【发布时间】:2017-01-26 02:55:41
【问题描述】:
这是代码:
<?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);
echo $info;
?>
我有这个输出:
{ "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
但我需要这个(仅城市或任何孤立值)
Mountain View
【问题讨论】:
-
$json = json_decode($info, true);回声 $json['city'];
-
我尝试使用 json_decode,但我的结果是一样的。完整的信息,不仅仅是城市。也许是 json_decode 的服务器问题?有别的办法吗?没有json_decode?span>