从他们的IP中获取访问者的国家

2022-08-30 06:09:55

我想通过他们的IP获得访问者的国家...现在我正在使用这个(http://api.hostip.info/country.php?ip=...... )

这是我的代码:

<?php

if (isset($_SERVER['HTTP_CLIENT_IP']))
{
    $real_ip_adress = $_SERVER['HTTP_CLIENT_IP'];
}

if (isset($_SERVER['HTTP_X_FORWARDED_FOR']))
{
    $real_ip_adress = $_SERVER['HTTP_X_FORWARDED_FOR'];
}
else
{
    $real_ip_adress = $_SERVER['REMOTE_ADDR'];
}

$cip = $real_ip_adress;
$iptolocation = 'http://api.hostip.info/country.php?ip=' . $cip;
$creatorlocation = file_get_contents($iptolocation);

?>

好吧,它工作正常,但问题是,这返回了像美国或加拿大这样的国家代码,而不是像美国或加拿大这样的整个国家名称。

那么,除了 hostip.info 提供这个之外,有什么好的替代方案吗?

我知道我可以写一些代码,最终将这两个字母变成整个国家名称,但我懒得写一个包含所有国家的代码......

P.S:出于某种原因,我不想使用任何现成的CSV文件或任何可以为我获取此信息的代码,比如ip2country现成代码和CSV。


答案 1

试试这个简单的PHP函数。

<?php

function ip_info($ip = NULL, $purpose = "location", $deep_detect = TRUE) {
    $output = NULL;
    if (filter_var($ip, FILTER_VALIDATE_IP) === FALSE) {
        $ip = $_SERVER["REMOTE_ADDR"];
        if ($deep_detect) {
            if (filter_var(@$_SERVER['HTTP_X_FORWARDED_FOR'], FILTER_VALIDATE_IP))
                $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
            if (filter_var(@$_SERVER['HTTP_CLIENT_IP'], FILTER_VALIDATE_IP))
                $ip = $_SERVER['HTTP_CLIENT_IP'];
        }
    }
    $purpose    = str_replace(array("name", "\n", "\t", " ", "-", "_"), NULL, strtolower(trim($purpose)));
    $support    = array("country", "countrycode", "state", "region", "city", "location", "address");
    $continents = array(
        "AF" => "Africa",
        "AN" => "Antarctica",
        "AS" => "Asia",
        "EU" => "Europe",
        "OC" => "Australia (Oceania)",
        "NA" => "North America",
        "SA" => "South America"
    );
    if (filter_var($ip, FILTER_VALIDATE_IP) && in_array($purpose, $support)) {
        $ipdat = @json_decode(file_get_contents("http://www.geoplugin.net/json.gp?ip=" . $ip));
        if (@strlen(trim($ipdat->geoplugin_countryCode)) == 2) {
            switch ($purpose) {
                case "location":
                    $output = array(
                        "city"           => @$ipdat->geoplugin_city,
                        "state"          => @$ipdat->geoplugin_regionName,
                        "country"        => @$ipdat->geoplugin_countryName,
                        "country_code"   => @$ipdat->geoplugin_countryCode,
                        "continent"      => @$continents[strtoupper($ipdat->geoplugin_continentCode)],
                        "continent_code" => @$ipdat->geoplugin_continentCode
                    );
                    break;
                case "address":
                    $address = array($ipdat->geoplugin_countryName);
                    if (@strlen($ipdat->geoplugin_regionName) >= 1)
                        $address[] = $ipdat->geoplugin_regionName;
                    if (@strlen($ipdat->geoplugin_city) >= 1)
                        $address[] = $ipdat->geoplugin_city;
                    $output = implode(", ", array_reverse($address));
                    break;
                case "city":
                    $output = @$ipdat->geoplugin_city;
                    break;
                case "state":
                    $output = @$ipdat->geoplugin_regionName;
                    break;
                case "region":
                    $output = @$ipdat->geoplugin_regionName;
                    break;
                case "country":
                    $output = @$ipdat->geoplugin_countryName;
                    break;
                case "countrycode":
                    $output = @$ipdat->geoplugin_countryCode;
                    break;
            }
        }
    }
    return $output;
}

?>

如何使用:

示例 1:获取访问者 IP 地址详细信息

<?php

echo ip_info("Visitor", "Country"); // India
echo ip_info("Visitor", "Country Code"); // IN
echo ip_info("Visitor", "State"); // Andhra Pradesh
echo ip_info("Visitor", "City"); // Proddatur
echo ip_info("Visitor", "Address"); // Proddatur, Andhra Pradesh, India

print_r(ip_info("Visitor", "Location")); // Array ( [city] => Proddatur [state] => Andhra Pradesh [country] => India [country_code] => IN [continent] => Asia [continent_code] => AS )

?>

示例 2:获取任何 IP 地址的详细信息。[支持 IPV4 和 IPV6]

<?php

echo ip_info("173.252.110.27", "Country"); // United States
echo ip_info("173.252.110.27", "Country Code"); // US
echo ip_info("173.252.110.27", "State"); // California
echo ip_info("173.252.110.27", "City"); // Menlo Park
echo ip_info("173.252.110.27", "Address"); // Menlo Park, California, United States

print_r(ip_info("173.252.110.27", "Location")); // Array ( [city] => Menlo Park [state] => California [country] => United States [country_code] => US [continent] => North America [continent_code] => NA )

?>

答案 2

您可以使用来自 http://www.geoplugin.net/ 的简单 API

$xml = simplexml_load_file("http://www.geoplugin.net/xml.gp?ip=".getRealIpAddr());
echo $xml->geoplugin_countryName ;


echo "<pre>";
foreach ($xml as $key => $value)
{
    echo $key , "= " , $value ,  " \n" ;
}
echo "</pre>";

使用的函数

function getRealIpAddr()
{
    if (!empty($_SERVER['HTTP_CLIENT_IP']))   //check ip from share internet
    {
      $ip=$_SERVER['HTTP_CLIENT_IP'];
    }
    elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR']))   //to check ip is pass from proxy
    {
      $ip=$_SERVER['HTTP_X_FORWARDED_FOR'];
    }
    else
    {
      $ip=$_SERVER['REMOTE_ADDR'];
    }
    return $ip;
}

输出

United States
geoplugin_city= San Antonio
geoplugin_region= TX
geoplugin_areaCode= 210
geoplugin_dmaCode= 641
geoplugin_countryCode= US
geoplugin_countryName= United States
geoplugin_continentCode= NA
geoplugin_latitude= 29.488899230957
geoplugin_longitude= -98.398696899414
geoplugin_regionCode= TX
geoplugin_regionName= Texas
geoplugin_currencyCode= USD
geoplugin_currencySymbol= $
geoplugin_currencyConverter= 1

它使您有很多可以玩的选项

谢谢

:)


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