Question

正如标题所说，我正在尝试制作一个天气应用程序，它可以吸引用户的位置并获取当地的天气。我被困在第二个不会执行的函数和console.log（）;没有提供任何输出。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

以这种方式更改网址：

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

你错过了http

Answer 2

Weather API的网址不正确，您需要将http://与API一起使用，否则将其视为相对网址。

使用

var url = "http://api.openweathermap.org/data/2.5/weather?q=" + ...

＆＃13;

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

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

第二个函数，getJSON没有执行

2 个答案: