2

在这段代码中，我试图为用户提供本地位置和温度，但不知何故，温度显示的摄氏温度较低，而且下面的更新也不像我尝试过的那样，它是 4-5 小时前数据，低于 10 摄氏度如果温度是 22(Celsius) 小时后，它显示为 3(Celsius) 工作示例在 codepen http://codepen.io/cannelflow/full/RrymYo/

var x = document.getElementById("demo");
var y = document.getElementById("demo1");
window.onload = getLocation();
//window.onload=getWeather();
function getLocation() {
    if (navigator.geolocation) {
        navigator.geolocation.getCurrentPosition(showPosition, showError);
    } else {
        x.innerHTML = "Geolocation is not supported by this browser.";
    }
}
//Location For Display
function showPosition(position) {
    var loc = { lat: position.coords.latitude, lon: position.coords.longitude };
    getWeather(loc);
    var baseURL = "https://maps.googleapis.com/maps/api/geocode/json?latlng=";
    var fullURL = baseURL + loc.lat + "," + loc.lon;
    $.ajax({
        url: fullURL,
        success: function (display) {
            x.innerHTML = display.results[1].formatted_address;
        }
    });

}
//Location For Weather
function getWeather(loc) {
    var baseURL = "http://api.openweathermap.org/data/2.5/weather?lat=";
    var appid = "064129b86c99c35c42d531db251b99e3";
    var fullURL = baseURL + loc.lat + "&lon=" + loc.lat + "&appid=" + appid + "&units=metric";
    //http://api.openweathermap.org/data/2.5/weather?lat=21.2600668&lon=81.5989561&appid=064129b86c99c35c42d531db251b99e3&units=metric
    $.ajax({
        url: fullURL,
        success: function (display1) {
            y.innerHTML = display1.main.temp;
        }
    });
}


function showError(error) {
    switch (error.code) {
        case error.PERMISSION_DENIED:
            x.innerHTML = "User denied the request for Geolocation."
            break;
        case error.POSITION_UNAVAILABLE:
            x.innerHTML = "Location information is unavailable."
            break;
        case error.TIMEOUT:
            x.innerHTML = "The request to get user location timed out."
            break;
        case error.UNKNOWN_ERROR:
            x.innerHTML = "An unknown error occurred."
            break;
    }
}
<body>
    <section>
        <div class="container-fluid text-center">
            <br />
            <!-- <h1><button class="btn btn-danger" onclick="getLocation()">Click Me To Get Your Location!</button></h1> -->
            <h1 class="text-primary" id="demo1"></h1>
            <br />
            <h1 class="text-primary" id="demo"></h1>
        </div>
    </section>
</body>

4

3 回答 3

1

你有一个错字：

var fullURL = baseURL + loc.lat + "&lon=" + loc.lat + "&appid=" + appid + "&units=metric";

应该

var fullURL = baseURL + loc.lat + "&lon=" + loc.lon + "&appid=" + appid + "&units=metric";
于 2016-01-29T06:38:02.537 回答
0

修改它喜欢

function getWeather(loc) {
    var baseURL = "http://api.openweathermap.org/data/2.5/weather?lat=";
    var appid = "064129b86c99c35c42d531db251b99e3";
    //var fullURL = baseURL + loc.lat + "&lon=" + loc.lat + "&appid=" + appid + "&units=metric";
    //http://api.openweathermap.org/data/2.5/forecast?lat=35&lon=139&appid=44db6a862fba0b067b1930da0d769e98&units=metric
    $.ajax({
        url: baseURL,
        type: 'get',
        dataType: 'JSONP',
        data: { lat: loc.lat, lon: loc.lon, units: 'metric', APPID: appid },
        success: function (data) {
            y.innerHTML = data['main']['temp'] + " °C";
        }
    });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script>

它奏效了

于 2016-01-29T09:49:25.287 回答
0

您在查询字符串中有错字。更好的选择是用户jQuery.param从对象构建查询字符串，因为它更容易阅读，因此更不容易出错。

function getWeather(loc) {
  var baseURL = "http://api.openweathermap.org/data/2.5/weather?";
  return $.ajax({
    url: baseURL + $.param({
      appid: "064129b86c99c35c42d531db251b99e3",
      lon: loc.lon,
      lat: loc.lat,
      units: "metric"
    }),
    success: function(display1) {
      y.innerHTML = display1.main.temp;
    }
  });
}
于 2016-01-29T07:55:32.583 回答