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

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;
}

其他回答

使用以下服务

1) http://api.hostip.info/get_html.php?ip=12.215.42.19

2)

$json = file_get_contents('http://freegeoip.appspot.com/json/66.102.13.106');
$expression = json_decode($json);
print_r($expression);

3) http://ipinfodb.com/ip_location_api.php

Try

  <?php
  //gives you the IP address of the visitors
  if (!empty($_SERVER['HTTP_CLIENT_IP'])) {
      $ip = $_SERVER['HTTP_CLIENT_IP'];}
  else if (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
      $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
  } else {
      $ip = $_SERVER['REMOTE_ADDR'];
  }

  //return the country code
  $url = "http://api.wipmania.com/$ip";
  $country = file_get_contents($url);
  echo $country;

  ?>

我知道这是旧的,但我尝试了其他一些解决方案在这里,他们似乎是过时的或只是返回null。我是这么做的。

使用http://www.geoplugin.net/json.gp?ip=,不需要任何类型的注册或支付服务。

function get_client_ip_server() {
  $ipaddress = '';
if (isset($_SERVER['HTTP_CLIENT_IP']))
  $ipaddress = $_SERVER['HTTP_CLIENT_IP'];
else if(isset($_SERVER['HTTP_X_FORWARDED_FOR']))
  $ipaddress = $_SERVER['HTTP_X_FORWARDED_FOR'];
else if(isset($_SERVER['HTTP_X_FORWARDED']))
  $ipaddress = $_SERVER['HTTP_X_FORWARDED'];
else if(isset($_SERVER['HTTP_FORWARDED_FOR']))
  $ipaddress = $_SERVER['HTTP_FORWARDED_FOR'];
else if(isset($_SERVER['HTTP_FORWARDED']))
  $ipaddress = $_SERVER['HTTP_FORWARDED'];
else if(isset($_SERVER['REMOTE_ADDR']))
  $ipaddress = $_SERVER['REMOTE_ADDR'];
else
  $ipaddress = 'UNKNOWN';

  return $ipaddress;
}

$ipaddress = get_client_ip_server();

function getCountry($ip){
    $curlSession = curl_init();
    curl_setopt($curlSession, CURLOPT_URL, 'http://www.geoplugin.net/json.gp?ip='.$ip);
    curl_setopt($curlSession, CURLOPT_BINARYTRANSFER, true);
    curl_setopt($curlSession, CURLOPT_RETURNTRANSFER, true);

    $jsonData = json_decode(curl_exec($curlSession));
    curl_close($curlSession);

    return $jsonData->geoplugin_countryCode;
}

echo "County: " .getCountry($ipaddress);

如果你想要更多关于它的信息,这是一个完整的Json返回:

{
  "geoplugin_request":"IP_ADDRESS",
  "geoplugin_status":200,
  "geoplugin_delay":"2ms",
  "geoplugin_credit":"Some of the returned data includes GeoLite data created by MaxMind, available from <a href='http:\/\/www.maxmind.com'>http:\/\/www.maxmind.com<\/a>.",
  "geoplugin_city":"Current City",
  "geoplugin_region":"Region",
  "geoplugin_regionCode":"Region Code",
  "geoplugin_regionName":"Region Name",
  "geoplugin_areaCode":"",
  "geoplugin_dmaCode":"650",
  "geoplugin_countryCode":"US",
  "geoplugin_countryName":"United States",
  "geoplugin_inEU":0,
  "geoplugin_euVATrate":false,
  "geoplugin_continentCode":"NA",
  "geoplugin_continentName":"North America",
  "geoplugin_latitude":"37.5563",
  "geoplugin_longitude":"-99.9413",
  "geoplugin_locationAccuracyRadius":"5",
  "geoplugin_timezone":"America\/Chicago",
  "geoplugin_currencyCode":"USD",
  "geoplugin_currencySymbol":"$",
  "geoplugin_currencySymbol_UTF8":"$",
  "geoplugin_currencyConverter":1
}

尝试这个简单的一行代码,您将从他们的ip远程地址获得访客的国家和城市。

$tags = get_meta_tags('http://www.geobytes.com/IpLocator.htm?GetLocation&template=php3.txt&IpAddress=' . $_SERVER['REMOTE_ADDR']);
echo $tags['country'];
echo $tags['city'];

您可以使用此代码片段。

 if($a = unserialize(file_get_contents('http://www.geoplugin.net/php.gp?ip='.$_SERVER['REMOTE_ADDR']))){
        $countrycode= $a['geoplugin_countryCode'];
        if ($countrycode=='UK' ){
            header( 'Location: https://www.example.co.uk/' ) ;exit;
        }
    }