Question

I want to include a php file in one another and decode the JSON that I get.

Here´s my code:

<?php 

$jsonstring = include('php/get_recipe_byID.php');


$obj = json_decode($jsonstring);
print_r($obj->Data);

?>

$jsonstring is

{"Data":{"Recipes":{"Recipe_10":{"ID":"10","TITLE":"Pferde\u00e4pfel","TEXT":"Sammeln und Essen","COUNT_PERSONS":"4","DURATION":"60","USER_ID":"1","DATE":"1000-01-01 00:00:00"}}},"Message":null,"Code":200} 

The error I get is

Notice: Trying to get property of non-object in /var/www/recipe_search.php on line 118

which is the row with print_r($obj->Data);

How can I handle that?

Answer 1

It is because you include the file and not use the file directly. You should be using file_get_contents instead.

However!

I have some times had issues with using json_decode and its turning into objects, and i prefer using its associative array options instead. Try using

$obj = json_decode($jsonstring);
print_r($obj['Data']);