laravel 如何在 Guzzle 通话中跳过登录屏幕

How to get past login screen on Guzzle call

提问by Lynx

I have to send information to an external website using cURL. I set up Guzzle on my Laravel application. I have the basics set up, but according to the documentation of the website, there is an action that's required for the username and password. How can I pass the 'action' along with the credentials needed to log in and get access?

我必须使用 cURL 将信息发送到外部网站。我在 Laravel 应用程序上设置了 Guzzle。我已经设置了基础知识，但是根据网站的文档，需要对用户名和密码进行操作。如何传递“操作”以及登录和访问所需的凭据？

The website states:

该网站指出：

curl [-k] –dump-header <header_file> -F “action=login” -F “username=<username>” -F “password=<password>” https://<website_URL>

curl [-k] –dump-header <header_file> -F “action=login” -F “username=<username>” -F “password=<password>” https://<website_URL>

My controller:

我的控制器：

    $client = new \GuzzleHttp\Client();

    $response = $client->get('http://website.com/page/login/', array(
        'auth' => array('username', 'password')
    ));

    $xml = $response;
    echo $xml;

The website will load on the echo, but it will only pull up the login screen. I need those credentials to bypass the login screen (with a successful login) to get to the portion of information I need for cURL.

该网站将加载到 上echo，但只会显示登录屏幕。我需要这些凭据来绕过登录屏幕（成功登录）以获取 cURL 所需的部分信息。

回答by Jeremiah Winsley

curl -Fsubmits a POST request instead of a GET request. So you'll need to modify your code accordingly, something like

curl -F提交 POST 请求而不是 GET 请求。所以你需要相应地修改你的代码，比如

$client = new \GuzzleHttp\Client();

$response = $client->post('http://website.com/page/login/', [
    'body' => [
        'username' => $username,
        'password' => $password,
        'action' => 'login'
    ],
    'cookies' => true
]
);

$xml = $response;
echo $xml;

See http://guzzle.readthedocs.org/en/latest/quickstart.html#post-requests, http://curl.haxx.se/docs/manpage.html#-F

见http://guzzle.readthedocs.org/en/latest/quickstart.html#post-requests，http://curl.haxx.se/docs/manpage.html#-F

Edit:

编辑：

Just add ['cookies' => true]to requests in order to use the auth cookie associated with this GuzzleHttp\Client(). http://guzzle.readthedocs.org/en/latest/clients.html#cookies

只需添加['cookies' => true]到请求中即可使用与此关联的 auth cookie GuzzleHttp\Client()。http://guzzle.readthedocs.org/en/latest/clients.html#cookies

$response2 = $client->get('http://website.com/otherpage/', ['cookies' => true]);

回答by Samsquanch

I was having trouble getting @JeremiahWinsley's answer to work on newer version of Guzzle so I've updated their code to work as of Guzzle 5.x.

我在获得@JeremiahWinsley 的回答以处理较新版本的 Guzzle 时遇到了麻烦，所以我已经更新了他们的代码以从 Guzzle 5.x 开始工作。

Three major changes are required

需要三大改变

• Using form_paramsinstead of bodyto prevent the error "Passing in the "body" request option as an array to send a POST request has been deprecated."
• Changing the cookies to use the CookieJarobject
• Use ->getBody()->getContents()to get the body of the request

• 使用form_params代替body来防止错误“不推荐将“body”请求选项作为数组传入以发送 POST 请求。”
• 更改 cookie 以使用CookieJar对象
• 使用->getBody()->getContents()得到请求体

Here is the updated code:

这是更新后的代码：

$client = new \GuzzleHttp\Client();
$cookieJar = new \GuzzleHttp\Cookie\CookieJar();

$response = $client->post('http://website.com/page/login/', [
    'form_params' => [
        'username' => $username,
        'password' => $password,
        'action' => 'login'
    ],
    'cookies' => $cookieJar
]
);

$xml = $response->getBody()->getContents();
echo $xml;

And to continue using cookies in future requests, pass in the cookieJarto the request:

并且要在以后的请求中继续使用 cookie，请将 传递cookieJar给请求：

$response2 = $client->get('http://website.com/otherpage/', ['cookies' => $cookieJar]);

回答by Rasel Ahmed

I was having trouble getting @JeremiahWinsley's and @Samsquanch's answer to work on newer version of Guzzle. So I've updated the code to work as of Guzzle 6.x.

我在获得 @JeremiahWinsley 和 @Samsquanch 的答案时遇到了麻烦，以便在较新版本的 Guzzle 上工作。所以我更新了代码以从 Guzzle 6.x 开始工作。

Guzzle 6.x. documents: http://docs.guzzlephp.org/en/stable/index.html

狂饮6.x。文件：http: //docs.guzzlephp.org/en/stable/index.html

Here is the updated code:

这是更新后的代码：

use GuzzleHttp\Client;
use GuzzleHttp\Cookie\CookieJar;

try {
        $client = new Client();
        $cookieJar = new CookieJar();

        $response = $client->request('POST', 'http://website.com/page/login/', [
            'form_params' => [
                'username' => '[email protected]',
                'password' => '123456'
            ],
            'cookies' => $cookieJar
        ]);

        $response2 = $client->request('GET', 'http://website.com/otherpage/', [
            'cookies' => $cookieJar
        ]);

        if ($response2->getStatusCode() == 200) {
            return $response2->getBody()->getContents();
        } else {
            return "Oops!";
        }
    } catch (\Exception $exception) {
        return 'Caught exception: ', $exception->getMessage();
    }