Merge these 2 MySQL query into one - php

Question

I explain you the problem. I have two tables : "listing" who contain all my lists with different informations and particulary one : a city ID. I have a second table "city" who contain the complete list of all cities of my countries with many informations : latitude, longitude, city name, etc.
I want to show all listing ordered by distance between actual point (geolocalised, I have lat/lng) and cities in the listing table.
Actualy, I do this with 2 MySQL queries, and it's work :
$formula = "(6366*acos(cos(radians($latitude))*cos(radians(`latitude`))*cos(radians(`longitude`) -radians($longitude))+sin(radians($latitude))*sin(radians(`latitude`))))";
$q = "SELECT *,$formula AS dist FROM ".$this->db_city." WHERE $formula<='$distance' ORDER by dist ASC";
$q = $this->db->query($q);
$resultSQL = $q->result_array();
And this one :
if ($localization != ''){
if ($whereClause == ''){
//$whereClause .= ' WHERE address_city LIKE "'.$localization.'"';
$loc = '';
foreach ($localization as $key => $value) {
if ($loc == ''){
$loc .= '"'.$value.'"';
}else{
$loc .= ', "'.$value.'"';
}
}
$whereClause .= ' WHERE address_city IN ('.$loc.')';
}else{
//$whereClause .= ' && address_city LIKE "'.$localization.'"';
$loc = '';
foreach ($localization as $key => $value) {
if ($loc == ''){
$loc .= '"'.$value.'"';
}else{
$loc .= ', "'.$value.'"';
}
}
$whereClause .= ' && address_city IN ('.$loc.')';
}
}
$q = "SELECT * FROM ".$this->db_listing.$whereClause." ORDER BY created_stamp DESC";
It's work, but the problem is I don't have the "dist" parameter accessible in this second query, so I can't order by dist. The solution for that is to merge the first big query (the one with the formula) with the second.
I tried with LEFT JOIN, but it didn't work. Can you help me please ? Thanks !

Answer 1

Try this, it will going to return you all the cities those with in the radius of $distance from provided longitude and latitudes.
"SELECT *,(((acos(sin(($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) AS distance FROM location WHERE (((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) <= $distance"
Thanks.