Question

如何在远程脚本（target_url.php）中检索POSTFEILD var内容以在同一个远程脚本中使用它？

我正在做以下这样的事情。它不返回任何错误。

//Have tried multiple ways to setup the POSTFEILD argument, such as:

$data = array('var'=>'varcontents');
$post_arg = http_build_query($data) . "\n";

//and

$post_arg = 'var ='  . urlencode($varcontents);


//create cURL connection
$ch = curl_init('http://www.remotedomain.com/target_url.php');

//set options
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);

//set data to be posted
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_arg);

//perform the request
$result = curl_exec($ch);

//show information regarding the request
print_r(curl_getinfo($ch));
echo curl_errno($ch) . '-' . curl_error($ch);

Answer 1

刚遇到同样的问题，这是我找到的解决方案，我想我会与你分享：

使用

$varcontent = file_get_contents("php://input");

你的target_url.php中的

应该可以解决这个问题

Answer 2

使用

<?php
$post_arg = http_build_query( array('var'=>'varcontents') );
//create cURL connection
$ch = curl_init('http://localhost/target_url.php');
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_arg);
$result = curl_exec($ch);
var_dump($result);

作为“客户”和

<?php
var_dump($_POST);

as target_url.php，输出为

string(186) "<pre class='xdebug-var-dump' dir='ltr'>
<b>array</b>
  'var' <font color='#888a85'>=&gt;</font> <small>string</small> <font color='#cc0000'>'varcontents'</font> <i>(length=11)</i>
</pre>"

（提醒我停用xdebug选项;-)），所以在target_url.php中可以通过$ _POST访问参数。

Answer 3

你的卷曲设置似乎不太对劲。试试这个：

$post_data = http_build_query(array('var'=>'varcontents'));
$url = 'http://www.remotedomain.com/target_url.php';

$ch = curl_init();
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data);
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
$buffer = curl_exec($ch);
curl_close($ch);

在你的target_url.php中，als已经说过：

$varcontent = $_POST['var'];

在远程脚本中访问CURLOPT_POSTFIELD变量？

3 个答案: