获取JSON数据

Im使用一个api将数据存储到JSON文件中。这些数据是从用户在我的网站上填写的表格中收集的。其插入方式如下:

$pers_payload = array(
    'gender' => 'Unknown', //or Male / Female
    'first_name' => $_POST['billing_first_name'],
    'family_name'   => $_POST ['billing_last_name'],
    'email' => $_POST['billing_email'],
    'linked_as_contact_to_organization' => array(
        array(
            'organization_id' => $organization_id, // add the person as a contact to the newly created organization
            'work_email' => $_POST['billing_email'],
            'work_phone' => $_POST['billing_phone']
            )
        ),
    'visiting_address' => array(
        'country_code'          =>  'NL'
        ), // can be extented with other address data
    'postal_address' => array(
        'country_code'      =>  $_POST['billing_country'] 
    )   // can be extented with other address data
);

然后:

$person = $SimplicateApi->makeApiCall('POST','/crm/person',json_encode($pers_payload));

现在,我想获得数据,而不是发布。我试着得到这样的数据:

$SimplicateApi->makeApiCall('GET','/crm/organization?q[name]=*my name*');

我不知道这是不是正确的方法,因为它不起作用,所以很明显它不是。

无论如何,我试图通过PHP实现的是,我想收集 name 现有人员的价值。此数据存储在 /api/v2/crm/person.json

Api文档(我读过但不太理解) http://api.simplicate.nl/