Question

我有一个api和一个我正在开发的客户端，无论是在laravel还是当我尝试使用guzzle连接到api时我得到了错误18。

在我的api控制器中，我有这个：

public function index()
{
    $users = User::orderBy('username', 'asc');

    return Response::json(array(
        'error' => false,
        'users' => $users->get()->toArray()),
        200
    );

}

如果我做curl --user admin@admin.com:password http://myapi.api/api/v1/users

我在控制台上正确获取了我需要的信息：

 {"error":false,"users":[{"id":"1","firstname":"","lastname":"","username":"admin@admin.com","created_at":"2014-10-17 15:35:10","updated_at":"2014-10-17 15:35:10","client_id":"0","enterprise_id":"0","usertype_id":"0"},{"id":"2","firstname":"","lastname":"","username":"seconduser","created_at":"2014-10-17 15:35:10","updated_at":"2014-10-17 15:35:10","client_id":"0","enterprise_id":"0","usertype_id":"0"}]}

网址甚至可以在浏览器中运行（我获得了一个用于身份验证的弹出窗口，之后我在浏览器上获得了相同的输入）。所以它只能通过guzzle来失败。

现在我的客户端安装了Guzzle，我正在尝试这个：

$client = new GuzzleHttp\Client();
$user='admin@admin.com';
$pass='password';

$res = $client->get('http://myapi.api/api/v1/users', array(
    'auth' =>  array('admin@admin.com', 'password')
));

$users=$res->json();  
$users=$users['users'];

我收到错误：

[curl] (#18) See http://curl.haxx.se/libcurl/c/libcurl-errors.html for an explanation of cURL errors [url] http://myapi.api/api/v1/users

我做错了什么？

编辑：在我得到的命令中使用-v：

$ curl -v --user admin@admin.com:password http://myapi.api/api/v1/users
* Hostname was NOT found in DNS cache
*   Trying 127.0.0.1...
* Connected to myapi.api (127.0.0.1) port 80 (#0)
* Server auth using Basic with user 'admin@admin.com'
> GET /api/v1/users HTTP/1.1
> Authorization: Basic YWRtaW5AYWRtaW4uY29tOnBhc3N3b3Jk
> User-Agent: curl/7.37.1
> Host: myapi.api
> Accept: */*
>
< HTTP/1.1 200 OK
< Date: Mon, 20 Oct 2014 13:08:56 GMT
* Server Apache/2.2.29 (Unix) mod_fastcgi/2.4.6 mod_wsgi/3.4 Python/2.7.8 PHP/5.5.17 mod_ssl/2.2.29 OpenSSL/0.9.8za DAV/2 mod_perl/2.0.8 Perl/v5.20.0 is not blacklisted
< Server: Apache/2.2.29 (Unix) mod_fastcgi/2.4.6 mod_wsgi/3.4 Python/2.7.8 PHP/5.5.17 mod_ssl/2.2.29 OpenSSL/0.9.8za DAV/2 mod_perl/2.0.8 Perl/v5.20.0
< X-Powered-By: PHP/5.5.17
< Cache-Control: no-cache
< Set-Cookie: laravel_session=eyJpdiI6IjRPMk9TT0ZnZklTMG1uWlFDancyMWc9PSIsInZhbHVlIjoic1V1RjA5aVBJdFNLM0JLclNROEE1a0dCeHNEMWhVNFVReTlUOHdidE44WEJzRnB4WFkxdWo0V0ozcXFVSW9LYzZiZzZSSlFCNXNTTjl2Mzh4TlFtTUE9PSIsIm1hYyI6IjhlYzY0YjFkNTQzNjk5ZGMxNDk3YmY4ZjU4YTYzYzM4YzgxZjg1MzlhMWUxNWVjYWE4ZThlMmU0N2RjNWFkZGMifQ%3D%3D; expires=Mon, 20-Oct-2014 15:08:57 GMT; Max-Age=7200; path=/; httponly
< Transfer-Encoding: chunked
< Content-Type: application/json
<
* Connection #0 to host myapi.api left intact
{"error":false,"users":[{"id":"1","firstname":"","lastname":"","username":"admin@admin.com","created_at":"2014-10-17 15:35:10","updated_at":"2014-10-17 15:35:10","client_id":"0","enterprise_id":"0","usertype_id":"0"},{"id":"2","firstname":"","lastname":"","username":"seconduser","created_at":"2014-10-17 15:35:10","updated_at":"2014-10-17 15:35:10","client_id":"0","enterprise_id":"0","usertype_id":"0"}

编辑：这个问题可能过于具体，即使我没有找到解决方案，并且已经采用了不同的方式，因为规格已经改变，我会留在这里，以防它可以帮助某人。

我对答案的猜测是正如AndréDaniel在评论中所说的那样，认证破坏了某些东西，而guzzle并没有给我原来的json，从而产生错误。

Answer 1

关于你的问题，你说这个URL通过命令行工作：

 http://myapi.api/api/v1/users/1

但是您通过guzzle

调用此URL

 http://myapi.api/api/v1/users

所以你确定你试图通过guzzle调用的路由通过命令行工作吗？

Answer 2

我终于改变了方法，因为我们需要一种不同的身份验证方法，因此这个问题不再与我的案例相关。

Answer 3

我知道这已经很老了，但这就是我与Guzzle一起工作的原因：

  switch (button.tag) {

        case != 1: //error occurs here


        default:
           //
        }

您甚至可以将授权数据存储在Guzzle实例上，如下所示：

$client = new GuzzleHttp\Client();
$user = 'admin@admin.com';
$pass = 'password';

$res = $client->get('http://myapi.api/api/v1/users', [
    'headers' => [
        'Authorization' => 'Basic ' . base64_encode($user . ":" . $pass),
    ],
]);

$users = $res->json();  
$users = $users['users'];

与Laravel的Guzzle卷曲错误18

3 个答案: