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问题描述

我想为这个 PHP 代码做一个样式,如下图:

    <?php

function get_ip() {
    if(isset($_SERVER['HTTP_CLIENT_IP'])) {
        return $_SERVER['HTTP_CLIENT_IP'];
    }
    elseif (isset($_SERVER['HTTP_X_FORWARDED_FOR'])) {
        return $_SERVER['HTTP_X_FORWARDED_FOR'];

    }
    else {
        return (isset($_SERVER['REMOTE_ADDR']) ? $_SERVER['REMOTE_ADDR'] : '');
    }       
}
$ip=get_ip();
$query=@unserialize(file_get_contents('http://ip-api.com/php/'.$ip));
if($query && $query['status'] == 'success') {
    echo "ISP:".$query['isp']."<br/>";
    echo "COUNTRY:".$query['country']."<br/>";
    echo "COUNTRY CODE:".$query['countryCode']."<br/>";
    echo "REGION NAME:".$query['regionName']."<br/>";
    echo "CITY:".$query['city']."<br/>";
    echo "ZIP:".$query['zip']."<br/>";
    echo "LATITUDE:".$query['latitude']."<br/>";
    echo "LONGITUDE:".$query['longitude']."<br/>";
    echo "TIMEZONE:".$query['timezone']."<br/>";
    echo "ORG:".$query['org']."<br/>";
    echo "AS:".$query['as']."<br/>";
}

?>

我想要这样的结果:

在此处输入图像描述

标签: phpstyling

解决方案


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