Question

我正在编写一个URL抓取工具（只是名称和说明），我正在尝试处理301重定向。

现在，我检查标题，如果不是200，我会尝试找到要重定向到标题内的位置。我的问题出现了，因为尽管我在那里看到了，但是array_search没有返回Location值的键。

这是代码段：

if(strpos($url_headers[0], "200") !== false){
        echo "in here";
        return $url;
    }else{
        print_r($url_headers);
        //look for location
        $location_key = array_search("Location: ", $url_headers);
        echo "Location Key: " . $location_key;
        $redirect_string = $url_headers[$location_key];
        $clean_url = str_replace("Location: ", "", $redirect_string);
        return $clean_url;
    }

这个输出是：

Array ( [0] => HTTP/1.0 301 Moved Permanently [1] => Location: http://www.google.com/ [2] => Content-Type: text/html; charset=UTF-8 [3] => Date: Wed, 13 Feb 2013 03:30:00 GMT [4] => Expires: Fri, 15 Mar 2013 03:30:00 GMT [5] => Cache-Control: public, max-age=2592000 [6] => Server: gws [7] => Content-Length: 219 [8] => X-XSS-Protection: 1; mode=block [9] => X-Frame-Options: SAMEORIGIN [10] => HTTP/1.0 200 OK [11] => Date: Wed, 13 Feb 2013 03:30:00 GMT [12] => Expires: -1 [13] => Cache-Control: private, max-age=0 [14] => Content-Type: text/html; charset=ISO-8859-1 [15] => Set-Cookie: PREF=ID=fe86e29432d4e240:FF=0:TM=1360726200:LM=1360726200:S=Wg8VEU7kc7UtcKc-; expires=Fri, 13-Feb-2015 03:30:00 GMT; path=/; domain=.google.com [16] => Set-Cookie: NID=67=KH8Zu8EpKjrhje8nD0lk_868mqvQr9pGwsAsaUuPDD_PRUgohJHoOkdlyYEHWmohUtndyENDJ0oZq8pC1aqOg20anXpUn5btQX5GYM6kYlgMhYxIPajtGp9KymmMDO1Y; expires=Thu, 15-Aug-2013 03:30:00 GMT; path=/; domain=.google.com; HttpOnly [17] => P3P: CP="This is not a P3P policy! See http://www.google.com/support/accounts/bin/answer.py?hl=en&answer=151657 for more info." [18] => Server: gws [19] => X-XSS-Protection: 1; mode=block [20] => X-Frame-Options: SAMEORIGIN ) Location Key: {"error":"invalid_url","error_code":null}

我做错了什么？在抓取用户提供的链接时，是否有更优雅的方法来处理重定向？

Answer 1

$url_headers[0] = 'HTTP/1.0 200';
if(strpos($url_headers[0], "200") > 0){
  echo "here";
} else {
  //look for location
   $location_key = getLocation($url_headers);
   echo "Location Key: " . $location_key;
}

function getLocation($data) {
    $url = false;
    foreach($data as $key => $value) {
        if (preg_match("/Location:/", $value)) {
             echo "A match was found.";
            //$url = $matches[1];
            $url = $data[$key];
            break;
        }
    }
    return $url;
}

Answer 2

strpos如果找不到匹配则返回false，因此您需要执行

if( ! strpos($url_headers[0], "200"))

Answer 3

        $ch = curl_init();
        curl_setopt($ch, CURLOPT_URL, $url);
        curl_setopt($ch, CURLOPT_HEADER, true);
        curl_setopt($ch, CURLOPT_FOLLOWLOCATION, false);
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
        $a = curl_exec($ch);
        if(preg_match('#Location: (.*)#', $a, $r)){
         $l = trim($r[1]);
         return $l;
        }else{
            return $url;
        }

这大部分都有效，但仍然无法重定向到https（由于某种原因，他们需要双重定向？）

（通过http://zzz.rezo.net/HowTo-Expand-Short-URLs.html）

当数组在Array中时，PHP Array Search返回false

3 个答案: