javascript - OpenWeatherMap.ORG API - $.get JSON 不起作用，我没有收到任何数据

Question

我正在尝试使用提供的 api 连接到 openweather.org 的测试代码。

如果我使用浏览器访问该网址:

http://api.openweathermap.org/data/2.5/weather?id=2172797&APPID=35000cdad97645316c048563e4183021

然后，我得到了正确的 Json:{"coord":{"lon":145.77,"lat":-16.92},"weather":[{"id":803,"main":"云","description":" splinter 的云","图标":"04n"}],"基地":"站","主要":{"温度":289.26,"压力":1013,"湿度":93,"temp_min":289.26,"temp_max": 289.26},"风":{"速度":1.61,"度":116.5},"雨":{"3小时":0.03},"云":{"全部":76},"dt":1474367584 ,"sys":{"type":3,"id":10843,"message":0.1585,"country":"AU","日出":1474315673,"日落":1474359164},"id":2172797 ,"名称":"凯恩斯","鳕鱼":200}

问题是当我使用jquery的$.getJSON时，我看不到任何数据。

这是为什么呢？怎么解决？

JS:

$(document).ready(function(){

  var api = 'http://api.openweathermap.org/data/2.5/weather?id=2172797&APPID=35000cdad97645316c048563e4183021';

   $.getJSON(api, {format:'json'},function(data){console.log(data.coord.lon)});


});

代码笔:https://codepen.io/elivanrock/pen/zKoYEj?editors=1011

提前致谢!

Answer 1

您可以使用 JSONp 从 openweathermap.com 获取数据，只需这样添加回调函数即可:

http://api.openweathermap.org/data/2.5/weather?id=2172797&APPID=35000cdad97645316c048563e4183021&callback=myfunc

示例如下:

$.ajax({
    url: "http://api.openweathermap.org/data/2.5/weather",
    jsonp: "callback",
    dataType: "jsonp",
    data: {
        id: "2172797",
        APPID: "35000cdad97645316c048563e4183021"
    },
    success: function( response ) {
        console.log( response ); // server response
        $('.current').html('<img src="http://openweathermap.org/img/w/' + response.weather[0].icon + '.png" /> ' + response.weather[0].main);
    }
});

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