# Find features within given coordinates and distance using MySQL

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?

• Why not PostGIS ? If you are starting geo project you can still change your stack Jan 14, 2013 at 7:32
• stackoverflow.com/a/40272394/1281385 Should be useful in speeding this query up (if needed) Oct 26, 2016 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, 2020 at 8:03

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( lat ) )
* sin( radians( lat ) )
)
) AS distance
FROM markers
HAVING distance < 30
ORDER BY distance
LIMIT 0 , 20;
``````
• Your answer should be ordered 1st. Feb 5, 2019 at 12:06
• @AmitShah If you think that you could ping the asker (@shihabK who has not been active on the site for nearly 6 years) and/or vote for meta.stackexchange.com/questions/268666/…
– PolyGeo
Feb 6, 2019 at 23:33
• This should be the accepted answer. Sep 28, 2019 at 13:08
• Worked like a charm! Jan 24, 2021 at 22:03

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( lat ) )
* sin( radians( lat ) )
)
) AS distance
FROM markers
HAVING distance < 30
ORDER BY distance
LIMIT 0 , 20;
``````

• 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? Nov 29, 2014 at 2:09
• Found: I had the lng variable empty! sorry! Nov 29, 2014 at 10:38
• Exactly what i wanted, but whats the query performance overload for thousands of records? and how about the accuracy? Feb 5, 2019 at 12:01
• much better to replace it to 6371392.896 for searching by meters Feb 6, 2019 at 22:32
• Is't possible to reduce the distance result like in my case from 2.71250308462937983478013848070986568927764892578125 to 2.71 ? Feb 28, 2020 at 19:45

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
``````
• 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 Oct 24, 2020 at 15:53
• @Ricardo correct, you would put the spatial function within the order by clause. Be sure that you also have a spatial index as well. Oct 28, 2020 at 6:11

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(
);

RETURN dist;

END
``````
• answer will be upvoted if you fix codestyle. it's right way Feb 6, 2019 at 23:31

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`
``````
• It is easier but ignores the fact that the earth is curved. Jun 3, 2015 at 13:35
• Need a formula to be accurate. Maybe this would be good on short distances only :D Dec 27, 2015 at 16:49
• Ya'll gonna launch a missile, or something? Jun 11, 2019 at 14:12
• @DennisBraga - if so, then perhaps this question is off-topic, better suited to http://globalthermonuclearwar.stackexchange.com...? Aug 22, 2019 at 9:37