PHP-这些是什么?如何响应某些值?
我希望创建一个基本的IP跟踪器,我试图找出如何响应此变量的某些方面,干杯:PHP-这些是什么?如何响应某些值?,php,variables,web,Php,Variables,Web,我希望创建一个基本的IP跟踪器,我试图找出如何响应此变量的某些方面,干杯: {"ip":"MY IP","country_code":"GB","country_name":"United Kingdom","region_code":"ENG","region_name":"England","city":"MY CITY","zip_code":"MY CODE","time_zone":"Europe/London","latitude":----,"longitude":-----,"m
{"ip":"MY IP","country_code":"GB","country_name":"United Kingdom","region_code":"ENG","region_name":"England","city":"MY CITY","zip_code":"MY CODE","time_zone":"Europe/London","latitude":----,"longitude":-----,"metro_code":-}
这是我正在阅读的代码:
$location = file_get_contents('http://freegeoip.net/json/'.$_SERVER['REMOTE_ADDR']);
print_r($location);
{
"ip": "77.99.179.98",
"country_code": "GB",
"country_name": "United Kingdom",
"region_code": "H9",
"region_name": "London, City of",
"city": "London",
"zipcode": "",
"latitude": 51.5142,
"longitude": -0.0931,
"metro_code": "",
"areacode": ""
}
干杯 响应某事做某事
//Example
$locationObject = json_decode($location);
echo $locationObject->ip;
//Prints out: 77.99.179.98
更多信息
回应某事
//Example
$locationObject = json_decode($location);
echo $locationObject->ip;
//Prints out: 77.99.179.98
更多信息
这是
JSON
。使用json\u decode
将其转换为对象并相应地使用。它是json
。使用json\u decode
将其转换为对象并相应使用。谢谢!你已经解决了我的问题。很高兴能帮忙!谷歌有一点关于JSON的知识,对它有一些了解真是太好了:)谢谢!你已经解决了我的问题。很高兴能帮忙!谷歌对JSON有点了解,对它有一些了解真是太好了:)