Weather forecast from IP address

ipapi ~ ipapi.co

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Dec 30, 2016

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Recipe to get weather forecast from an IP address

Ingredients

• IP address to location API : ipapi.co (free, no key required)
• Weather API : openweathermap.org (free, but requires key)

Ruby

require 'net/http'
require 'json'$ip = '208.67.222.222'$latlong = Net::HTTP.get(URI('https://ipapi.co/' + $ip + '/latlong/')).split(",")$weather = JSON.parse(Net::HTTP.get(URI('http://api.openweathermap.org/data/2.5/weather?lat='+$latlong[0]+'&lon='+$latlong[1]+'&appid='+$API_KEY)))

Python

from requests import getip = '208.67.222.222'latlong = get('https://ipapi.co/{}/latlong/'.format(ip)).text.split(',')weather = get('http://api.openweathermap.org/data/2.5/weather?lat={}&lon={}&appid=API_KEY'.format(latlong[0], latlong[1])).json()

PHP

$ip = '208.67.222.222';$latlong = explode(",", file_get_contents('https://ipapi.co/' . $ip . '/latlong/'));$weather = file_get_contents('http://api.openweathermap.org/data/2.5/weather?lat=' . $latlong[0] . '&lon=' . $latlong[1] . '&appid=API_KEY');echo $weather;$json = json_decode($weather);

Javascript

// jQuery required for this examplevar ip = '208.67.222.222'$.get('https://ipapi.co/'+ip+'/latlong/', function(response){
  var latlong = response.split(',');
  console.log(latlong);  $.get('http://api.openweathermap.org/data/2.5/weather?lat=' + latlong[0] + '&lon=' + latlong[1] + '&appid=API_KEY', function(wResponse){
    console.log(weather);
  })
})