Geo Location Radius Search Using PHP and MySQL

SELECT latitude, longitude, SQRT(POW(69.1 * (latitude - $lat), 2) +POW(69.1 * ($long - longitude) * COS(latitude / 57.3), 2)) AS distanceFROM TableName HAVING distance < 25 ORDER BY distance;

Where

$lat = User latitude

$long = User longitude

You have two option to find job by location you required 1. Search users lat long2. Job users Posted Lat long (it may be job location or job posted user location from database)3. Query will be like - SELECT zip,Round(((ACOS(SIN('$lat' * PI() / 180) * SIN(latitude * PI() / 180) + COS('$lat' * PI() / 180) * COS(latitude * PI() / 180) * COS(('$lon'-longitude) * PI() / 180)) * 180 / PI()) * 60 * 1.1515),(2)) AS distance FROM Jobs Having distance <= 30 Else 1. You have to call all jobs data in single query.     foreach($joblist as $job){        $milesresult = $this->calculateDistance($user_lat,$user_lon,$job['latitude'],$jobr['longitude']);        $miles = explode("-",$vendor['miles']);         $vendor_max_miles = $miles[1];        }2. PHP function for lat long      function calculateDistance($lat1, $lon1, $lat2, $lon2, $unit) {            $theta = $lon1 - $lon2;            $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));            $dist = acos($dist);             $dist = rad2deg($dist);              $miles = $dist * 60 * 1.1515;            $unit = strtoupper($unit);            if ($unit == "K") {              return ($miles * 1.609344);            } else if ($unit == "N") {                return ($miles * 0.8684);              } else {                  return $miles;                }        }

I have gone with below -

SELECT id,       name,       lat,       lng,       ROUND((6371 * acos(                       cos(radians($lat)) * cos(radians(lat)) * cos(radians(lng) - radians($lng)) +                       sin(radians($lat)) * sin(radians(lat)))), (2)) AS distanceFROM jobsHAVING distance < 50ORDER BY distance;

I have done benchmarking with data and found this slightly faster than Mukesh's answer and 2x better than @jision's answer.