小编sub*_*rui的帖子

当我试图重定向到其他网站时,我收到此错误:

遇到PHP错误

严重性:警告

消息:parse_url(/%22**)[function.parse-url]:无法解析URL

文件名:core/URI.php

行号:219

遇到了错误

所提交的网址带有不被接受的字符.

这是我在URI.php中的所有代码

private function _detect_uri()
{
    if ( ! isset($_SERVER['REQUEST_URI']) OR ! isset($_SERVER['SCRIPT_NAME']))
    {
        return '';
    }

    $uri = $_SERVER['REQUEST_URI'];
    if (strpos($uri, $_SERVER['SCRIPT_NAME']) === 0)
    {
        $uri = substr($uri, strlen($_SERVER['SCRIPT_NAME']));
    }
    elseif (strpos($uri, dirname($_SERVER['SCRIPT_NAME'])) === 0)
    {
        $uri = substr($uri, strlen(dirname($_SERVER['SCRIPT_NAME'])));
    }

    // This section ensures that even on servers that require the URI to be in the query string (Nginx) a correct
    // URI is found, and also fixes the …

我正在尝试从Contrie接收有关Alexa热门网站的信息,我希望收到:

• 网站位置;
• 网址;

对于我已经获得的URL,但是当我为网站位置添加标签时,某些东西不起作用,这是我的代码:

<?php

for ($z=0;$z<2;$z++) {
$html=file_get_contents('http://www.alexa.com/topsites/countries;'.$z.'/PT');
preg_match_all(
    '/<div class="count">.*?<\/div>.*?<a href="\/siteinfo\/.*?">(.*?)<\/a>/s',
    $html,
    $array, //array with sites
    PREG_SET_ORDER
);

for ($i=1;$i<count($array);$i++) {
    echo "<pre>"; print_r($array); echo "</pre>"; 
}
} 


?>

我明白了:

Array
(
[0] => Array
    (
        [0] => 
1



google.pt
        [1] => google.pt
    )

[1] => Array
    (
        [0] => 
2