我想通过他们的IP吸引游客…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,它正常工作,但问题是,这返回国家代码,如US或CA.,而不是整个国家名称,如美国或加拿大。

那么,除了hostip.info提供的这些,还有什么好的选择吗?

我知道我可以编写一些代码,最终将这两个字母转换为整个国家的名称,但我实在懒得编写包含所有国家的代码……

注:出于某种原因,我不想使用任何现成的CSV文件或任何代码,将抓取这个信息为我,像ip2country现成的代码和CSV。


当前回答

你可以使用http://ipinfo.io/获取ip地址的详细信息。

<?php
    function ip_details($ip)
    {
    $json = file_get_contents("http://ipinfo.io/{$ip}");
    $details = json_decode($json);
    return $details;
    }

    $details = ip_details(YoUR IP ADDRESS); 

    echo $details->city;
    echo "<br>".$details->country; 
    echo "<br>".$details->org; 
    echo "<br>".$details->hostname; /

    ?>

其他回答

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;
}

?>

使用方法:

Example1:获取访问者IP地址的详细信息

<?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 )

?>

例2:获取任意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 )

?>

我尝试了Chandra的答案,但我的服务器配置不允许file_get_contents()

PHP警告:file_get_contents() URL文件访问在服务器配置中被禁用

我修改了Chandra的代码,使它也适用于使用cURL的服务器:

function ip_visitor_country()
{

    $client  = @$_SERVER['HTTP_CLIENT_IP'];
    $forward = @$_SERVER['HTTP_X_FORWARDED_FOR'];
    $remote  = $_SERVER['REMOTE_ADDR'];
    $country  = "Unknown";

    if(filter_var($client, FILTER_VALIDATE_IP))
    {
        $ip = $client;
    }
    elseif(filter_var($forward, FILTER_VALIDATE_IP))
    {
        $ip = $forward;
    }
    else
    {
        $ip = $remote;
    }
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, "http://www.geoplugin.net/json.gp?ip=".$ip);
    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
    $ip_data_in = curl_exec($ch); // string
    curl_close($ch);

    $ip_data = json_decode($ip_data_in,true);
    $ip_data = str_replace('&quot;', '"', $ip_data); // for PHP 5.2 see stackoverflow.com/questions/3110487/

    if($ip_data && $ip_data['geoplugin_countryName'] != null) {
        $country = $ip_data['geoplugin_countryName'];
    }

    return 'IP: '.$ip.' # Country: '.$country;
}

echo ip_visitor_country(); // output Coutry name

?>

希望能有所帮助;-)

我正在使用ipinfodb.com api,并得到你正在寻找的东西。

它是完全免费的,你只需要注册他们获得你的api密钥。你可以包括他们的php类从他们的网站下载,或者你可以使用url格式检索信息。

以下是我正在做的:

我在我的脚本中包含了他们的php类,并使用下面的代码:

$ipLite = new ip2location_lite;
$ipLite->setKey('your_api_key');
if(!$_COOKIE["visitorCity"]){ //I am using cookie to store information
  $visitorCity = $ipLite->getCity($_SERVER['REMOTE_ADDR']);
  if ($visitorCity['statusCode'] == 'OK') {
    $data = base64_encode(serialize($visitorCity));
    setcookie("visitorCity", $data, time()+3600*24*7); //set cookie for 1 week
  }
}
$visitorCity = unserialize(base64_decode($_COOKIE["visitorCity"]));
echo $visitorCity['countryName'].' Region'.$visitorCity['regionName'];

这是它。

我有一个简短的答案,我在一个项目中使用过。 在我的回答中,我认为你有访问者IP地址。

$ip = "202.142.178.220";
$ipdat = @json_decode(file_get_contents("http://www.geoplugin.net/json.gp?ip=" . $ip));
//get ISO2 country code
if(property_exists($ipdat, 'geoplugin_countryCode')) {
    echo $ipdat->geoplugin_countryCode;
}
//get country full name
if(property_exists($ipdat, 'geoplugin_countryName')) {
    echo $ipdat->geoplugin_countryName;
}

我们可以使用geobytes.com来获得使用用户IP地址的位置

$user_ip = getIP();
$meta_tags = get_meta_tags('http://www.geobytes.com/IPLocator.htm?GetLocation&template=php3.txt&IPAddress=' . $user_ip);
echo '<pre>';
print_r($meta_tags);

它将返回这样的数据

Array(
    [known] => true
    [locationcode] => USCALANG
    [fips104] => US
    [iso2] => US
    [iso3] => USA
    [ison] => 840
    [internet] => US
    [countryid] => 254
    [country] => United States
    [regionid] => 126
    [region] => California
    [regioncode] => CA
    [adm1code] =>     
    [cityid] => 7275
    [city] => Los Angeles
    [latitude] => 34.0452
    [longitude] => -118.2840
    [timezone] => -08:00
    [certainty] => 53
    [mapbytesremaining] => Free
)

函数获取用户IP

function getIP(){
if (isset($_SERVER["HTTP_X_FORWARDED_FOR"])){
    $pattern = "/^(([1-9]?[0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5]).){3}([1-9]?[0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])$/";
    if(preg_match($pattern, $_SERVER["HTTP_X_FORWARDED_FOR"])){
            $userIP = $_SERVER["HTTP_X_FORWARDED_FOR"];
    }else{
            $userIP = $_SERVER["REMOTE_ADDR"];
    }
}
else{
  $userIP = $_SERVER["REMOTE_ADDR"];
}
return $userIP;
}