PHP GeoLocation

russellharrower · 12-25-2009, 09:43 AM

Hi guys,
trying to work out the following
I have created a list of GeoLocations and I would like to create a simple system that tells me if a person is inside the area of the GeoLocation I have set for e.g.

If a person it in the area of
144.30'39.82/-38.16'22.00 by with of 144.31'12.99/-38.16'16.20
by
144.30'41.77/-38.16'37.09 by with of 144.31'12.74/-38.16'33.05

so if a person is in that area I want to show that info.

I have the script that collects the GeoLocation but not away to know if a person is in an area.

EDIT: Here is the picture of the area I want.

russellharrower · 12-25-2009, 11:38 AM

What I was thinking would be an if statement something like

<?php
if($user >144.39.82/38.22.0 && $user < 144.12.99/38.16.20)
{
echo "your in the area."
}

Dr John · 12-25-2009, 12:41 PM

You'd have to test the latitude and the longitude separately, and use >= and <=

russellharrower · 12-25-2009, 02:07 PM

Can I ask what the difference between > and >=

russellharrower · 12-25-2009, 02:10 PM

Also I noticed you said "You'd have to test the latitude and the longitude separately"
Can I ask what about the diagonal line at the bottom? is their away a PHP script would know what numbers go their to make that line? Sorry I am not sure how to make that sound any clearer.

Dr John · 12-25-2009, 02:29 PM

> means greater than
>= means greater than or equal too
< less than
<= less than or equal to

need those >= etc to get places on any of the boundary lines.

If you have the $lat and $long as separate variables from the geolocation info, and you also have
$lowerLat - the bottom of the box
$upperLat - the top of the box
$westernLong - left side
$easternLong - right side
for your box limits (assuming you have collected these values for your box from somewhere else)
then it's

if ( $lat>=$lowerLat and $lat<=$upperLat and $long>=$westernLong and $long<=$eastermlong)
{
echo "in the box";
and whatever else you want;
}
else
{
echo "out the box";
and whatever else you want;
}

Village Idiot · 12-25-2009, 08:17 PM

Its not quite that simple since he is not working with a rectangle. You can only check for bounds like that when all the lines are either parallel or perpendicular to the graph's axis. Since your are working with a four sides poligon, you have to check against the equation of each line. What he needs is to find if a given point is within any four points. This has come down to a math problem. Since I am not very familiar with GPS (and want to use small non-decimal numbers) we will be doing this on a standard graph.

When you are given four points on a graph, you can make four lines to form a four-sided poligon out of it. See the graph below. Given the points, you need to make a graph out of them. The first thing you need to do is find which points will compose the bottom, left, top and right lines, the points will meet the following criteria:
Bottom points: The two points with the lowest Y
Left points: The two points with the lowest X
Top points: The two points with the highest Y
Right points: The two points with the highest X

Given the points (-2,-1), (2,0), (-2,4), (2,3), (2,0), that would be
Bottom: (-2,-1) and (2,0)
Left: (-2,-1) and (-2,4)
Top: (-2,4) and (2,3)
Right: (2,3) and (2,0)

Now that we have our points, we can get the graph, if you'll remember algebra, the equation for a line is "y=mx+b" or y = slope(x)+y-intercept. So what we need to find is the slope and the y-intercept. To find the slope, use (y2-y1)/(x2-x1), to find the Y intercept set X and Y to either of the two points and solve. This would give us the following:

Bottom: y=(1/4)x+(1/2)
Left: x=-2
Top: y=3
Right: x=2

Shortcut: If the line is top or bottom and both Y are the same, its equation is y=(Y point), if the line is right or left and both X are the same, the equation is x=(X point).

Now that we have our boundaries, we can check against them. This works the same way:

Code:

if xPoint >(left function of Y) AND xPoint < (right function of Y) AND yPoint > (bottom function of X) AND yPoint < (top function of X)
It is in
else
It is not in
end if

I don't have time to elaborate further, I'll do that sometime tonight.Most of this was just teaching the algebra behind it, once you know that the process is really simple.

Jarod B · 12-26-2009, 05:38 AM

Quote:

Originally Posted by russellharrower

Hi guys,
trying to work out the following
I have created a list of GeoLocations and I would like to create a simple system that tells me if a person is inside the area of the GeoLocation I have set for e.g.

