Node.js and PHP method to obtain geographical location based on IP_javascript skills

1. Node.js implementation code

Copy code The code is as follows:

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

/**
* Obtain address information based on ip
*/
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) {                                                                                                                                    ;                                                                                         ; });
}
}).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));
})

Request result:

Copy code

The code is as follows: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": ""
}

2. PHP implementation code

Copy code

The code is as follows:$ip = "220.181.111.85";$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:" . $result- >city . "
";
print_r($result);

?>

Request result:

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The code is as follows:City: XuzhoustdClass Object( [ret] => 1

[start ] => 49.68.0.0

[end] => 49.68.255.255
[country] => China
[province] => Jiangsu
[city] => Xuzhou
[district] =>