*cos(radians(b.long_dec – c.long_dec))))*69, 2) AS distance FROM postcode_names a,postcode_data b,postcode_data c WHERE c.Postcode = ‘NR32′ AND a.Postcode = b.Postcode AND round(degrees(acos(sin(radians(b.lat_dec)) * sin(radians(c.lat_dec)) + cos(radians(b.lat_dec)) * cos(radians(c.lat_dec)) * cos(radians(b.long_dec – c.long_dec))))*69, 2) HAVING (distance
11. Tony Fox says:

    Jan 19, 2005 at 8:57 pm

    The end of the script will not show because of the less than symbol

    add at the end “less than” 120)
12. Dan Karran says:

    Jul 05, 2005 at 7:50 am

    Thanks. I found your page via Google and implemented the code in SQL (before realising someone else had done the work for me in your comments). I’m using it to calculate the distance between places I’ve taken photographs, and displaying the distances alongside a map using the new Google Maps API interface.

    http://www.dankarran.com/photography/map/?town=castletown
13. Mark Waterhouse says:

    Aug 13, 2005 at 9:29 pm

    Hi
    Does anyone have a copy of the postcode sql database? Its no longer available at the URL above.

    Thanks
    Mark
14. nik says:

    Sep 29, 2005 at 1:40 am

    postcode data and sql file at the attached url

    http://www.easypeasy.com/guides/article.php?article=64