If a person it in the area of
144.30'39.82/-38.16'22.00 by with of 144.31'12.99/-38.16'16.20
by
144.30'41.77/-38.16'37.09 by with of 144.31'12.74/-38.16'33.05

so if a person is in that area I want to show that info.

I have the script that collects the GeoLocation but not away to know if a person is in an area.

EDIT: Here is the picture of the area I want.

Try this, I worked with something like this once with graph dimensions once. I dont remember it exactly, but from how I've always consider it was to imagine it as a square. So point 1, 2, 3, and 4 are the cases...

PHP Code:


			
if( (144.30.39.82 <= -38.16.22.00) && /*Point #1*/
    (144.31.12.99 <= -38.16.16.20) && /*Point #2*/
    (144.30.41.77 <= -38.16.37.09) && /*Point #3*/
    (144.31.12.74 <= -38.16.33.05)    /*Point #4*/ ) {
    print("You are within this area");
}

Village Idiot · 12-26-2009, 06:07 PM

Quote:

Originally Posted by Jarod B

Try this, I worked with something like this once with graph dimensions once. I dont remember it exactly, but from how I've always consider it was to imagine it as a square. So point 1, 2, 3, and 4 are the cases...

PHP Code:


			
if( (144.30.39.82 <= -38.16.22.00) && /*Point #1*/
    (144.31.12.99 <= -38.16.16.20) && /*Point #2*/
    (144.30.41.77 <= -38.16.37.09) && /*Point #3*/
    (144.31.12.74 <= -38.16.33.05)    /*Point #4*/ ) {
    print("You are within this area");
}

As I explained in my previous post, that is not a rectangle so you can not do that. You have to check it against the equation of each line opposed to the points themselves. Cheking it your way also only works when the rectangle's sides are either perpendicular or parallel with the axis.

russellharrower · 12-27-2009, 06:02 AM

Quote:

Originally Posted by Village Idiot

Its not quite that simple since he is not working with a rectangle. You can only check for bounds like that when all the lines are either parallel or perpendicular to the graph's axis. Since your are working with a four sides poligon, you have to check against the equation of each line. What he needs is to find if a given point is within any four points. This has come down to a math problem. Since I am not very familiar with GPS (and want to use small non-decimal numbers) we will be doing this on a standard graph.

When you are given four points on a graph, you can make four lines to form a four-sided poligon out of it. See the graph below. Given the points, you need to make a graph out of them. The first thing you need to do is find which points will compose the bottom, left, top and right lines, the points will meet the following criteria:
Bottom points: The two points with the lowest Y
Left points: The two points with the lowest X
Top points: The two points with the highest Y
Right points: The two points with the highest X

Given the points (-2,-1), (2,0), (-2,4), (2,3), (2,0), that would be
Bottom: (-2,-1) and (2,0)
Left: (-2,-1) and (-2,4)
Top: (-2,4) and (2,3)
Right: (2,3) and (2,0)

Now that we have our points, we can get the graph, if you'll remember algebra, the equation for a line is "y=mx+b" or y = slope(x)+y-intercept. So what we need to find is the slope and the y-intercept. To find the slope, use (y2-y1)/(x2-x1), to find the Y intercept set X and Y to either of the two points and solve. This would give us the following:

Bottom: y=(1/4)x+(1/2)
Left: x=-2
Top: y=3
Right: x=2

Shortcut: If the line is top or bottom and both Y are the same, its equation is y=(Y point), if the line is right or left and both X are the same, the equation is x=(X point).

Now that we have our boundaries, we can check against them. This works the same way:

Code:

if xPoint >(left function of Y) AND xPoint < (right function of Y) AND yPoint > (bottom function of X) AND yPoint < (top function of X)
It is in
else
It is not in
end if

I don't have time to elaborate further, I'll do that sometime tonight.Most of this was just teaching the algebra behind it, once you know that the process is really simple.

First of all I want to thank you all,

But need to ask, (cos my math teacher did not teach us algebra, Can you make a small PHP example please.

Cos I don't understand the xPoint>(left function ...

