symfony使用findall读取出来的数据如何转成json

$repository = $this->getDoctrine()->getRepository('AppBundle:User');$all = $repository->findAll();

array (size=2)  0 =>     object(AppBundle\Entity\User)[248]      private 'id' => int 1      private 'name' => string 'A Foo Bar' (length=9)      private 'pass' => string '19.99' (length=5)  1 =>     object(AppBundle\Entity\User)[251]      private 'id' => int 2      private 'name' => string 'Two Fot Bar' (length=11)      private 'pass' => string '40.00' (length=5)

返回的数据是这样的， 现在这个 如何转成 json数据
我现在用 return new JsonResponse($all); [{}]

回复讨论(解决方案)

你需要给 AppBundle\Entity\User 实现 JsonSerializable 接口
如

class T implements JsonSerializable {  private $id;  private $name;  private $pass;  function __construct($id, $name, $pass) {    $this->id = $id;    $this->name = $name;    $this->pass = $pass;  }  function jsonSerialize() {    return array(      'id' => $this->id,      'name' => $this->name,      'pass' => $this->pass,      );  }}$d[] = new T(1, 'a', 'p');$d[] = new T(2, 'b', 'p');echo json_encode($d);

[{"id":1,"name":"a","pass":"p"},{"id":2,"name":"b","pass":"p"}]
否则只能是 [{},{}]
因为返回的属性是私有的

你需要给 AppBundle\Entity\User 实现 JsonSerializable 接口
如

class T implements JsonSerializable {  private $id;  private $name;  private $pass;  function __construct($id, $name, $pass) {    $this->id = $id;    $this->name = $name;    $this->pass = $pass;  }  function jsonSerialize() {    return array(      'id' => $this->id,      'name' => $this->name,      'pass' => $this->pass,      );  }}$d[] = new T(1, 'a', 'p');$d[] = new T(2, 'b', 'p');echo json_encode($d);

[{"id":1,"name":"a","pass":"p"},{"id":2,"name":"b","pass":"p"}]
否则只能是 [{},{}]
因为返回的属性是私有的

多谢，就是因为属性是私有的。