How node. js and PHP get geographical locations based on IP

  • 2020-03-30 02:23:55
  • OfStack

Node.js implementation code

var http = require('http');
var util = require('util');


var getIpInfo = function(ip, cb) {
    var sina_server = 'http://int.dpool.sina.com.cn/iplookup/iplookup.php?format=json&ip=';
    var url = sina_server + ip;
    http.get(url, function(res) {
        var code = res.statusCode;
        if (code == 200) {
            res.on('data', function(data) {
                try {
                    cb(null, JSON.parse(data));
                } catch (err) {
                    cb(err);
                }
            });
        } else {
            cb({ code: code });
        }
    }).on('error', function(e) { cb(e); });
};

getIpInfo('220.181.111.85', function(err, msg) {
    console.log(' city : ' + msg.city);
    console.log('msg: ' + util.inspect(msg, true, 8));
})

Result of request:
 city :  xuzhou 
{
    "ret": 1,
    "start": "49.68.0.0",
    "end": "49.68.255.255",
    "country": " China ",
    "province": " jiangsu ",
    "city": " xuzhou ",
    "district": "",
    "isp": " telecom ",
    "type": "",
    "desc": ""
}

Two, PHP implementation code
<?

$ip = "220.181.111.85";
$url = "http://int.dpool.sina.com.cn/iplookup/iplookup.php?format=json&ip=$ip";
$data = file_get_contents($url);
$result = json_decode($data);
echo " city :" . $result->city . "<br>";
print_r($result);

?>

Result of request:
 city : xuzhou 
stdClass Object
(
    [ret] => 1
    [start] => 49.68.0.0
    [end] => 49.68.255.255
    [country] =>  China 
    [province] =>  jiangsu 
    [city] =>  xuzhou 
    [district] => 
    [isp] =>  telecom 
    [type] => 
    [desc] =>
)


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