need help creating a mysql query - php

Question

Let's assume I a have a table called "users" which has the following relevant columns:
username
zipcode
Let's also assume I have a table called "shops" which has the following relevant columns:
shop_name
zipcode
Let's also assume I have a table called "preferred_shops" which has the following relevant columns:
shop_name
username
Let's also assume I have a table called "zipData" which has the following relevant columns:
zipcode
lat
lon
Currently I can run this code with success:
$lat="49.886436"; // Value should be obtained from the zipData table by doing a "SELECT FROM zipData WHERE zipcode=(users_zip_code)" or similar query
$lon="-97.14553"; // Value should be obtained from the zipData table by doing a "SELECT FROM zipData WHERE zipcode=(users_zip_code)" or similar query
$radius=5;
$query="SELECT zipcode FROM zipData
WHERE (POW((69.1*(lon-\"$lon\")*cos($lat/57.3)),\"2\")+
POW((69.1*(lat-\"$lat\")),\"2\"))<($radius*$radius)";
The query above will successfully display all the other zip codes that are within the given radius of the supplied lat/lon but I have no interest in knowing what other zip codes there are... I want to know what shops are within the radius.
Any help you can give would be greatly appreciated.
==== Edited because I realized it is actually a little more complex ====
This is what I need to do...
Retrieve the user's zipcode from "users" table
Find the user's lat/lon from the zipData table
Find all the shops within a given radius that is also in a "preferred_shops" table with the given user in the "username" column
Problems to consider: the zipcode in "zipData" has no white space but the zipcodes in "users" and "shops" have white space for cosmetic reasons... such as Canadian Postal codes which are in the "A1A 1A1" format
==== Edited to post solution ====
The system says I am not allowed to answer my own question. That sounds strange since with the help of others I already found the answer. As a work around I am editing the original post. Here is the solution...
Ok so I figured it out (with the help of the people that replied)... This is what I did.
IT LIKELY COULD BE A LOT SHORTER AND CLEANER SO PLEASE FEEL FREE TO MAKE IT BETTER IF YOU KNOW A BETTER WAY.
$query = "SELECT * FROM preferred_shops
WHERE `username`='{$_SESSION['account_name']}'
";
$result = mysql_query($query);
if (mysql_errno())
{
die( "ERROR ".mysql_errno($link) . ": " . mysql_error($link) );
}
$num_rows = mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
// Get user's preferences and postal code
$query2 = "SELECT * FROM seekers
WHERE `username`='{$_SESSION['account_name']}'
";
$result2 = mysql_query($query2);
if (mysql_errno())
{
die( "ERROR ".mysql_errno($link) . ": " . mysql_error($link) );
}
$num_rows2 = mysql_num_rows($result2);
$row2 = mysql_fetch_array($result2);
$radius=$row2['radius']; // Didn't mention that is column was in the table but that didn't matter... the value could have come from anywhere.
// Get user's lat/lon
// Remove white space from the postal code
$query3="SELECT * FROM zipData WHERE zipcode=replace('{$row2['postal']}',' ','')";
$result3 = mysql_query($query3);
if (mysql_errno())
{
die( "ERROR ".mysql_errno($link) . ": " . mysql_error($link) );
}
$num_rows3 = mysql_num_rows($result3);
$row3 = mysql_fetch_array($result3);
$lat=$row3["lat"];
$lon=$row3["lon"];
$query4="SELECT shop_name FROM zipData,shops
WHERE (POW((69.1*(lon-\"$lon\")*cos($lat/57.3)),\"2\")+POW((69.1*(lat-\"$lat\")),\"2\"))<($radius*$radius)
AND replace(shops.zipcode,' ','') = zipData.zipcode
AND shops.shop_name={$row['shop_name']}
";
$result4 = mysql_query($query4);
if (mysql_errno())
{
die( "ERROR ".mysql_errno($link) . ": " . mysql_error($link) );
}
$num_rows4 = mysql_num_rows($result4);
$num_jobs=$num_rows4;
$i=0;
while($row4 = mysql_fetch_array($result4))
{
$shopArray[$i]=$row4["shop_name"];
$i++;
}
var_dump($shopArray);
}

Answer 1

You need to join to the shops table
SELECT shop_name FROM zipData,shops
WHERE (POW((69.1*(lon-\"$lon\")*cos($lat/57.3)),\"2\")+
POW((69.1*(lat-\"$lat\")),\"2\"))<($radius*$radius) and
shops.zipcode = zipdata.zipcode

Answer 2

select shopname from shops where zipcode in ( <your other working query> )