I would be very greatful.
Thanks

cachepl0x · 12-27-2009, 06:55 AM

X == Horizontal, and Y == Vertical. :P

Salathe · 12-27-2009, 12:07 PM

I have a function which determines whether a point is within a polygon (of any shape) based on decimal lat/lon pairs (rather than degrees, mins, seconds as you appear to be using) if that might be of any use?

Village Idiot · 12-28-2009, 08:57 PM

Quote:

Originally Posted by russellharrower

First of all I want to thank you all,

But need to ask, (cos my math teacher did not teach us algebra, Can you make a small PHP example please.

Cos I don't understand the xPoint>(left function ...

I would be very greatful.
Thanks

You would probably be better off going on Google and looking up tutorials on mathematical functions than having me explaining it. But here is basically what I am doing (this is not all technically mathematically correct, but it is an easy way to get the job done). Function of Y means that you replace all the X variables in the equation with the given variable. Function of X means that you replace all the Y variables with the given value. As it ends up, I didn't give you any values to plug in, so that example wouldn't work.

You actually stuck me with some inspiration to write a script for this. Its been a long time since I've wrote anything in PHP so that was actually rather fun. The issue I've faced many times is that I know how to do most common tasks, this was actually a problem solving challenge so I did really enjoy making this.

PHP Code:


			
<?
class point
{
    function __construct($xParam,$yParam)
    {
        $this->x=$xParam;
        $this->y=$yParam;
    }

    public $x;
    public $y;
};

class equation
{
    function __construct($point1,$point2,$debug=false)
    {
        //Find the slope of the two points
        
        //If the line is horizontal
        if(($point2->y-$point1->y) == 0)
        {
            if($debug==true)
            
            $slope=0;
            $this->isHorizontal = true;        
            $this->isVertical = false;
            $this->y=$point2->y;
        }
        
        //If the line is vertical
        else if(($point2->x-$point1->x) == 0)
        {
            if($debug==true)
            $slope=0;
            $this->isVertical = true;
            $this->isHorizontal = false;
            $this->x=$point2->x;
        }
        else
        {
            if($debug==true)
            $slope = ($point2->y-$point1->y)/($point2->x-$point1->x);
            
            //Since vertical lines have no y intercept and horosontal lines don't need one to be stated, we only need to compute this if the line is at an angle
            $yInt=$point1->y-($slope*$point1->x);
            
            $this->isVertical = false;
            $this->isHorizontal = false;
        }
        $this->m = $slope;
        $this->b = $yInt;
    }
    
    public function solveForY($xParam)
    {
        if($this->isHorizontal == true)
        {
            return $this->y;
        }
        return $this->m*$xParam+$this->b;
    }
    public function solveForX($yParam)
    {
        //echo "x=$this->x m=$this->m yParam=$yParam b=$this->b ";
        if(isVertical == true)
        {
            return $this->x;
        }
        return ($yParam-$this->b)/$this->m;
    }

    public $m;
    public $b;
    
    
    private $isVertical;
    private $isHorizontal;
    public $x;
    public $y;
};

function isPointIn4Lines(&$pTopLine, &$pRightLine, &$pBottomLine, &$pLeftLine, &$testPoint)
{
    
    if(
        $testPoint->y < $pTopLine->solveForY($testPoint->x) &&// If the point lies below the top point
        $testPoint->y > $pBottomLine->solveForY($testPoint->x) &&// if the point lies above the bottom point
        $testPoint->x > $pLeftLine->solveForX($testPoint->y) && // if the point lies right of the left line
        $testPoint->x < $pRightLine->solveForX($testPoint->y) // If the point lies left of the right line
      )
    {
        return "true";
    }
    return "false";
}

//Application time
$point1 = new point(-2,3);
$point2 = new point(-2,0);
$point3 = new point(3,-4);
$point4 = new point(3,3);

$testPoint1 = new point(1,1);
$testPoint2 = new point(-1,-2);

//I don't feel like writing up a sorting function, you know how do ID them so do that yourself.
$topLeft = $point1;
$topRight = $point4;
$bottomLeft = $point2;
$bottomRight = $point3;



//Generate the lines
$topLine = new equation($topLeft,$topRight,true);
$rightLine = new equation($topRight,$bottomRight,true);
$bottomLine = new equation($bottomLeft,$bottomRight,true);
$leftLine = new equation($bottomLeft,$topLeft,true);

echo isPointIn4Lines($topLine,$rightLine,$bottomLine,$leftLine,$testPoint1) . " and " . isPointIn4Lines($topLine,$rightLine,$bottomLine,$leftLine,$testPoint2);
//will output "true and false"

?>

What this does it based off of four points, generates four lines. Then it tests to see if the given test points are within these four lines.

russellharrower · 12-30-2009, 05:08 AM

Quote:

Originally Posted by Salathe

I have a function which determines whether a point is within a polygon (of any shape) based on decimal lat/lon pairs (rather than degrees, mins, seconds as you appear to be using) if that might be of any use?

If you could that would be great.
thanks

russellharrower · 12-30-2009, 10:47 AM

I found a script that works the way I needed it to, I thought I would post it below for anyone else.
{code}
<?php
class pointLocation {
var $pointOnVertex = true; // Check if the point sits exactly on one of the vertices

function pointLocation() {
}

function pointInPolygon($point, $polygon, $pointOnVertex = true) {
$this->pointOnVertex = $pointOnVertex;

// Transform string coordinates into arrays with x and y values
$point = $this->pointStringToCoordinates($point);
$vertices = array();
foreach ($polygon as $vertex) {
$vertices[] = $this->pointStringToCoordinates($vertex);
}

// Check if the point sits exactly on a vertex
if ($this->pointOnVertex == true and $this->pointOnVertex($point, $vertices) == true) {
return "inside";
}

// Check if the point is inside the polygon or on the boundary
$intersections = 0;
$vertices_count = count($vertices);

for ($i=1; $i < $vertices_count; $i++) {
$vertex1 = $vertices[$i-1];
$vertex2 = $vertices[$i];
if ($vertex1['y'] == $vertex2['y'] and $vertex1['y'] == $point['y'] and $point['x'] > min($vertex1['x'], $vertex2['x']) and $point['x'] < max($vertex1['x'], $vertex2['x'])) { // Check if point is on an horizontal polygon boundary
return "boundary";
}
if ($point['y'] > min($vertex1['y'], $vertex2['y']) and $point['y'] <= max($vertex1['y'], $vertex2['y']) and $point['x'] <= max($vertex1['x'], $vertex2['x']) and $vertex1['y'] != $vertex2['y']) {
$xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x'];
if ($xinters == $point['x']) { // Check if point is on the polygon boundary (other than horizontal)
return "boundary";
}
if ($vertex1['x'] == $vertex2['x'] || $point['x'] <= $xinters) {
$intersections++;
}
}
}
// If the number of edges we passed through is even, then it's in the polygon.
if ($intersections % 2 != 0) {
return "inside";
} else {
return "outside";
}
}

function pointOnVertex($point, $vertices) {
foreach($vertices as $vertex) {
if ($point == $vertex) {
return true;
}
}

}

function pointStringToCoordinates($pointString) {
$coordinates = explode(" ", $pointString);
return array("x" => $coordinates[0], "y" => $coordinates[1]);
}

}

/*** Example ***/
//$pointLocation = new pointLocation();
//$points = array("30 19", "0 0", "10 0", "30 20", "11 0", "0 11", "0 10", "30 22", "20 20");
//$polygon = array("10 0", "20 0", "30 10", "30 20", "20 30", "10 30", "0 20", "0 10", "10 0");
//foreach($points as $key => $point) {
// echo "$key ($point) is " . $pointLocation->pointInPolygon($point, $polygon) . "<br>";
//}
{/code}

