javascript - getJSON 未执行的第二个函数

Question

正如标题所说，我正在尝试制作一个天气应用程序，它可以获取用户的位置并获取他们本地的天气。我陷入了第二个无法执行的函数和 console.log(); 上不提供任何输出。任何帮助将不胜感激。

$(document).ready(function () {
    function getLocation() {
        $.getJSON("http://ipinfo.io", function (response) {
            var cc = response.country;
            var city = response.city;
            var state = response.region;
            $(".city").html(city + "," + state);
            console.log(cc);
            var url = "api.openweathermap.org/data/2.5/weather?q=" + city + "," + cc + "&APPID=1d0e324c03cd19ecf0abf20ac2708666";
            console.log(city);
            console.log(url);
            getWeather(url);

        });
    }
    function getWeather(url) {
        $.getJSON(url, function (response) {

            console.log(url);
            $(".temp").html(Math.round(response.main.temp));

        });
    }
    getLocation();

});

Answer 1

按如下方式更改网址:

var url = "http://api.openweathermap.org/data/2.5/weather?q="+city+","+cc+"&APPID=1d0e324c03cd19ecf0abf20ac2708666";

你缺少http