23

I have a MySQL table with user name, latitude and longitude of the user.

I would like to get a list of users who are inside the circle or square of a given latitude and longitude with given distance. For example, my input Lat = 78.3232 and Long = 65.3234 and distance = 30 miles. I would like to get the list of users who are inside 30 miles distance from the point 78.3232 and 65.3234.

Is it possible to solve this with single query? Or can you give me a hint start solving this query?

3
  • Why not PostGIS ? If you are starting geo project you can still change your stack Jan 14 '13 at 7:32
  • stackoverflow.com/a/40272394/1281385 Should be useful in speeding this query up (if needed)
    – exussum
    Oct 26 '16 at 21:30
  • I don't think he calculates it correctly. I am trying to calculate the distance between airports Bourgas (BOJ - 42.416668, 27.283333) and Varna (VAR - 43.237260, 27.829096). With this request ( 6371 * acos ( cos (radians (43.237260)) * cos (radians (lat)) * cos (radians (lon) - radians (27.829096)) + sin (radians (43.237260)) * sin (radians (lat)) ) ) AS distance the result is 101,52135729600961 KM and in google maps the distance is 78.88 km. ![enter image description here](i.stack.imgur.com/FGnCC.png) May 12 '20 at 8:03
43

Mapperz's answer is invalid. Sinus must be calculated from latitude and NOT from longitude. So corect SQL statement is:

SELECT
    id, (
      3959 * acos (
      cos ( radians(78.3232) )
      * cos( radians( lat ) )
      * cos( radians( lng ) - radians(65.3234) )
      + sin ( radians(78.3232) )
      * sin( radians( lat ) )
    )
) AS distance
FROM markers
HAVING distance < 30
ORDER BY distance
LIMIT 0 , 20;
4
38

The SQL statement that will find the closest 20 locations that are within a radius of 30 miles to the 78.3232, 65.3234 coordinate. It calculates the distance based on the latitude/longitude of that row and the target latitude/longitude, and then asks for only rows where the distance value is less than 30 miles, orders the whole query by distance, and limits it to 20 results. To search by kilometers instead of miles, replace 3959 with 6371.

SELECT
  id, (
    3959 * acos (
      cos ( radians(78.3232) )
      * cos( radians( lat ) )
      * cos( radians( lng ) - radians(65.3234) )
      + sin ( radians(78.3232) )
      * sin( radians( lat ) )
    )
  ) AS distance
FROM markers
HAVING distance < 30
ORDER BY distance
LIMIT 0 , 20;

This is using the Google Maps API v3 with a MySQL backend which your already have.

https://developers.google.com/maps/articles/phpsqlsearch_v3#findnearsql

6
  • I'm getting a syntax error in my select using this, "#1582 - Incorrect parameter count in the call to native function 'radians' what could it be?
    – bluantinoo
    Nov 29 '14 at 2:09
  • Found: I had the lng variable empty! sorry!
    – bluantinoo
    Nov 29 '14 at 10:38
  • Exactly what i wanted, but whats the query performance overload for thousands of records? and how about the accuracy?
    – Amit Shah
    Feb 5 '19 at 12:01
  • 1
    much better to replace it to 6371392.896 for searching by meters Feb 6 '19 at 22:32
  • Is't possible to reduce the distance result like in my case from 2.71250308462937983478013848070986568927764892578125 to 2.71 ?
    – Ali Adil
    Feb 28 '20 at 19:45
2

It might be base to create a function .. so you can reuse it other other areas. Also would make your query a bit cleaner... At least that is my 2 cents.

DELIMITER $$

CREATE FUNCTION calcDistance(lat FLOAT, lng FLOAT, pnt_lat FLOAT, pnt_lng FLOAT)
    RETURNS FLOAT
BEGIN

    DECLARE dist FLOAT;
    SET dist =
          3959 * acos(
                cos(radians(pnt_lat))
                * cos(radians(lat))
                * cos(radians(lng) - radians(pnt_lng))
            + sin(radians(pnt_lat))
                    * sin(radians(lat))
          );

    RETURN dist;

END
1
  • answer will be upvoted if you fix codestyle. it's right way Feb 6 '19 at 23:31
2

It's 2020, you should be using the built in spatial function of the RDBS. In this case you are after the ST_Distance function.

-- geo column in users table is of Geometry data type, with a spatial index
SET @g1 = ST_SRID(POINT(78.3232, 65.3234), 4326);
SELECT * from users WHERE ST_Distance(users.geo, @g1, 'foot') < 158400; -- 5280 feet per mile
2
  • What about if apart from that, we also want to sort them by distance? First one, closest one. Do we use ORDER BY...? Thanks a lot @JohnC
    – Ricardo
    Oct 24 '20 at 15:53
  • 1
    @Ricardo correct, you would put the spatial function within the order by clause. Be sure that you also have a spatial index as well.
    – John C
    Oct 28 '20 at 6:11
0

Here is my variant of query, seems a little bit easier (http://dexxtr.com/post/83498801191/how-to-determine-point-inside-circle-using-mysql)

SELECT 
    *
FROM 
    `locator`
WHERE
    SQRT(POW(X(`center`) - 49.843317 , 2) + POW(Y(`center`) - 24.026642, 2)) * 100 < `radius`
4
  • 5
    It is easier but ignores the fact that the earth is curved. Jun 3 '15 at 13:35
  • Need a formula to be accurate. Maybe this would be good on short distances only :D
    – Jethro
    Dec 27 '15 at 16:49
  • Ya'll gonna launch a missile, or something? Jun 11 '19 at 14:12
  • 2
    @DennisBraga - if so, then perhaps this question is off-topic, better suited to http://globalthermonuclearwar.stackexchange.com...?
    – ashleedawg
    Aug 22 '19 at 9:37

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