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

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;

  ?>

其他回答

由CPAN的Perl社区维护的ip->国家数据库有一个维护良好的平面文件版本

访问这些文件不需要数据服务器,数据本身大约是515k

Higemaru编写了一个PHP包装器来处理这些数据:PHP -ip-country-fast

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
}

不确定这是否是一个新的服务,但现在(2016)在php中最简单的方法是使用geoplugin的php web服务:http://www.geoplugin.net/php.gp:

基本用法:

// GET IP ADDRESS
if (!empty($_SERVER['HTTP_CLIENT_IP'])) {
    $ip = $_SERVER['HTTP_CLIENT_IP'];
} else if (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
    $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
} else if (!empty($_SERVER['REMOTE_ADDR'])) {
    $ip = $_SERVER['REMOTE_ADDR'];
} else {
    $ip = false;
}

// CALL THE WEBSERVICE
$ip_info = unserialize(file_get_contents('http://www.geoplugin.net/php.gp?ip='.$ip));

它们还提供了一个现成的类:http://www.geoplugin.com/_media/webservices/geoplugin.class.php.tgz?id=webservices%3Aphp&cache=cache

这可以接受吗? 纯PHP

$country = geoip_country_code_by_name($_SERVER['REMOTE_ADDR']);
print $country;

裁判:https://www.php.net/manual/en/function.geoip-country-code-by-name.php