发布数据并显示源代码 - Post data & show source code

Post data & show source code

我有疑问，是否可以先将数据发布到站点，然后获取我发布数据的站点的源代码？

我试过这个：

$html = file_get_contents('http://test.com/test.php?test=test');

但是？test=test 是 $_GET....

所以是的，我希望有人能帮助我！ :)
提前感谢（对不起英语不好！

您可以使用此函数的第三个参数： 上下文

$postdata = http_build_query(
    array(
        'var1' => 'some content',
        'var2' => 'doh'
    )
);
$opts = array(
    'http' => array(
        'method' => "POST",
        'header' => "Connection: close'r'n".
                        "Content-Length: ".strlen($postdata)."'r'n",
        'content' => $postdata
  )
);
$context = stream_context_create($opts);
$result = file_get_contents('http://example.com/submit.php', false, $context);

编辑 - 小错误应该是：

$opts = array(
    'http' => array(
        'method' => "POST",
        'header' => "Connection: close'r'n".
                    "Content-type: application/x-www-form-urlencoded'r'n".
                    "Content-Length: ".strlen($postdata)."'r'n",
        'content' => $postdata
  )
);

您无法获取源代码，但可以获取服务器提供的页面/输出。正如你提到的file_get_contents()，你可以用它来发送一个 POST 请求，但它看起来像这样。

// Create map with request parameters
$params = array ('surname' => 'Filip', 'lastname' => 'Czaja');
// Build Http query using params
$query = http_build_query ($params);
// Create Http context details
$contextData = array ( 
                'method' => 'POST',
                'header' => "Connection: close'r'n".
                            "Content-Length: ".strlen($query)."'r'n",
                'content'=> $query );
// Create context resource for our request
$context = stream_context_create (array ( 'http' => $contextData ));
// Read page rendered as result of your POST request
$result =  file_get_contents (
                  'http://www.sample-post-page.com',  // page url
                  false,
                  $context);
// Server response is now stored in $result variable so you can process it

示例来自： http://fczaja.blogspot.se/2011/07/php-how-to-send-post-request-with.html