PHP & JSON Parsing

Question

I have read a few Q&As on here and additional articles, but cannot get this script to display the relevant values from the JSON return (which are long and lat from this Google Maps call).

<?php

// Address for Google to search
$address = 'London,UK';

// Get the map json data from Google Maps using the $address variable
$googleCall = 'http://maps.googleapis.com/maps/api/geocode/json?address=' . $address .'&sensor=false';

$json = file_get_contents($googleCall);
//header("Content-type: application/json");
//echo $json;

echo $json->results->geometry->location->lat;
echo $json->results->geometry->location->lng;
?>

I am sure I am 99% there, just cannot see where the error is.

json_decode is missing in your code.

Mahn
– Mahn

2013-06-09 14:46:15 +00:00
Commented Jun 9, 2013 at 14:46 — Mahn
– Mahn, Commented Jun 9, 2013 at 14:46

Fabio · Accepted Answer · 2013-06-09 14:51:38Z

2

You can decode your json as an associative array and the access to all data with scopes

$address = 'London,UK';
$googleCall = 'http://maps.googleapis.com/maps/api/geocode/json?address=' . $address .'&sensor=false';

$json = file_get_contents($googleCall);
$array = json_decode($json, true); //note second parameter on true as we need associative array

echo $array['results'][0]['geometry']['location']['lat'] . '<br>';
echo $array['results'][0]['geometry']['location']['lng'];

This will output

51.5112139
-0.1198244

answered Jun 9, 2013 at 14:51

Fabio

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9 Comments

Dave Chen Over a year ago

You can access all data without making it an associative array.

Learner-Driver Over a year ago

Out of interest, most articles didnt wrap the values in ''. Should it be either way or is this the 'right' way.

Fabio Over a year ago

@PHP-L-Driver without quotes php will throw notice about using undefined constants

Fabio Over a year ago

@PHP-L-Driver you are welcome dude. If you feel answer was correct and it helped you out please consider to accept it, so you will also help people in the future with similar problems. To accept you just need to click on the large green tick mark (✔) under the answer's score.

Dave Chen Over a year ago

Just curious though, why the ,true? I know it makes it an associative array, but why?

|

tess3ract · Accepted Answer · 2013-06-09 14:47:44Z

0

try this out:

<?php

// Address for Google to search
$address = 'London,UK';

// Get the map json data from Google Maps using the $address variable
$googleCall = 'http://maps.googleapis.com/maps/api/geocode/json?address=' . $address .'&sensor=false';

$json = file_get_contents($googleCall);

$json = json_decode($json);
//header("Content-type: application/json");
//echo $json;

echo $json->results->geometry->location->lat;
echo $json->results->geometry->location->lng;
?>

answered Jun 9, 2013 at 14:47

tess3ract

1837 bronze badges

Comments

Ali Akbar Azizi · Accepted Answer · 2013-06-09 14:55:41Z

0

use json_decode here documentation http://php.net/manual/en/function.json-decode.php

<?php

// Address for Google to search
$address = 'London,UK';

// Get the map json data from Google Maps using the $address variable
$googleCall = 'http://maps.googleapis.com/maps/api/geocode/json?address=' . $address .'&sensor=false';

$json = file_get_contents($googleCall);

$json = json_decode($json);


echo $json->results[0]->geometry->location->lat;
echo $json->results[0]->geometry->location->lng;
?>

edited Jun 9, 2013 at 14:55

answered Jun 9, 2013 at 14:47

Ali Akbar Azizi

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2 Comments

Learner-Driver Over a year ago

Thanks for the help. I had similar code in before, but it didn't display a result. The same is with your code above. Am I reading the JSON data incorrectly?

Ali Akbar Azizi Over a year ago

@PHP-L-Driver i update my answer , i found maps.googleapis.com/maps/api/geocode/…‌ return data in result[0]

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