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PHP variable question

Posted on 2014-10-27
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Last Modified: 2014-10-27
When I hardcore lat and long in this request all works as expected and there is a JSON result.

https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=38.3403232756246,-77.0765647362513&radius=500&types=&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw

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But when I place $lat and $long in a variable and do this request:
https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=".$lat.",".$long."&radius=500&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw

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There seems to be no output. I can't quite understand why because when I echo $lat and $long the result for $lat and $long is there.
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Question by:seopti
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11 Comments
 
LVL 61

Expert Comment

by:Julian Hansen
ID: 40406524
We need to see more code.

The following code works
<?php
$lat='38.3403232756246';
$long='-77.0765647362513';
$result = "https://maps.googleapis.com/maps/api/place/nearbysearch/json?location={$lat},{$long}&radius=500&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw";
echo $result;

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Produces the required result.
If you are using numeric values for $lat / $lang you might get a truncated answer - for instance
<?php
$lat=38.3403232756246;
$long=-77.0765647362513;
$result = "https://maps.googleapis.com/maps/api/place/nearbysearch/json?location={$lat},{$long}&radius=500&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw";
echo $result;

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produces the following
https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=38.340323275625,-77.076564736251&radius=500&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw

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Note the truncated values.

Aside: A note on formatting strings - I find embedding variables in strings like so

"This is a {$variable} in the string"

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Is more readable and easier to type than
"This is a " . $variable . " in the string"

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Author Comment

by:seopti
ID: 40406567
Thank you for the detailed answer. I tried the solution you posted but no success. It always only works with the hardcoded lat and long not with $lat and $long.

This is the code which works fine but only with hardcoded values.

$address_url = "http://www.geocodefarm.com/api/forward/json/c3140ba389a289bd42c51ebe2d72ccc67fb4633c/$address";


$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $address_url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_PROXYPORT, 3128);
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
$response = curl_exec($ch);
curl_close($ch);
 

$response_a = json_decode($response);

//Print the provided address in Human readable/ complete Postal address;
echo $address_returned = $response_a->geocoding_results->ADDRESS->address_returned.'<br />';
 
//Print the Latitude and Longitude of the address
echo $lat = $response_a->geocoding_results->COORDINATES->latitude.'<br />';
echo $long = $response_a->geocoding_results->COORDINATES->longitude;



	
		$api_key = "AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw";
		foreach ($categoryArray as $category => $title) { ?>
			<div class="<?php echo (array_key_exists($category, $loadedCatArray) ? "" : "hidden"); ?>" id="<?php echo $category; ?>_items">
				<div class="data_title"><?php echo $title; ?></div>
				<ol class="data_items">
				<?php
			$bizArray = file_get_contents("https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=38.3403232277480,-77.0765647362513&radius=500&types=".$category."&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw");


		
				$bizArray = json_decode($bizArray, true);


.....

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0
 
LVL 111

Expert Comment

by:Ray Paseur
ID: 40406585
In the most recent code snippet on line 1, $address is an undefined variable.  Are we maybe still missing some of the code?

When I fix the parse errors and dump the response from the API, this is what I see:
<b>Notice</b>:  Undefined variable: address in <b>/home/iconoun/public_html/demo/temp_seopti.php</b> on line <b>4</b><br />
string(1202) "{
    "geocoding_results": {
        "LEGAL_COPYRIGHT": {
            "copyright_notice": "Copyright (c) 2013 Grabenhofen Corporation (Colorado, USA) - All Rights Reserved.",
            "result_distribution": "UNPERMITTED, UNLICENSED",
            "copyright_logo": "https:\/\/www.geocodefarm.com\/assets\/img\/logo.png",
            "terms_of_service": "https:\/\/www.geocodefarm.com\/policies\/terms-of-service\/",
            "privacy_policy": "https:\/\/www.geocodefarm.com\/policies\/privacy-policy\/"
        },
        "STATUS": {
            "access": "KEY_VALID, ACCESS_GRANTED",
            "status": "FAILED, NO_RESULTS"
        },
        "ACCOUNT": {
            "name": "Thomas Kowalewski",
            "email": "support@apollofind.com",
            "api_key": "c3140ba389a289bd42c51ebe2d72ccc67fb4633c",
            "monthly_due": "75.00",
            "next_due": "2014-11-25 23:06:13",
            "distribution_license": "NONE, UNLICENSED",
            "usage_limit": "100000",
            "used_today": "1329",
            "remaining_queries": "98671"
        },
        "STATISTICS": {
            "load_time": "0.41",
            "https_ssl": "DISABLED, INSECURE"
        }
    }
}"

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Author Comment

by:seopti
ID: 40406597
Yes $address is defined, I didn't post it.

	$address = $thiStuff['bizAddr'].','.$thisStuff['bizCity'].','.$thisStuff['bizState'].','.$thisStuff['bizZip'];

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0
 

Author Comment

by:seopti
ID: 40406614
I think the problem is:

This will echo $lat and $long after a successful geocoding without any problems.

echo $lat = $response_a->geocoding_results->COORDINATES->latitude.'<br />';
echo $long = $response_a->geocoding_results->COORDINATES->longitude;

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But when $lat and $long is used in this API Call it will not.

https://maps.googleapis.com/maps/api/place/nearbysearch/json?location={$lat},{$long}&radius=500&types=".$category."&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw

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0
 
LVL 111

Expert Comment

by:Ray Paseur
ID: 40406625
I think the problem is that we do not know what variables you have in that script, and neither do you!

Here's a good way to begin to understand the inner workings.  This function is your friend.  Use it every time you are not 100% sure of the contents of any variable - especially variables from external sources like $_GET and $_POST.
http://php.net/manual/en/function.var-dump.php

When you use var_dump() on a variable that may contain HTML or other markup, you may need to use your browser's "view source" to see the true contents of the variable.
0
 
LVL 111

Expert Comment

by:Ray Paseur
ID: 40406639
This seems to work correctly.
http://iconoun.com/demo/temp_seopti.php

<?php // demo/temp_seopti.php


// SEE http://www.experts-exchange.com/Programming/Languages/Scripting/PHP/Q_28545160.html


// USE THIS SETTING TO BE SURE YOU'RE NOT ACCIDENTALLY RELYING ON AN UNDEFINED VARIABLE
error_reporting(E_ALL);

// ASSIGN VALUES TO VARIABLES
$lat  = '38.930451';
$long = '-77.148134';

// CREATE THE URL, USING THE VARIABLES WE JUST ASSIGNED
$url = "https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=$lat,$long&radius=500&types=&key=AIzaSyBSCZt0IMSPZjkoyiEIj08953OBI3qVZBw";

// FETCH THE RESPONSE AND DISPLAY THE DATA
$doc = file_get_contents($url);

echo '<pre>';
var_dump($url);
var_dump($doc);

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0
 
LVL 61

Expert Comment

by:Julian Hansen
ID: 40406675
There is no difference between

$x = 31.111111;
$x = "This is the latittude : {$lat}";

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And

$x = "This is the latitude : 31.111111";

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There is something you are not showing us.

Can't see where in your code snippet you are using the string you say is not working/
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Author Comment

by:seopti
ID: 40407097
I'm so dumb, found out the error thanks to Ray, I had error reportung turned off before.

The problem was with this variable:
$lat = $response_a->geocoding_results->COORDINATES->latitude.'<br />';

$lat had a <br> assigned. $lat was then used in the API request which broke the whole line.
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LVL 111

Accepted Solution

by:
Ray Paseur earned 2000 total points
ID: 40407108
error_reporting(E_ALL) and var_dump() have saved me countless hours!
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