Finding all agents in database from longitude and latitude

I will give you a idea how it will work you can make the exact thing then:
1. read the values of $lat and $lon from the user where he is.
2. make the sql like below.
3. 'C' is the constant value which you can calculate and put directly. How?
    i. find the relation how lat,lon actuate in real distances, or
    ii. take lat,lon of two points on map at 10 miles apart and calculate:
      C = ((lat1 - lat2) * (lat1 - lat2) + (lon1 - lon2) * (lon1 - lon2);


SELECT * from users WHERE ((lat - $lat) * (lat - $lat) + (lon - $lon) * (lon - $lon)) > C;

i belive you can take forward from here

yes its < and > thanks XzKto.

but C*C is again a Constant :) so why give that burden to the sql server :) we can calculate and give it directly. if you dont like C we can make it D :) where D = C * C :)

Err:

yes its < and NOT > thanks XzKto.

but C*C is again a Constant :) so why give that burden to the sql server :) we can calculate and give it directly. if you dont like C we can make it D :) where D = C * C :)

That's great code people, I'm trying to add to my form.  Could anyone show me how to have a drop down in php with say:

10
20
30
40
50

(which is the miles), and then integrate this into my SQL search?

Many thanks

Damian

Following is an approx code:
I have left the db connection and fetch for you. You also have to fill the actual user's co-ordinates:

 

<?php
if(isset($_GET['submit'])){
$miles = $_GET['miles'];
$miles_latlon = 0.00020882756 * $miles * $miles; //Approx Value
$user_longitude = 41;//'fill actual 
$user_latitude = -71;//fill actual

$sql='SELECT username
        FROM `USERS`
        WHERE    ((longitude-'.$user_longitude.')*(longitude-'.$user_longitude.')
                    +
                    (latitude-'.$user_latitude.')*(latitude-'.$user_latitude.'))<'. $miles_latlon).';';

//Connect to DB and fetch result

}

?>
<form method='get">
<select name="miles">
<option value="10"></option>
<option value="20"></option>
<option value="30"></option>
<option value="40"></option>
<option value="50"></option>
</select>
<input type="submit" name="submit" value="submit"/>
</form>

Select allOpen in new window

This article teaches the principles, along with tested code samples:
https://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_4276-What-is-near-me-Proximity-calculations-using-PHP-and-MySQL.html

Hi,
    I have added this to my code:

$sql='SELECT *
        FROM `members`
        WHERE    ((longitude-'.$user_longitude.')*(longitude-'.$user_longitude.')
                    +
                    (latitude-'.$user_latitude.')*(latitude-'.$user_latitude.'))<'.($miles_latlon).';';

if (!mysql_query($sql))
                          {
                                die('Error: ' . mysql_error());                          
                 }
                 
echo $sql;

However, in the echo of the SQL return, this is all I see:

SELECT * FROM `members` WHERE ((longitude--3.0134983)*(longitude--3.0134983) + (latitude-53.8306789)*(latitude-53.8306789))<0.020882756;

It just contains the select statement with all my values?  What I need it to do is list the people i.e. firstname.

What have I done wrong?

Regards,

Damian

See the comment on line 14:

//Connect to DB and fetch result

Ray,
    That's what this line is for:

$sql='SELECT *
        FROM `members`
        WHERE    ((longitude-'.$user_longitude.')*(longitude-'.$user_longitude.')
                    +
                    (latitude-'.$user_latitude.')*(latitude-'.$user_latitude.'))<'.($miles_latlon).';';

if (!mysql_query($sql))
                          {
                                die('Error: ' . mysql_error());                          
                 }
                 
echo $sql;

???

this link give info on how to connect to db and fetch the results

http://www.webcheatsheet.com/PHP/connect_mysql_database.php

The first code snippet in the article shows how to connect, select and query a data base.
https://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_4276-What-is-near-me-Proximity-calculations-using-PHP-and-MySQL.html

See lines 75 through 90.

Hai,

Just find the attachment and go throw on "GoogleApi.class"

you can get different ideas to make the map more better,

GoogleMapAPI() function is the main one.
GoogleMapAPI.class.php

liveaspankaj,
    There seems to be an issue with your code......  if I take the SQL statement:

$sql='SELECT username
        FROM `USERS`
        WHERE    ((longitude-'.$user_longitude.')*(longitude-'.$user_longitude.')
                    +
                    (latitude-'.$user_latitude.')*(latitude-'.$user_latitude.'))<'. $miles_latlon).';';

it doesn't return any values, so I manually entered the values for the longitude and latitude, as well as the distance on the end and it works, both in SQL directly, and in PHP, however as soon as I change the :

'.$user_latitude.'

it starts to fail.  I can confirm this variable does have a value.

How can I output the SQL string to see what's being formed?

Thanks

If a map is what the asker wants, this article teaches how to do that.
https://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_3350-Using-the-Google-Maps-API-in-PHP.html

An example of using that technology is on this page:
http://landonbaseball.com/site_visitors.php

"How can I output the SQL string to see what's being formed?"

If the SQL string is in a varible named, for example, $sql you can use this:

var_dump($sql);

Thanks very much everyone, all sorted now!  Great work.....