javascript - PHP:带有API的真实IP地址

Question

你好，我使用这个API:http://ipapi.co/json要获取本地主机(127.0.0.1)中的国家、城市和真实IP地址，工作完美，但是当我将代码放入我的网站时，我的代码不显示客户端的地址，而是显示服务器的地址，我该如何解决这个问题？ p>

我可以使用 $_SERVER['REMOTE_ADDR']; 获取客户端 IP 地址，但我需要此 api 来获取国家和城市:http://ipapi.co/json .

Function TTT() {
    $ip = false;
    $city = false;
    $region = false;
    $country = false;
    $country_code = false;

    if($json = @file_get_contents("http://www.geoplugin.net/json.gp")) {
      $obj = json_decode($json);
      if(isset($obj->geoplugin_request) && $obj->geoplugin_request != false) {
        $ip = "IP: ". $obj->geoplugin_request. "</br>";
      }
      if(isset($obj->geoplugin_city) && $obj->geoplugin_city != false) {
        $city = "City: ". $obj->geoplugin_city. "</br>";
      }
      if(isset($obj->geoplugin_city) && $obj->geoplugin_city != false) {
        $region = "Region: ". $obj->geoplugin_city. "</br>";
      }
      if(isset($obj->geoplugin_countryName) && $obj->geoplugin_countryName != false) {
        $country = "Country: ". $obj->geoplugin_countryName. "</br>";
      }
      if(isset($obj->geoplugin_countryCode) && $obj->geoplugin_countryCode != false) {
        $country_code = "Country Code: ". $obj->geoplugin_countryCode. "</br>";
      }

      return $ip. $city. $region. $country. $country_code;
    }
}

echo TTT();

Answer 1

使用http://ipapi.co/json处的API

$obj = json_decode($json);

将此 JSON 放入关联数组中:

{
    "ip": "00.00.00.00",
    "city": "Atlanta",
    "region": "Georgia",
    "country": "US",
    "postal": "31532",
    "latitude": 34.461,
    "longitude": -85.9877,
    "timezone": "America/New_York"
}

像这样访问城市和地区:

if(isset($obj->city) && $obj->city != false) {
    $city = "City: ". $obj->city. "</br>";
  }
if(isset($obj->region) && $obj->region != false) {
    $region = "Region: ". $obj->region. "</br>";
  }

您似乎没有使用最初声明的 API。