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我知道这可能很简单,但我卡住了,无法找到解决方案。我只是想创建一个简单的嵌套JSON对象,看起来像以下:如何在PHP5中创建一个嵌套的json对象
{
"user": {"firstname":"foo","lastname":"bar","email":"[email protected]"}
}
到目前为止,我可以创建内JSON如下:
class user_profile {
private $firstname = '';
private $lastname ='';
private $email = '';
public function __construct($first, $last, $email){
$this->firstname = $first;
$this->lastname = $last;
$this->email = $email;
}
public function expose() {
return get_object_vars($this);
}
}
$up = new user_profile('foo','bar','[email protected]');
echo json_encode($up->expose());
我尝试添加一个数组:
echo json_encode(array('user',$up->expose()), JSON_FORCE_OBJECT);
但其结果是:
{
"0":"user","1": {"firstname":"foo","lastname":"bar","email":"[email protected]"}
}
如何创建外部“用户”部分?