Question

这是代码：

{ "businessName" : "", "businessWebsite" : "", "city" : "Mountain View", "continent" : "North America", "country" : "United States", "countryCode" : "US", "ipName" : "google-public-dns-a.google.com", "ipType" : "Residential", "isp" : "Google", "lat" : "37.3860", "lon" : "-122.0838", "org" : "Google Inc.", "query" : "8.8.8.8", "region" : "California", "status" : "success" } 1

我有这个输出：

Mountain View

但我需要这个（只有城市或任何孤立的价值）

{{1}}

Answer 1

您只需要解码JSON然后访问类属性：

$ipInfo = json_decode($info);
$city = $ipInfo->city;

Answer 2

json_decode

      $json = json_decode($info, true); 
      echo $json['city'];

或者：

<?php
     $user_ip = getenv('REMOTE_ADDR');
     $geo = json_decode(file_get_contents("http://extreme-ip-lookup.com/json/$user_ip"));
     $country = $geo->country;
     $city = $geo->city;
     $ipType = $geo->ipType;
     $businessName = $geo->businessName;
     $businessWebsite = $geo->businessWebsite;

     echo "Location $city";
?>

Answer 3

使用参数true查看json_decode以获取数组。

<?php
$ip = $_SERVER['REMOTE_ADDR'];
$url = "http://extreme-ip-lookup.com/json/".$ip;

$curl = curl_init();
curl_setopt($curl, CURLOPT_URL,$url);
$info = curl_exec($curl);
curl_close($curl);

$array = json_decode($info, true);
echo $array['city'];

从CURL JSON获取特定值

3 个答案: