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@OllieJones
Last active September 25, 2023 14:09
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Fast nearest-location finder for SQL (MySQL, PostgreSQL, SQL Server)
SELECT zip, primary_city,
latitude, longitude, distance
FROM (
SELECT z.zip,
z.primary_city,
z.latitude, z.longitude,
p.radius,
p.distance_unit
* DEGREES(ACOS(LEAST(1.0, COS(RADIANS(p.latpoint))
* COS(RADIANS(z.latitude))
* COS(RADIANS(p.longpoint - z.longitude))
+ SIN(RADIANS(p.latpoint))
* SIN(RADIANS(z.latitude))))) AS distance
FROM zip AS z
JOIN ( /* these are the query parameters */
SELECT 42.81 AS latpoint, -70.81 AS longpoint,
50.0 AS radius, 111.045 AS distance_unit
) AS p ON 1=1
WHERE z.latitude
BETWEEN p.latpoint - (p.radius / p.distance_unit)
AND p.latpoint + (p.radius / p.distance_unit)
AND z.longitude
BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
) AS d
WHERE distance <= radius
ORDER BY distance
LIMIT 15
SELECT z.zip,
z.primary_city,
z.latitude, z.longitude,
p.distance_unit
* DEGREES(ACOS(LEAST(1.0, COS(RADIANS(p.latpoint))
* COS(RADIANS(z.latitude))
* COS(RADIANS(p.longpoint) - RADIANS(z.longitude))
+ SIN(RADIANS(p.latpoint))
* SIN(RADIANS(z.latitude))))) AS distance_in_km
FROM zip AS z
JOIN ( /* these are the query parameters */
SELECT 42.81 AS latpoint, -70.81 AS longpoint,
50.0 AS radius, 111.045 AS distance_unit
) AS p ON 1=1
WHERE z.latitude
BETWEEN p.latpoint - (p.radius / p.distance_unit)
AND p.latpoint + (p.radius / p.distance_unit)
AND z.longitude
BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
ORDER BY distance_in_km
LIMIT 15
@joaomarcos96
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@LostInDarkMath in my case I had to use HAVING to take advantage of the indices.

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