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使用php cURL和json通过ip在线服务获取用户位置

ip location curl json php

0

WebSon · 技术社区 · 7 年前

<?php 
$ip = $_SERVER['REMOTE_ADDR'];
$apiurl = "https://api.2ip.ua/geo.json?ip=$ip";
//--
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "$apiurl");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$contents = curl_exec ($ch);
//--
curl_close ($ch);

//- **************-//
$pieces = explode('"', $contents);
$country = $pieces['7'];
$city = $pieces['11'];
$city2 = $pieces['13'];

echo "Your IP is :" . $ip  . " and country " .$country. " and city " .$city;


 ?>

2 回复 | 直到 7 年前

1

Fernando 7 年前

别用爆炸!,api的答案是JSON,而您必须这样做:

$data = json_decode($contents, true);
$country = $data['country'];
$city = $data['city'];

顺便说一句,您还可以使用支持ipv4和ipv6的多个检测的代码来改进IP检测

if (isset($_SERVER['REMOTE_HOST'])) {
        $ip = $_SERVER['REMOTE_HOST'];
    } elseif (isset($_SERVER['HOST'])) {
        $ip = $_SERVER['HOST'];
    } elseif (isset($_SERVER['REMOTE_ADDR'])) {
        $ip = $_SERVER['REMOTE_ADDR'];
    } else {
        $host = '';
    }
    $ip = str_replace("::ffff:","",$ip);//::ffff:127.127.127.127 en caso de jugo ipv6

2

1

max 7 年前

首先,不要使用explode,而是使用json\u decode,然后您的API的SSL证书有问题,所以请使用http。

<?php
$ip = $_SERVER['REMOTE_ADDR'];
$apiurl = "http://api.2ip.ua/geo.json?ip=";
$contents = file_get_contents($apiurl . $ip);
$json = json_decode($contents);
$country = $json->country;
$city = $json->city;
echo "Your IP is :" . $ip  . " and country " . $country . " and city " . $city;