Convert JSON object to associative php array

Question

I am using AngularJS with PHP on server side to access database. To make a POST method I write this request:

var req = {
      method: 'POST',
      url: 'action.php',
      data:{'tblname': 'user',
      'conditions' : {
            'select' : 'user_name',
            'where' : {
                 'user_category' : 'admin'
             },
       'order_by' : 'user_name'
       }   
};

In PHP I want to convert my JSON data object into a php associative array.

$request_data = json_decode(file_get_contents("php://input"));
$conditions = json_decode($request_data->conditions,true);

I used json_decode but it seems its not converting the JSON object to associative php array. I want the JSON object to be converted to the following PHP array:

$conditions = array(
        "select" => "user_name",
        "where" =>
            array("user_category" => "admin") ,
        "order_by" => "user_name"
);

Assuming that $request_data contains what you expect it to; you have already decoded the data, so it is no longer a string but an object. You should either decode the whole thing as an array or cast the conditions section / object to an array. — jeroen
– jeroen, Commented Nov 2, 2018 at 9:47
I replaced $conditions = json_decode($request_data->conditions,true); with $conditions = (array) $request_data->conditions; and now it works. Thanks @jeroen and @splash58 — EddyG
– EddyG, Commented Nov 2, 2018 at 9:55

yivi · Accepted Answer · 2018-11-02 10:55:39Z

1

You are trying to json_decode data that has already been decoded.

Once you do:

$request_data = json_decode(file_get_contents("php://input"), TRUE);

You've already have the information in an associative array. (The second parameter tells json_decode() that you want your result as an associative array and not as an object).

The next step is as simple as:

$conditions = $request_data['conditions'];

answered Nov 2, 2018 at 10:55

yivi

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