Find all nearest zip codes with lat and long from a particular latitude and longitude in sql query - mysql

Question

I've a database table in which there are 4 columns.
1. id
2. person_name
3. country
4. zip_code
Now I want all the zip codes with their real latitude and longitude which come in a given radius of 10 mile from a given lat long.
suppose my latitude and longitudes are (19.24947300,72.85681400) and distance is 10 mile, then what SQL query should I make to return all nearest zip codes and their lat longs.
I only have the following query
SELECT ((ACOS(SIN($lat * PI() / 180) * SIN(lat * PI() / 180) + COS($lat * PI()
/ 180) * COS(lat * PI() / 180) * COS(($lon – lon) * PI() / 180)) * 180 / PI()) * 60 *
1.1515) AS `distance` FROM `members` HAVING `distance`<=’10’ ORDER BY `distance` ASC
But it requires the lat longs of all zip codes in the table but I want them in run time.

Answer 1

I'm not sure I understand what you mean by
But it requires the lat longs of all zip codes in the table but I want them in run time.
Since you want a SQL query, I assume that you have a table with every zip code's latitude and longitude.
You probably shouldn't do all that math in your SQL query. I would first pull every zip code that is in a 10 mile x 10 mile square centered in your queried position and then, if you really need just those that are positively in a 10 mile radius, check in your program (I assume you are using PHP, since you have variables starting with $).
SELECT zip, latitude, longitude FROM zip_positions
WHERE latitude > 19.24947300 - DISTANCE and latitude < 19.24947300 + DISTANCE
AND longitude > 72.85681400 - DISTANCE and longitude < 72.85681400 + DISTANCE
where DISTANCE is a rough equivalent to 10 miles in latitude/longitude that you would have to calculate.
Then you would, if you need it, check in PHP (or whatever language you're using) every result to see if it is indeed less than 10 miles from the point queried. I can't help you as to how to precisely convert latitude and longitude to distance in miles.

Answer 2

I assume this abandoned question was resolved, but let me share my opinions here in case someone else is wrestling with this one. I present 3 options for letting closet zip from DB using lat long as the lookup:
DB LOOKUP OPTION 1 (QUICK AND DIRTY):
SET #lat = 41.675868;
SET #long = -73.864484;
SET #tol = .05;
SELECT * FROM mytable.ZipCodes
WHERE latitude >= #lat AND latitude <= #lat + #tol
AND longitude >= #long AND longitude <= #long + #tol
ORDER BY (ABS(latitude-longitude) - ABS(#lat-#long)) ASC
LIMIT 1
OPTION #2 (VARIATION ON #1)
$qry = "SELECT zipcode,cityname,stateabbr FROM dealer-site.ZipCodes WHERE (latitude >= {$lat} AND latitude <= {$lat}+1) AND (ABS(longitude) >= ABS({$long}) AND ABS(longitude) < ABS({$long})+1) ORDER BY latitude ASC LIMIT 1”;
OPTION #3 BEST METHOD CALCULATE . I WOULD TRY TO USE THIS:
SET #latitude = 41.675868;
SET #longitude = -73.864484;
SELECT
*
FROM
(SELECT
*,
(((ACOS(SIN((#latitude * PI() / 180)) *
SIN((latitude * PI() / 180)) + COS((#latitude * PI() / 180)) * COS((latitude * PI() / 180)) * COS(((#longitude - longitude) * PI() / 180)))) * 180 / PI()) * 60 * 1.1515 * 1.609344) AS distance
FROM
mytable.ZipCodes) ZipCodes
WHERE
distance <= 5
LIMIT 10
PHP MYSQL (LEGACY CALL) :
$qry = "SELECT zipcode
FROM (SELECT *,
(((ACOS(SIN(({$lat} * PI() / 180)) *
SIN((`latitude` * PI() / 180)) + COS(({$lat} * PI() / 180)) * COS((`latitude` * PI() / 180)) *
COS((({$long} - `longitude`) * PI() / 180)))) * 180 / PI()) * 60 * 1.1515 * 1.609344)
AS distance
FROM
`MyDB`.ZipCodes) ZipCodes
WHERE distance <= 5
LIMIT 1
";