php file_get_contents 当 url 不存在时
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file_get_contents when url doesn't exist
提问by sami
I'm using file_get_contents() to access a URL.
我正在使用 file_get_contents() 访问 URL。
file_get_contents('http://somenotrealurl.com/notrealpage');
If the URL is not real, it return this error message. How can I get it to error gracefully so that I know that the page doesn't exist and act accordingly without displaying this error message?
如果 URL 不真实，则返回此错误消息。如何让它优雅地出错，以便我知道该页面不存在并采取相应措施而不显示此错误消息？
file_get_contents('http://somenotrealurl.com/notrealpage')
[function.file-get-contents]:
failed to open stream: HTTP request failed! HTTP/1.0 404 Not Found
in myphppage.php on line 3
for example in zend you can say: if ($request->isSuccessful())
例如在 Zend 中，您可以说： if ($request->isSuccessful())
$client = New Zend_Http_Client();
$client->setUri('http://someurl.com/somepage');
$request = $client->request();
if ($request->isSuccessful()) {
//do stuff with the result
}
回答by ynh
You need to check the HTTP response code:
您需要检查HTTP 响应代码：
function get_http_response_code($url) {
$headers = get_headers($url);
return substr($headers[0], 9, 3);
}
if(get_http_response_code('http://somenotrealurl.com/notrealpage') != "200"){
echo "error";
}else{
file_get_contents('http://somenotrealurl.com/notrealpage');
}
回答by Orbling
With such commands in PHP, you can prefix them with an @to suppress such warnings.
使用 PHP 中的此类命令，您可以为它们添加前缀@以抑制此类警告。
@file_get_contents('http://somenotrealurl.com/notrealpage');
file_get_contents()returns FALSEif a failure occurs, so if you check the returned result against that then you can handle the failure
file_get_contents()FALSE如果发生失败则返回，因此如果您根据返回的结果检查返回的结果，则可以处理失败
$pageDocument = @file_get_contents('http://somenotrealurl.com/notrealpage');
if ($pageDocument === false) {
// Handle error
}
回答by Grzegorz
Each time you call file_get_contentswith an http wrapper, a variable in local scope is created: $http_response_header
每次file_get_contents使用 http 包装器调用时，都会在本地范围内创建一个变量：$http_response_header
This variable contains all HTTP headers. This method is better over get_headers()function since only one request is executed.
此变量包含所有 HTTP 标头。这种方法优于get_headers()函数，因为只执行一个请求。
Note: 2 different requests can end differently. For example, get_headers()will return 503 and file_get_contents() would return 200. And you would get proper output but would not use it due to 503 error in get_headers() call.
注意：2 个不同的请求可以以不同的方式结束。例如，get_headers()将返回 503，file_get_contents() 将返回 200。您将获得正确的输出，但由于 get_headers() 调用中的 503 错误而不会使用它。
function getUrl($url) {
$content = file_get_contents($url);
// you can add some code to extract/parse response number from first header.
// For example from "HTTP/1.1 200 OK" string.
return array(
'headers' => $http_response_header,
'content' => $content
);
}
// Handle 40x and 50x errors
$response = getUrl("http://example.com/secret-message");
if ($response['content'] === FALSE)
echo $response['headers'][0]; // HTTP/1.1 401 Unauthorized
else
echo $response['content'];
This aproach also alows you to have track of few request headers stored in different variables since if you use file_get_contents() $http_response_headeris overwritten in local scope.
这种方法还允许您跟踪存储在不同变量中的几个请求标头，因为如果您使用 file_get_contents() $http_response_header在本地范围内被覆盖。
回答by nikc.org
While file_get_contentsis very terse and convenient, I tend to favour the Curl library for better control. Here's an example.
虽然file_get_contents非常简洁和方便，但我倾向于使用 Curl 库来更好地控制。这是一个例子。
function fetchUrl($uri) {
$handle = curl_init();
curl_setopt($handle, CURLOPT_URL, $uri);
curl_setopt($handle, CURLOPT_POST, false);
curl_setopt($handle, CURLOPT_BINARYTRANSFER, false);
curl_setopt($handle, CURLOPT_HEADER, true);
curl_setopt($handle, CURLOPT_RETURNTRANSFER, true);
curl_setopt($handle, CURLOPT_CONNECTTIMEOUT, 10);
$response = curl_exec($handle);
$hlength = curl_getinfo($handle, CURLINFO_HEADER_SIZE);
$httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE);
$body = substr($response, $hlength);
// If HTTP response is not 200, throw exception
if ($httpCode != 200) {
throw new Exception($httpCode);
}
return $body;
}
$url = 'http://some.host.com/path/to/doc';
try {
$response = fetchUrl($url);
} catch (Exception $e) {
error_log('Fetch URL failed: ' . $e->getMessage() . ' for ' . $url);
}
回答by tfont
Simple and functional (easy to use anywhere):
简单而实用（易于在任何地方使用）：
function file_contents_exist($url, $response_code = 200)
{
$headers = get_headers($url);
if (substr($headers[0], 9, 3) == $response_code)
{
return TRUE;
}
else
{
return FALSE;
}
}
Example:
例子：
$file_path = 'http://www.google.com';
if(file_contents_exist($file_path))
{
$file = file_get_contents($file_path);
}
回答by Kuijkens
To avoid double requests as commented by Orblingon the answer of ynhyou could combine their answers. If you get a valid response in the first place, use that. If not find out what the problem was (if needed).
为了避免重复请求作为评论说Orbling上的答案YNH你可以结合自己的答案。如果您首先得到有效的响应，请使用它。如果没有找出问题所在（如果需要）。
$urlToGet = 'http://somenotrealurl.com/notrealpage';
$pageDocument = @file_get_contents($urlToGet);
if ($pageDocument === false) {
$headers = get_headers($urlToGet);
$responseCode = substr($headers[0], 9, 3);
// Handle errors based on response code
if ($responseCode == '404') {
//do something, page is missing
}
// Etc.
} else {
// Use $pageDocument, echo or whatever you are doing
}
回答by alniks
You may add 'ignore_errors' => true to options:
您可以将 'ignore_errors' => true 添加到选项中：
$options = array(
'http' => array(
'ignore_errors' => true,
'header' => "Content-Type: application/json\r\n"
)
);
$context = stream_context_create($options);
$result = file_get_contents('http://example.com', false, $context);
In that case you will be able to read a response from the server.
在这种情况下，您将能够从服务器读取响应。
回答by SixSense
$url = 'https://www.yourdomain.com';
Normal
普通的
function checkOnline($url) {
$headers = get_headers($url);
$code = substr($headers[0], 9, 3);
if ($code == 200) {
return true;
}
return false;
}
if (checkOnline($url)) {
// URL is online, do something..
$getURL = file_get_contents($url);
} else {
// URL is offline, throw an error..
}
Pro
亲
if (substr(get_headers($url)[0], 9, 3) == 200) {
// URL is online, do something..
}
Wtf level
重量级
(substr(get_headers($url)[0], 9, 3) == 200) ? echo 'Online' : echo 'Offline';
