我如何处理file_get_contents()函数在PHP警告?

我写了这样的PHP代码

$site="http://www.google.com";
$content = file_get_content($site);
echo $content;

但是当我从$site中删除“http://”时，我得到以下警告:

警告: file_get_contents (www.google.com) (函数。file-get-contents]:失败打开水流:

我试过，试过，但没成功。

当前回答

您可以使用这个脚本

$url = @file_get_contents("http://www.itreb.info");
if ($url) {
    // if url is true execute this 
    echo $url;
} else {
    // if not exceute this 
    echo "connection error";
}

2012-02-27 07:57:12

其他回答

function custom_file_get_contents($url) {
    return file_get_contents(
        $url,
        false,
        stream_context_create(
            array(
                'http' => array(
                    'ignore_errors' => true
                )
            )
        )
    );
}

$content=FALSE;

if($content=custom_file_get_contents($url)) {
    //play with the result
} else {
    //handle the error
}

2014-02-24 00:28:28

您可以使用这个脚本

$url = @file_get_contents("http://www.itreb.info");
if ($url) {
    // if url is true execute this 
    echo $url;
} else {
    // if not exceute this 
    echo "connection error";
}

2012-02-27 07:57:12

这将尝试获取数据，如果它不起作用，它将捕获错误，并允许您在捕获范围内执行所需的任何操作。

try {
    $content = file_get_contents($site);
} catch(\Exception $e) {
    return 'The file was not found';
}

2016-07-28 16:37:04

就像这样:

public function get($curl,$options){
    $context = stream_context_create($options);
    $file = @file_get_contents($curl, false, $context);
    $str1=$str2=$status=null;
    sscanf($http_response_header[0] ,'%s %d %s', $str1,$status, $str2);
    if($status==200)
        return $file        
    else 
        throw new \Exception($http_response_header[0]);
}

2019-04-23 16:36:32

我解决了所有的问题，这是工作的所有环节

public function getTitle($url)
    {
        try {
            if (strpos($url, 'www.youtube.com/watch') !== false) {
                $apikey = 'AIzaSyCPeA3MlMPeT1CU18NHfJawWAx18VoowOY';
                $videoId = explode('&', explode("=", $url)[1])[0];
                $url = 'https://www.googleapis.com/youtube/v3/videos?id=' . $videoId . '&key=' . $apikey . '&part=snippet';

                $ch = curl_init();

                curl_setopt($ch, CURLOPT_HEADER, 0);
                curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
                curl_setopt($ch, CURLOPT_URL, $url);
                curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
                curl_setopt($ch, CURLOPT_VERBOSE, 0);
                curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
                $response = curl_exec($ch);
                curl_close($ch);

                $data = json_decode($response);
                $value = json_decode(json_encode($data), true);

                $title = $value['items'][0]['snippet']['title'];
            } else {
                set_error_handler(
                    function () {
                            return false;
                    }
                );
                if (($str = file_get_contents($url)) === false) {
                    $title = $url;
                } else {
                    preg_match("/\<title\>(.*)\<\/title\>/i", $str, $title);
                    $title = $title[1];
                    if (preg_replace('/[\x00-\x1F\x7F-\xFF]/', '', $title))
                        $title = utf8_encode($title);
                    $title = html_entity_decode($title);
                }
                restore_error_handler();
            }
        } catch (Exception $e) {
            $title = $url;
        }
        return $title;
    }

2021-10-09 09:12:24

我如何处理file_get_contents()函数在PHP警告?

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