我如何检查一个URL是否存在(不是404)在PHP?


当前回答

$headers = @get_headers($this->_value);
if(strpos($headers[0],'200')===false)return false;

所以任何时候你接触一个网站,得到200个以上的东西,它会工作

其他回答

$url = 'http://google.com';
$not_url = 'stp://google.com';

if (@file_get_contents($url)): echo "Found '$url'!";
else: echo "Can't find '$url'.";
endif;
if (@file_get_contents($not_url)): echo "Found '$not_url!";
else: echo "Can't find '$not_url'.";
endif;

// Found 'http://google.com'!Can't find 'stp://google.com'.

很快:

function http_response($url){
    $resURL = curl_init(); 
    curl_setopt($resURL, CURLOPT_URL, $url); 
    curl_setopt($resURL, CURLOPT_BINARYTRANSFER, 1); 
    curl_setopt($resURL, CURLOPT_HEADERFUNCTION, 'curlHeaderCallback'); 
    curl_setopt($resURL, CURLOPT_FAILONERROR, 1); 
    curl_exec ($resURL); 
    $intReturnCode = curl_getinfo($resURL, CURLINFO_HTTP_CODE); 
    curl_close ($resURL); 
    if ($intReturnCode != 200 && $intReturnCode != 302 && $intReturnCode != 304) { return 0; } else return 1;
}

echo 'google:';
echo http_response('http://www.google.com');
echo '/ ogogle:';
echo http_response('http://www.ogogle.com');

以上所有解决方案+额外的糖。(终极AIO解决方案)

/**
 * Check that given URL is valid and exists.
 * @param string $url URL to check
 * @return bool TRUE when valid | FALSE anyway
 */
function urlExists ( $url ) {
    // Remove all illegal characters from a url
    $url = filter_var($url, FILTER_SANITIZE_URL);

    // Validate URI
    if (filter_var($url, FILTER_VALIDATE_URL) === FALSE
        // check only for http/https schemes.
        || !in_array(strtolower(parse_url($url, PHP_URL_SCHEME)), ['http','https'], true )
    ) {
        return false;
    }

    // Check that URL exists
    $file_headers = @get_headers($url);
    return !(!$file_headers || $file_headers[0] === 'HTTP/1.1 404 Not Found');
}

例子:

var_dump ( urlExists('http://stackoverflow.com/') );
// Output: true;
function urlIsOk($url)
{
    $headers = @get_headers($url);
    $httpStatus = intval(substr($headers[0], 9, 3));
    if ($httpStatus<400)
    {
        return true;
    }
    return false;
}

这是一个解决方案,只读取源代码的第一个字节…如果file_get_contents失败,返回false…这也适用于远程文件,如图像。

 function urlExists($url)
{
    if (@file_get_contents($url,false,NULL,0,1))
    {
        return true;
    }
    return false;
}