我不知道自己做错了什么,但我已经为此付出了几天的努力。
这是我从命令行发出的cURL请求:
curl -i -H "Accept: text/html" http://laravel.project/api/v1/users/4
返回:
HTTP/1.1 200 OK
Server: nginx/1.6.2
Content-Type: application/json
Transfer-Encoding: chunked
Connection: keep-alive
Cache-Control: no-cache
Date: Sun, 29 Mar 2015 10:33:36 GMT
Set-Cookie: laravel_session=eyJpdiI6ImNPTkZIYVJZSVRKaHBOZTR3SWh0dHc9PSIsInZhbHVlIjoiblpZYVJlN2dBY1ljejNIYUQycXpsNXRWd1I5a3JSRG8wSWdDOWlHVTMrYUcrdDBNVmVuZUNkOGtJb2M4bXFpTFF3cUdoTFZOVXBWXC82Q1luSGd5bjJBPT0iLCJtYWMiOiI0ZTEwOWQxMmVhMzY2NjI1Yzc1MTBmZmRmYjUyOGQwNDlhYzRjOTNiM2FiOWIyN2E1YjA0OTM4YTUxZmNmMzMzIn0%3D; expires=Sun, 29-Mar-2015 12:33:36 GMT; Max-Age=7200; path=/; httponly
{
"data":{
"id":4,
"name":"Helena",
"email":"hh@gmail.com",
"created_at":"2015-03-26 21:13:16",
"updated_at":"2015-03-26 21:13:16"
}
}
所以一切看起来都很好:内容类型设置正确,响应为JSON。
但是现在看看如果我在PHP中使用带有curl的API会发生什么:
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $final_url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "GET");
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Accept: application/json'));
$result = curl_exec($ch);
return json_decode($result);
我得到的回复是:
{#165
+"data": {#167
+"id": 4
+"name": "Helena"
+"email": "hh@gmail.com"
+"created_at": "2015-03-26 21:13:16"
+"updated_at": "2015-03-26 21:13:16"
}
}
而且,如果我返回不带json_decode
的$result
,我得到的是:
"{
"data":{
"id":4,
"name":"Helena",
"email":"hh@gmail.com",
"created_at":"2015-03-26 21:13:16",
"updated_at":"2015-03-26 21:13:16"
}
}"
正确的回答,但要加引号。我在PHP文档中读到curl_opt_returntranfer
以字符串的形式返回结果,但我不是这个星球上唯一一个只想获得JSON的人。
这是我的ApiController
类:
class ApiController extends Controller {
// Base controller for API Controllers
protected $statusCode = 200;
protected function respond($data)
{
return Response::json([
'data' => $data,
]);
}
protected function respondNotFound($message = 'No data found')
{
return Response::json([
'error' => [
'message' => $message,
'status_code' => $this->getStatusCode(),
]
]);
}
}
这是我的UserController
:
class UserController extends ApiController {
public function show($user)
{
if ($user == null)
return $this->setStatusCode(404)->respondNotFound('User not found');
return $this->respond($user);
}
}
如果我返回$result而不使用json_decode,我会得到这样的结果:正确的响应,但内部引用
不,你凭什么这么想?我想问题是你是如何打印的,你很可能是用var_export($result)
、var_dump($result)
或echo json_encode($result);
打印的,这是在添加引号。如果你只想要json,只需用echo $result;
回显它,不需要额外的处理,只需按原样回显字符串,它已经是json了。
我认为这将解决您的问题:
curl -i -H "Accept: text/html" http://laravel.project/api/v1/users/4 | tr -d '"'
$response = (string)$result;
$resp_arr = explode("<!DOCTYPE",$response);
$obj = json_decode(trim($japi_arr[0]));
if(isset($obj[0]))
{
$rsp_id = $obj[0]->id;
$rsp_name = $obj[0]->name;