sql - How to avoid multiple column in correlated sub query assignment in MySQL update -

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i'm attempting assign closest location community based on community postcode , using haversine formula sql described here. need return single scalar value can't seem avoid having second calculated distance value needed determine closest location. help.

update community c     join postcode p on p.id = c.postcode_id     join (         select 100.0 radius, 111.045 distance_unit     ) set c.location_id = (     select l.id,         a.distance_unit             * degrees(acos(cos(radians(p.latitude))             * cos(radians(l.latitude))             * cos(radians(p.longitude - l.longitude))             + sin(radians(p.latitude))             * sin(radians(l.latitude)))) distance     location l     l.latitude         between p.latitude  - (a.radius / a.distance_unit)             , p.latitude  + (a.radius / a.distance_unit)     , l.longitude         between p.longitude - (a.radius / (a.distance_unit * cos(radians(p.latitude))))             , p.longitude + (a.radius / (a.distance_unit * cos(radians(p.latitude))))     having distance <= a.radius     order distance     limit 1 )

using structure have, need move distance calculation where , order by clauses:

set c.location_id = (     select l.id     location l     l.latitude         between p.latitude  - (a.radius / a.distance_unit)             , p.latitude  + (a.radius / a.distance_unit)     , l.longitude         between p.longitude - (a.radius / (a.distance_unit * cos(radians(p.latitude))))             , p.longitude + (a.radius / (a.distance_unit * cos(radians(p.latitude))))     , a.distance_unit             * degrees(acos(cos(radians(p.latitude))             * cos(radians(l.latitude))             * cos(radians(p.longitude - l.longitude))             + sin(radians(p.latitude))             * sin(radians(l.latitude)))) <= a.radius     order a.distance_unit             * degrees(acos(cos(radians(p.latitude))             * cos(radians(l.latitude))             * cos(radians(p.longitude - l.longitude))             + sin(radians(p.latitude))             * sin(radians(l.latitude))))      limit 1 )

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