将php API数组收集到json

Question

Using Harvest php API http://mdbitz.com/harvest-api/examples/ and my harvest php array prints following data:

$myresult = $harvestAPI->getUser($client_id);
$data = $myresult->get( "data" );
print_r($data);

data:

Harvest_User Object ( [_root:protected] 
=> user [_convert:protected] 
=> 1 [_values:protected] 
=> Array ( [id] 
=> 999 [email] 
=> user@example.com [created-at] 
=> 2014-06-12T20:00:00Z [is-admin] 
=> false [first-name] 
=> John [last-name] 
=> Smith [timezone] 
=> Mountain Time (US & Canada) [is-contractor] 
=> false [telephone] 
=> [is-active] 
=> true [has-access-to-all-future-projects] 
=> false [default-hourly-rate] 
=> 200.0 [department] 
=> Development [dev] 
=> false [updated-at] 
=> 2015-06-15T18:00:00Z [cost-rate] 
=> 100.0 ) )

but when using json class to conver it to json, it gives me empty {} any idea whats happening ?

$dataJSON = json_encode($data);
print_r($dataJSON);

Answer 1

It does not work because all the properties are protected. While print_r is a special debugging function and can display protected properties, json_encode can only read public properties. As your Harvest_User object doesn't have any public properties, the JSON object is empty.

The Harvest_User class inherits from Harvest_Abstract , which again implements the magic __get and __set methods. Unfortunately, json_encode cannot use them, because it doesn't even know which variables to look for.

To solve your problem, you can write a helper class to transform your Harvest objects into plain PHP objects and then encode them to JSON.

You might also want to create a bug report for Harvest and ask them to implement Serializable and/or JsonSerializable .