dashixiong · 10-18-2012, 01:40 PM

Some conservatives have Coach Factory Outlet pushed that critique further, saying that Mr. Obama’s policies are too costly, often assist the wrong people Louis Vuitton Belts and could have the paradoxical effect of driving up college costs. The dispute turns not just on different Coach Factory Outlet assessments of how policies play out, but on differing philosophical views about the role of government. During Gucci Belts his time in office, Mr. Obama has sharply increased aid to low- and middle-income students, notably through the Pell Grant Coach Factory Outlet program, which grew from $14.6 billion given to 6 million students in 2008, to nearly $40 billion for Coach Factory Outlet almost 10 million students this year. His administration also made it easier to request aid, shortening the Coach Factory Online complex federal application and allowing people to transfer their financial information electronically from the Internal Coach Outlet Online Revenue Service database. But while many education experts laud his efforts, analysts of varying political Coach Outlet Online stripes have also questioned how much impact some of the president’s policies will have, noting that the prices Coach Online Outlet charged by colleges, and student borrowing, continue to climb.But behind the headlines about soaring costs, the Coach Factory Outlet Online reality is more complex and wildly uneven, because a growing number of students receive Coach Outlet Online financial aid, and only relatively high-income families pay those fast-rising sticker prices. Adjusted for Coach Factory Online inflation, the College Board calculates, the average net price changed little over the last decade at private Coach Factory Outlet schools, and rose only modestly at public ones.Defending federal spending, Arne Duncan, the secretary of Hermes Belts education, said that for more than 30 years, college prices had risen even when federal aid had not, leading him to believe Coach Factory Online there was zero correlation.

dashixiong · 10-22-2012, 09:28 AM

You’ve relativelyCoach Outlet recently arrived in New Delhi after living in two of Asia’s other great cities,Coach Outlet Store Online Tokyo and Hong Kong, for several years. Do these cities feel like they’re part of the same continent? Yes, and no. In terms Coach Factory Onlineof infrastructure, they couldn’t be more different. Getting regularCoach Outlet power and water at my house in New Delhi is never a sure thing, even though Coach Purse Outlet OnlineI’m paying the same rent that I paid in Tokyo and almost the same electricity prices. Both Hong Kong and Tokyo are also crowded places,Coach Factory Outlet Online but both cities are incredibly well planned and efficiently run. Efficient is not a word I would use to describe my Coach Bags Outlet Onlineday-to-day life in New Delhi. On the other hand, one thing that I think Hong Kong and New Delhi have in common isCoach Handbags Outlet a shared sense of optimism — a feeling that the best is yet to come. That’s definitely not the feeling you get in Tokyo,Coach Outlet Online or in the U.S. when I go home. It’s a big part of what I find addictive about living and working in this part of the world. You feel like you’re watching the future unfold.

dashixiong · 01-29-2013, 12:31 PM

Organizers said Coach Outlet Online was opportune because the battle’s 150-year anniversary is in December, and Fredericksburg Coach Factory Outlet has been preparing to mark the sesquicentennial. in the new agreement is that Coach Outlet Online revolutionary councils from 14 Syrian provinces now each have a representative, though not all live Coach Online Outlet in Syria. The hope is that will bind the coalition to those inside the country. Perhaps Coach Bags Outlet the most important body the new group is expected to form is a Revolutionary Military Council Coach Factory Online to oversee the splintered fighting organizations and to funnel both lethal and nonlethal Coach Factory Outlet military aid to the rebels. It should unite units of the Free Syrian Army, various militias Coach Outlet Store Online and brigades in each city and large groups of defectors. Before the ink was even dry on the Coach Outlet Store final draft, negotiators hoped that it would bring them the antiaircraft missiles they crave to Coach Factory Stores take on the Syrian Air Force. The United States and Britain have offered only Coach Handbags Outlet nonmilitary aid to the uprising. A similar attempt by the Syrian National Council to Coach Factory Store supervise the military never jelled. Organizers said funding was too haphazard. Eventually foreign Coach Factory Online governments like Qatar and Saudi Arabia, which are financing and arming the rebels, found Coach Factory Online their own favorite factions to deal with. Foreign leaders notably including Secretary of State Coach Outlet Hillary Rodham Clinton urged this unification largely so they could coordinate their Coach Factory Outlet efforts and aid through a group of technocrats. Once it receives international recognition, the Coach Outlet Store Online coalition is supposed to establish a temporary Coach Outlet Online military never jelled.

programmingservices · 01-31-2013, 10:45 AM

Thank You! I found this very useful for me. Contributions like this help to coding a good understanding about the subject and gain a detailed insight of it. Further Novices like me get to bud themselves to advance and stay upfront in the cold & competitive industry.

__________________
Hire PHP Programmer India | Mobile Programming Services | Outsourcing Programming to India

sara111 · 02-01-2013, 11:04 AM

We have best experience to identify the potential behind you. And we have best experience to find jobs for our clients according to their caliper.
staffing agency
staffing agencies
job recruiters
online recruitment service
candor jobs