PHP：“file_get_contents（）”从内容验证的URL返回NULL [英] PHP: "file_get_contents()" returning NULL from content-verified URL

问题描述

问题：

/ strong>

当我尝试通过 file_get_contents（）获取特定本地URL的内容时，它将返回一个空字符串'' length = 0）。

我尝试过的：

  $ url ='http：//localhost/bbq/index.php/company/get/id/2131/return/json'; 
 $ results = file_get_contents（$ url）; 
 echo'< pre>结果：< br />'，var_export（$ results），'< / pre> 
  

直接访问 http：//localhost/bbq/index.php/公司/ get / id / 2131 / return / json ，一切都很美好： {id：2131，value：Acme Anvil Corp。 / code>





为了检查自己，我用 $ url ='http://www.google.com' / code>，并且它按预期工作，然后我尝试单独的本地uri， http：// localhost /？phpinfo = 1 问题：

我是什么？缺少

[ADDITIONAL INFO] >
allow_url_fopen 是ON

>某些。

作为CODEIGNITER项目的安全措施，控制器脚本的第一行具有：

  if（！defined（'BASEPATH'））exit（'不允许直接脚本访问）; 
  

评论它没有提供任何改进，但我认为这可能值得一提。 / p>

此外，这里是一个HTTP响应头的转储。也许有人可以看到那里的东西。这一切看起来都很直接，但我真的不知道我应该寻找什么。

  array = 13）
 0 => string'HTTP / 1.1 200 OK'（length = 15）
 1 => string'Date：Thu，27 Dec 2012 19:58:47 GMT'（length = 35）
 2 =>字符串'服务器：Apache / 2.4.2（Win64）PHP / 5.4.3'（长度= 38）
 3 => string'X-Powered-By：PHP / 5.4.3'（length = 23）
 4 => string'Set-Cookie：csrf_cookie_name = 4ae46f9a7e612ae22a080a4e88bd349c; expires = Thu，27-Dec-2012 21:58:47 GMT; path = /'（length = 108）
 5 => string'Set-Cookie：PHPSESSID = ah70p0ldl5qptcauei1t8ihud3; path = /'（length = 56）
 6 => string'Expires：Thu，19 Nov 1981 08:52:00 GMT'（length = 38）
 7 => string'Cache-Control：no-store，no-cache，must-revalidate，post-check = 0，pre-check = 0'（length = 77）
 8 = string'Pragma：no-cache'（length = 16）
 9 => string'Refresh：0; url = http：//localhost/bbq/index.php/index.php/user/login'（length = 66）
 10 => string'Content-Length：0'（length = 17）
 11 => string'Connection：close'（length = 17）
 12 => string'Content-Type：text / html'（length = 23）
  
 
 
 
  [/ ADDITIONAL INFO]  
 
 
 
  BTW ：我正在运行：
 
 
 
 
 
 Win7 64上的WAMPSERVER 
 
 
 Apache 2.4.2 
 
 
 PHP 5.4.3 
 
 
 CodeIgniter 2.1.2 
 
 
 
 
解决方案
立即开始调试：
 ＃我们希望文本输出用于调试。或者这是工作，或者你需要在浏览器中查找
＃source（或者更好，用`curl`命令行客户端调试）
 header（'Content-Type：text / plain'）; 
 
＃对于快速和脏的调试 - 开始告诉是否值得
＃警告或通知。 （do log in production）
 error_reporting（〜0）; ini_set（'display_errors'，1）; 
 
 $ url ='http：// localhost / bbq / company / get / id / 2131 / return / json'; 
 $ results = file_get_contents（$ url）; 
 var_dump（$ http_response_header，$ results）; 
 
＃硬退出，刷新所有（潜在）缓冲区。 
 die（）; 
  
现在，对文件的请求必须至少返回一些文本输出，因为 var_dump 。您还可以在标题调用之下的开头添加一些标记输出，因此很清楚，该脚本已被调用，并且已开始运行。
 
 
 
但是更好的变体正在使用步骤调试器并创建具有断点的调试会话。我推荐Xdebug。
 
I'm betting that it'll end up being something simple ... but I can use a hand in spotting it.

Problem:

When I try to get the contents of a specific local URL via file_get_contents(), it comes back as an empty string '' (length=0).

What I've tried:
    $url = 'http://localhost/bbq/index.php/company/get/id/2131/return/json';
    $results = file_get_contents($url);
    echo'<pre>Results:<br />',var_export($results),'</pre>';
When directly visiting http://localhost/bbq/index.php/company/get/id/2131/return/json, everything echos beautifully: {"id":"2131","value":"Acme Anvil Corp."}

To check myself, I tried it with $url = 'http://www.google.com', and it worked as expected so then I tried a separate Local uri, http://localhost/?phpinfo=1, and it also worked just fine.

Question:

What am I missing???




[ADDITIONAL INFO]

allow_url_fopen is ON

I may have just found something.

As a safety measure in CODEIGNITER projects, the first line of the controller script has:
    if ( !defined('BASEPATH')) exit('No direct script access allowed');
Commenting it out didn't provide any improvements, but I thought it may be worth mentioning nonetheless.

Also, here's a dump of the HTTP Response Header. Maybe someone can see something in there. It all seems pretty straight forward to me, but I don't really know what I should be looking for.
array (size=13)
  0 => string 'HTTP/1.1 200 OK' (length=15)
  1 => string 'Date: Thu, 27 Dec 2012 19:58:47 GMT' (length=35)
  2 => string 'Server: Apache/2.4.2 (Win64) PHP/5.4.3' (length=38)
  3 => string 'X-Powered-By: PHP/5.4.3' (length=23)
  4 => string 'Set-Cookie: csrf_cookie_name=4ae46f9a7e612ae22a080a4e88bd349c; expires=Thu, 27-Dec-2012 21:58:47 GMT; path=/' (length=108)
  5 => string 'Set-Cookie: PHPSESSID=ah70p0ldl5qptcauei1t8ihud3; path=/' (length=56)
  6 => string 'Expires: Thu, 19 Nov 1981 08:52:00 GMT' (length=38)
  7 => string 'Cache-Control: no-store, no-cache, must-revalidate, post-check=0, pre-check=0' (length=77)
  8 => string 'Pragma: no-cache' (length=16)
  9 => string 'Refresh: 0;url=http://localhost/bbq/index.php/index.php/user/login' (length=66)
  10 => string 'Content-Length: 0' (length=17)
  11 => string 'Connection: close' (length=17)
  12 => string 'Content-Type: text/html' (length=23) 
[/ADDITIONAL INFO]

BTW, I'm running:


WAMPSERVER on Win7 64 
Apache 2.4.2  
PHP 5.4.3 
CodeIgniter 2.1.2

解决方案
Start to debug this now:
# we want text output for debugging. Either this works or you need to look into
# source in your browser (or even better, debug with the `curl` commandline client.)
header('Content-Type: text/plain');

# For quick and dirty debugging - start to tell about if something is worth
# to warn or notice about. (do logging in production)
error_reporting(~0); ini_set('display_errors', 1);

$url = 'http://localhost/bbq/company/get/id/2131/return/json';
$results = file_get_contents($url);
var_dump($http_response_header, $results);

# hard exit, flush all (potential) buffers.
die();
A request to the file must now at least give you back some text output because of the var_dump. You can also add some marker output at the beginning below the header call, so it's clear that the script has been called and it did start to run.

However a better variant is using a step-debugger and creating a debugging session with breakpoints. I recommend Xdebug.

                        
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