如何在 Guzzle 通话中跳过登录屏幕 [英] How to get past login screen on Guzzle call

问题描述

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"以及登录和获取访问权限所需的凭据?

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?

网站声明:

curl [-k] –dump-header -F 操作=登录" -F 用户名=<用户名>"-F 密码=<密码>"https://

我的控制器:

    $client = new GuzzleHttpClient();

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

    $xml = $response;
    echo $xml;

网站将在 echo 上加载，但只会拉出登录屏幕.我需要这些凭据来绕过登录屏幕(登录成功)以获取 cURL 所需的部分信息.

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.

推荐答案

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

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

$client = new GuzzleHttpClient();

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

$xml = $response;
echo $xml;

参见 http://guzzle.readthedocs.org/en/latest/quickstart.html#发布请求，http://curl.haxx.se/docs/manpage.html#-F

只需添加 ['cookies' =>true] 请求以使用与此 GuzzleHttpClient() 关联的身份验证 cookie.http://guzzle.readthedocs.org/en/latest/clients.html#cookies

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

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

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