Compare one query with multiple results in PHP - php

Question

I have two text input. Like This:
So, I have some kind of dynamic ajax search. I pass the input data and make two different mysql select. Something like this:
Table finalmap - SELECT 1
id -------- latitud-----longitud---
1 | 6.2523915 | -75.5737028 |
2 | 6.2640349 | -75.5990783 |
3 | 6.2642411 | -75.5999791 |
4 | 6.2638461 | -75.5982590 |
-------------------------------------
Table finalmap - SELECT 2
id -------- latitud-----longitud---
6 | 6.262669 | -75.596799 |
7 | 6.258019 | -75.598001 |
8 | 6.253668 | -75.599374 |
9 | 6.250724 | -75.602335 |
-------------------------------------
So, I want to compare every single "latitud and longitud field" with all the "latitud" and "longitud" fields of the SELECT2:
I have this Php, I have to make some improvements but can say that it worked:
<?php
$buscar = $_POST['b'];
$buscarcarrera = $_POST['c'];
$whatIWant = substr($buscar, strpos($buscar, "Calle") + 5);
$whatIWant2 = substr($buscarcarrera, strpos($buscarcarrera, "Carrera") + 5);
$vacio = "Calle50A";
$vacioc = "Carrera50A";
if (preg_match('/[A-Za-z]/', $whatIWant))
{
buscar($buscar, "", $buscarcarrera, "");
}
else
{
buscar($buscar, $vacio, $buscarcarrera, $vacioc);
}
function buscar($b, $exclusion, $buscarcarrera, $exclusion2)
{
$con = mysql_connect('localhost', 'root', '');
mysql_select_db('map', $con);
$sql = mysql_query("SELECT * FROM finalmap WHERE calle LIKE '%" . $b . "%' AND calle not in ('$exclusion')", $con);
$contar = mysql_num_rows($sql);
if ($contar == 0)
{
echo "No se han encontrado resultados para '<b>" . $b . "</b>'.";
}
else
{
while ($row = mysql_fetch_array($sql))
{
$nombre = $row['calle'];
$id = $row['id'];
$lat = $row['latitud'];
$lon = $row['longitud'];
}
}
$sql2 = mysql_query("SELECT * FROM finalmap WHERE calle LIKE '%" . $buscarcarrera . "%' AND calle not in ('$exclusion2')", $con);
$contar2 = mysql_num_rows($sql2);
if ($contar2 == 0)
{
echo "No se han encontrado resultados para '<b>" . $b . "</b>'.";
}
else
{
while ($row2 = mysql_fetch_array($sql2))
{
$nombre2 = $row2['calle'];
$id2 = $row2['id'];
$lat2 = $row2['latitud'];
$lon2 = $row2['longitud'];
}
}
}
function distance($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;
}
}
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "M") . " Miles<br />";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br />";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "N") . " Nautical Miles<br />";
?>
Then: how could compare each of the values ​​using the function to determine the proximity between the coordinates (using my distance() function)?. My problem is I don't know how to compare distance between each point of the 1st query with each point of the 2nd query, with every possible combinations.
I want to have something like this function compare (lat1,lon1,lat2,lon2);
(lat1,lon1,lat3,lon3),(lat1,lon1,lat4,lon4),(lat1,lon1,lat5,lon5),(lat1,lon1,lat6,lon6) and so on.
Thank you very much in advance for any help given

