php 如何使用file_get_contents在PHP中发布数据？
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How to post data in PHP using file_get_contents?
提问by Paras Chopra
I'm using PHP's function file_get_contents()to fetch contents of a URL and then I process headers through the variable $http_response_header.
我正在使用 PHP 的函数file_get_contents()来获取 URL 的内容，然后我通过变量处理标头$http_response_header。
Now the problem is that some of the URLs need some data to be posted to the URL (for example, login pages).
现在的问题是某些 URL 需要将一些数据发布到 URL（例如，登录页面）。
How do I do that?
我怎么做？
I realize using stream_context I may be able to do that but I am not entirely clear.
我意识到使用 stream_context 我也许能够做到这一点，但我并不完全清楚。
Thanks.
谢谢。
回答by Pascal MARTIN
Sending an HTTP POST request using file_get_contentsis not that hard, actually : as you guessed, you have to use the $contextparameter.
使用发送 HTTP POST 请求file_get_contents并不难，实际上：正如您猜测的那样，您必须使用$context参数。
There's an example given in the PHP manual, at this page : HTTP context options(quoting):
PHP 手册中给出了一个示例，在此页面上：HTTP 上下文选项（引用）：
$postdata = http_build_query(
array(
'var1' => 'some content',
'var2' => 'doh'
)
);
$opts = array('http' =>
array(
'method' => 'POST',
'header' => 'Content-Type: application/x-www-form-urlencoded',
'content' => $postdata
)
);
$context = stream_context_create($opts);
$result = file_get_contents('http://example.com/submit.php', false, $context);
Basically, you have to create a stream, with the right options (there is a full list on that page), and use it as the third parameter to file_get_contents-- nothing more ;-)
基本上，您必须使用正确的选项创建一个流（该页面上有一个完整列表），并将其用作第三个参数file_get_contents- 仅此而已;-)
As a sidenote : generally speaking, to send HTTP POST requests, we tend to use curl, which provides a lot of options an all -- but streams are one of the nice things of PHP that nobody knows about... too bad...
作为旁注：一般来说，要发送 HTTP POST 请求，我们倾向于使用 curl，它提供了很多选项——但流是 PHP 的优点之一，没人知道......太糟糕了。 .
回答by Macbric
An alternative, you can also use fopen
另一种选择，您也可以使用fopen
$params = array('http' => array(
'method' => 'POST',
'content' => 'toto=1&tata=2'
));
$ctx = stream_context_create($params);
$fp = @fopen($sUrl, 'rb', false, $ctx);
if (!$fp)
{
throw new Exception("Problem with $sUrl, $php_errormsg");
}
$response = @stream_get_contents($fp);
if ($response === false)
{
throw new Exception("Problem reading data from $sUrl, $php_errormsg");
}
回答by user2525449
$sUrl = 'http://www.linktopage.com/login/';
$params = array('http' => array(
'method' => 'POST',
'content' => 'username=admin195&password=d123456789'
));
$ctx = stream_context_create($params);
$fp = @fopen($sUrl, 'rb', false, $ctx);
if(!$fp) {
throw new Exception("Problem with $sUrl, $php_errormsg");
}
$response = @stream_get_contents($fp);
if($response === false) {
throw new Exception("Problem reading data from $sUrl, $php_errormsg");
}
