file_get_contents() 没有返回？

发布于 2024-09-02 13:05:35 字数 652 浏览 5 评论 0 原文

为什么我没有在 file_get_contents($myurl) 上得到返回，但我确实得到了 wget

updated:

的输出，我使用curl 来获取返回标头和结果，并发现它位于标头而不是在body

HTTP/1.1 200 Document follows
Transfer-Encoding: chunked
Content-type: text/plain
Server: XMS (724Solutions HTA XSAM_30_M2_B020 20070803.172831)
Date: Fri, 21 May 2010 10:48:31 GMT
Accept-Ranges: bytes

HTTP/1.1 404 ChargedPartyNotAvailable
Transfer-Encoding: chunked
Content-type: text/plain
Server: XMS (724Solutions HTA XSAM_30_M2_B020 20070803.172831)
Date: Fri, 21 May 2010 10:34:13 GMT
Accept-Ranges: bytes

我怎样才能只提取出“200 Document Follow”和“404 ChargedPartyNotAvailable”？

原文

why i not getting return on file_get_contents($myurl) but i do get output for wget

updated:

i use curl to get the return header and result and discovery that the it is on the header instead in of the body

HTTP/1.1 200 Document follows
Transfer-Encoding: chunked
Content-type: text/plain
Server: XMS (724Solutions HTA XSAM_30_M2_B020 20070803.172831)
Date: Fri, 21 May 2010 10:48:31 GMT
Accept-Ranges: bytes

HTTP/1.1 404 ChargedPartyNotAvailable
Transfer-Encoding: chunked
Content-type: text/plain
Server: XMS (724Solutions HTA XSAM_30_M2_B020 20070803.172831)
Date: Fri, 21 May 2010 10:34:13 GMT
Accept-Ranges: bytes

How can i extract out only "200 Document Follow" and "404 ChargedPartyNotAvailable" ?

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貪欢 2024-09-09 13:05:35

您是否启用了 allow_url_fopen你的php配置？

但是，我希望您会收到警告，如果没有的话 - 您是否显示错误/警告？您可以在脚本顶部临时添加 :

error_reporting(E_ALL);
ini_set('display_errors', true);

然后您可能会明白为什么 file_get_contents() 不起作用。

编辑

您可能只能使用get_headers() 如果您只对标题感兴趣。

Do you have allow_url_fopen enabled in your php configuration?

However I would expect that you would get a warning generated if not - do you have errors/warnings displayed? You could add temporarily at the top of your script :

error_reporting(E_ALL);
ini_set('display_errors', true);

and then you might see why file_get_contents() doesn't work.

Edit

You might just be able to use get_headers() if you're only interested in headers.

回复收藏 0 原文

她如夕阳 2024-09-09 13:05:35

您可以使用 PHP Curl 包来检索 URL 的内容

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$myUrl);
$result = curl_exec($ch);
curl_close($ch);

，或者如果您只想使用 file_get_contents，请检查您是否正确配置了 PHP.ini

http://www.php.net/manual/en/filesystem.configuration.php#ini.allow-url-fopen

正如这里提到的（http://php.net/manual/en/ function.file-get-contents.php)

URL 可以用作文件名
这个函数如果 fopen 包装器
已启用。请参阅 fopen() 来了解
有关如何指定的更多详细信息
文件名。

You can use PHP Curl package to retrieve the content of URL

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$myUrl);
$result = curl_exec($ch);
curl_close($ch);

Or if you want to use only file_get_contents, check if you configured PHP.ini correctly

http://www.php.net/manual/en/filesystem.configuration.php#ini.allow-url-fopen

As mentioned here (http://php.net/manual/en/function.file-get-contents.php)

A URL can be used as a filename with
this function if the fopen wrappers
have been enabled. See fopen() for
more details on how to specify the
filename.