Answer 1

Your logic is completely wrong and you are way complexifying what you are looking for. It s much simpler than you think.
First: in order to drastically speed up your script and queries as well as protecting it from any attacks, use PDO prepared statements and security protocols. Mysql none prepared statements are deprecated for more than a decade !!!
My experience (25 years of coding) shows me to never ever trust what s coming into your server so I am a little phobic about security....
This tutorial explains the first basic steps you need to do and explains you this script as it should be (copy past all codes as they are).
Note, Using a local IP address instead of Localhost will speed up mysql connections by 2 to 5 depending on dev platform (lamp,wamp etc...)
set execution parameters
ini_set('memory_limit', '64M'); // memory limit used by script
set_time_limit(0); // no time limit
Establish Your secured mysql connection
try {
$con = new PDO('mysql:host=127.0.0.1;dbname=map;charset=UTF8','root','password');
$con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
echo 'Connection failed: ' . $e->getMessage();
}
Establish a second level security protocol to check if it s really an ajax call
$isAjax = isset($_SERVER['HTTP_X_REQUESTED_WITH']) AND
strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) === 'xmlhttprequest';
if (!$isAjax) {
$user_error = 'Access denied - not an AJAX request from allowed computer...';
trigger_error($user_error, E_USER_ERROR);
exit;
}
Second: Use the distance function INSIDE your mysql query in order to properly sort the results you are looking for. Please note the mysql distance used here is different from the one you are currently using (faster and more precise).
It will gives you the distance between reference points from table 1 and all others from Table 2 in km, ordered by distance (closest as first row).
We suppose you get your results from your ajax like this:
if($isAjax){
$param4 = '%' . $_POST['b']. '%'; // MYSQL LIKE bind param buscar
$param5 = '(' . $_POST['c'] . ')'; //MYSQL NOT IN bind param buscarcarrera
}
We suppose you get your results from first query as this (simplified):
$sql1_array = array(
array('id' => 1, 'Lati' => 6.2523915, 'Longi' => -75.5737028),
array('id' => 2, 'Lati' => 6.2640349, 'Longi' => -75.5990783)
);
(note: always use "lati" and "longi" to avoid collision with PHP function like "long")
Your prepared statement:
$sql2 = "SELECT *, ( 6371 * acos( cos(radians(?))
* cos(radians(Latitude)) * cos(radians(Longitude) - radians(?)) +
sin( radians(?) ) * sin(radians(Latitude)))) AS distance
FROM finalmap WHERE calle LIKE ? AND calle NOT IN ?
ORDER BY distance LIMIT 10";
$stmt = $con->prepare($sql2);
Third: Execute a loop with each query1 result to get the distances compared with table 2. In the loop, you bind your parameters to your prepared stmt. This will drastically speed up your queries as connection with database is persistent and the query is prepared with PDO (compiled and in memory).
foreach ($sql1_array as $key => $value) {
$stmt->bindParam(1, $value['Lati'], PDO::PARAM_INT);
$stmt->bindParam(2, $value['Longi'], PDO::PARAM_INT);
$stmt->bindParam(3, $value['Lati'], PDO::PARAM_INT);
$stmt->bindParam(4, $param4, PDO::PARAM_STR);
$stmt->bindParam(5, $param5, PDO::PARAM_STR);
$stmt->execute();
$count = $stmt->rowCount();
if ($count >= 1) {
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
}
}
/* $result ordered by distance */
|**************************************|
|field 0 | field 1 | field 2 | distance|
|**************************************|
| 1 | a1 | a2 | 0.00123 | 1 meters
| 2 | b1 | b2 | 0.01202 | 12 meters
| 3 | c1 | c2 | 0.23453 | 234 meters
| 4 | d1 | d2 | 1.58741 | 1km and 587 meters
|**************************************|
I think you are all set ! Enjoy.

Answer 2

You should do something like this :
function buscar($b, $exclusion, $buscarcarrera, $exclusion2)
{
$nombre1 = array();
$id1= array();
$lat1= array();
$lon1= array();
$nombre2 = array();
$id2= array();
$lat2= array();
$lon2= array();
$con = mysql_connect('localhost', 'root', '');
mysql_select_db('map', $con);
$sql = mysql_query("SELECT * FROM finalmap WHERE calle LIKE '%" . $b . "%' AND calle not in ('$exclusion')", $con);
$contar = mysql_num_rows($sql);
if ($contar == 0)
{
echo "No se han encontrado resultados para '<b>" . $b . "</b>'.";
} else {
while ($row = mysql_fetch_array($sql))
{
if(array_key_exists('calle',$row) && array_key_exists('id',$row) && array_key_exists('latitud',$row) && array_key_exists('longitud',$row)) {
// ignore data with a missing field
$nombre1[] = $row['calle'];
$id1[] = $row['id'];
$lat1[] = $row['latitud'];
$lon1[] = $row['longitud'];
}
}
}
$sql2 = mysql_query("SELECT * FROM finalmap WHERE calle LIKE '%" . $buscarcarrera . "%' AND calle not in ('$exclusion2')", $con);
$contar2 = mysql_num_rows($sql2);
if ($contar2 == 0)
{
echo "No se han encontrado resultados para '<b>" . $b . "</b>'.";
} else {
while ($row2 = mysql_fetch_array($sql2))
{
if(array_key_exists('calle',$row2) && array_key_exists('id',$row2) && array_key_exists('latitud',$row2) && array_key_exists('longitud',$row2)) {
// ignore data with a missing field
$nombre2[] = $row2['calle'];
$id2[] = $row2['id'];
$lat2[] = $row2['latitud'];
$lon2[] = $row2['longitud'];
}
}
}
$nbData1 = count($nombre1);
$nbData2 = count($nombre2);
if($nbData1 > 0 && $nbData2 > 0) {
// there are data in both 1st and 2nd query
$distances = array();
// loop on all data from 1st query
for($i1 = 0; $i1 < $nbData1; $i1++) {
// loop on all data from 2nd query
for($i2 = $i1; $i2 < $nbData2; $i2++) {
// storing result of the distance
$distances[] = array('i1' => $i1,
'i2' => $i2,
'dist' => array('M' => distance($lat1[$i1], $lon1[$i1], $lat2[$i2], $lon2[$i2], 'M'),
'K' => distance($lat1[$i1], $lon1[$i1], $lat2[$i2], $lon2[$i2], 'K'),
'N' => distance($lat1[$i1], $lon1[$i1], $lat2[$i2], $lon2[$i2], 'N')
)
)
}
}
}
foreach($distances as $distance) {
// use each result
$i1 = $distance['i1'];
$lat1 = $distance['lat1'];
$lon1 = $distance['lon1'];
$i2 = $distance['i1'];
$lat2 = $distance['lat2'];
$lon2 = $distance['lon2'];
$dist = $distance['dist'];
$distM = $dist['M'];
$distK = $dist['K'];
$distN = $dist['N'];
echo "Have fun with theses data !\n";
}
}
So, storing the content of your rows.
Then parse all your data, to calculate all possible distances.
Then use the result as you wish.

Answer 3

This is much better to process distance calculation in the 2nd query, and this is faster, so the distance() function not needed. Also it seems that your formula is not correct.
$buscar = $_POST['b'];
$buscarcarrera = $_POST['c'];
$whatIWant = substr($buscar, strpos($buscar, "Calle") + 5);
$whatIWant2 = substr($buscarcarrera, strpos($buscarcarrera, "Carrera") + 5);
$vacio = "Calle50A";
$vacioc = "Carrera50A";
if (preg_match('/[A-Za-z]/', $whatIWant))
{
buscar($buscar, "", $buscarcarrera, "");
}
else
{
buscar($buscar, $vacio, $buscarcarrera, $vacioc);
}
function buscar($b, $exclusion, $buscarcarrera, $exclusion2)
{
$con = mysql_connect('localhost', 'root', '');
mysql_select_db('map', $con);
$sql = mysql_query("SELECT * FROM finalmap WHERE calle LIKE '%" . $b . "%' AND calle not in ('$exclusion')", $con);
$contar = mysql_num_rows($sql);
if ($contar == 0)
{
echo "No se han encontrado resultados para '<b>" . $b . "</b>'.";
}
else
{
while ($row = mysql_fetch_array($sql))
{
$nombre = $row['calle'];
$id = $row['id'];
$lat = $row['latitud'];
$lon = $row['longitud'];
$dLat = '((latitud-'.$lat.')*PI()/180)';
$dLong = '((longitud-'.$lon.')*PI()/180)';
$a = '(sin('.$dLat.'/2) * sin('.$dLat.'/2) + cos('.$lat.'*pi()/180) * cos(map_coords_1*pi()/180) * sin('.$dLong.'/2) * sin('.$dLong.'/2))';
$sub_sql = '2*atan2(sqrt('.$a.'), sqrt(1-'.$a.'))';
$results = mysql_query(
"SELECT DISTINCT
id, calle, " . $sub_sql . " AS distance FROM finalmap
WHERE
calle LIKE '%" . $buscarcarrera . "%' AND calle not in ('$exclusion2')
ORDER BY
distance
", $con);
if ($results == 0)
{
echo "No se han encontrado resultados para '<b>" . $buscarcarrera . "</b>'.";
}
else
{
while ($row2 = mysql_fetch_array($results)) {
echo "Calle: " . $row2['calle'] . " Miles - " . $row2['distance']*3956;
echo "Calle: " . $row2['calle'] . " Kilometers - " . $row2['distance']*6367;
echo "Calle: " . $row2['calle'] . " Nautical Miles - " . $row2['distance']*3435;
}
}
}
}
}

Answer 4

I'm giving it a first try.
$output_array = array();
$i = 0;
while ($row = mysql_fetch_array($sql))
{
$j = 0;
while ($row2 = mysql_fetch_array($sql2))
{
$out_array[$i][$j][0] = $row2['latitud'] - $row['latitud']
$out_array[$i][$j][1] = $row2['longitud'] - $row['longitud']
++$j;
}
++$i;
}
print "<pre>";
print_r($out_array);
print "</pre>";
Preview (with random numbers) :
Array
(
[0] => Array
(
[0] => Array
(
[0] => 6
[1] => 9
)
[1] => Array
(
[0] => 6
[1] => 9
)
[2] => Array
(
[0] => 6
[1] => 9
)
............
............
............
Array[0][0][0] == difference between 1st result latitud(table 1) AND 1st result latitud(table 2)
Array[0][1][0] == difference latitud (1st res table1) AND (2nd res table2)
Same case for longitud.

Answer 5

I think cross join of table will do your work.
Let say first table is Table1 and second is Table2.
So the query:
Select Table1.id, Table1.latitud, Table1.longitude,
Table2.id, Table2.latitud, Table2.longitude
from
Table1 , Table2
This will give you all combination which you required.

Answer 6

I think that INNER JOIN will work. I'm just confused that you will provide the table-1 long and lat value or not.So I 'm writing both the options.Let's say first table is table1 and second is table2
When you will provide the lat and long value of table 1SELECT * FROM table1 AS t1 INNER JOIN table2 AS t2 ON t1.latitude=t2.latitud OR t2.longitude=t1.longitude WHERE t1.latitude='#enter the value#' AND t1.longitude='#enter the value#'
It will provide the filter result of the equal value in both tableSELECT * FROM table1 AS t1 INNER JOIN table2 AS t2 ON t1.latitude=t2.latitud OR t2.longitude=t1.longitude
Let me know if needs anything else.Thanking you.

Answer 7

You can easily generate the necessary result set in SQL without having to write a bunch of PHP code though.
It would also make thing a lot easier if you would actually show what you want your results to look like as a table. I'm confused as to what you're actually trying to do. My assumption is that you want to match the two queries when either the latitudes match exactly OR the longitudes match exactly.
Update: I found the SQL buried in your PHP code. Something like this should work:
"SELECT t1.latitud AS lat1, t1.longitud AS lon1,
t2.latitud AS lat2, t2.longitud AS lon2
FROM finalmap AS t1
INNER JOIN finalmap AS t2 ON t1.latitud=t2.latitud OR t1.longitud=t2.longitud
WHERE t1.calle LIKE '%" . $b . "%' AND t1.calle not in ('$exclusion')
AND t2.calle LIKE '%" . $buscarcarrera . "%' AND t2.calle NOT IN ('exclusion2');"
If you want all results from SELECT 1 even if there are no matches in SELECT 2 change the INNER JOIN to a LEFT JOIN.
I also noticed you're checking if fields have values after the query is returned. You can do those checks in the WHERE clause on the SQL query that way you don't get any results you don't actually want. It just depends on what a "no-value" looks like (NULL? empty string? etc.) as to how you would check that.

Answer 8

You can take Cartesian product of both tables so that you can find all possible combinations.
Otherwise, take two loops like
for(int i=0;i < n; i++)
for(int j=0; j< m; j++)
someProcess(a[i][0], a[i][1], b[j][0], b[j][1]);
And this will give you all possible combinations.
Your function should then have logic for geocoding/reverse geocoding to calculate distance.
I hope this helps!!

Answer 9

I'd support this answer, but add the other major point:
Fourth: do not use distance calculation and sorting with customly coded haversine formula neither in PHP, nor in MySQL. MySQL has spatial functions support for a long time. Moreover, when dealing with GIS it's better to move to PostgreSQL + PostGIS extension
Fifth: Utilize MySQL or any other RDBMS spatial indexes when sorting by distance. I doubt you can code anything more efficient with your "every-to-every distance comparison" approach.
Solution: review this answer and apply the following queries to your DB table:
ALTER TABLE `finalmap` ENGINE=MYISAM; -- changing your storage engine to MyISAM. It supports spatial indexes in MySQL
ALTER TABLE `finalmap` ADD `pt` POINT NOT NULL; -- adding POINT() spatial datatype for zip cetner. Eventually, you may remove the old lat/lng decimal columns
UPDATE `finalmap` SET `pt` = POINT(latitud,longitud);
ALTER TABLE `finalmap` ADD SPATIAL INDEX(`pt`);
Then, assuming you get a coordinates in input (let's say (6.25,-75.6)) that you need to compare to all records in your table and find the closest one, you need:
SELECT *, ST_Distance_Sphere(`pt`,POINT('6.25','-75.6')) as `dst`
FROM `fnialmap`
ORDER BY `dst`
LIMIT 1;
And, finally, sixth: please name your columns and variables in English. It may be so that someone else supports your code in the future and they have to deal with "latitud" and "longitud"