Gri*_*ckz 6 html javascript api json
我正在编写一些JavaScript来从JSON中提取包含名称,经度,纬度和openweather API调用的信息.我需要的是将API信息从API调用中获取到HTML页面中,以便您可以获取每个信息的天气预报.我有两个元素分开工作,但无法弄清楚如何让它们一起工作.
请帮忙？:-)
来自d.weather的API天气预报
api.openweathermap.org/data/2.5/forecase?lat=50.8609&lon=-0.08014&&units=metric
用于提取openweather JSON数据的HTML页面
<html>
<head>
<title>Weather</title>
<meta charset="utf-8">
<script src="http://code.jquery.com/jquery-1.7.min.js" ></script>
<script src="http://code.jquery.com/ui/1.7.0/jquery-ui.js" ></script>
<script>
function getWeather(callback) {
var weather = 'http://api.openweathermap.org/data/2.5/forecast?lat=51.5072&lon=0.1275&units=metric';
$.ajax({
dataType: "jsonp",
url: weather,
success: callback
});
}
// get data:
getWeather(function (data) {
console.log('weather data received');
console.log(data.list[0].weather[0].description);
console.log(data.list[0].weather[0].main);
});
getWeather(function (data) {
document.write('weather data received');
document.write('<br>');
document.write(data.list[0].weather[0].description);
document.write('<br>');
document.write(data.list[0].weather[0].main);
document.write('<br>');
document.write(data.list[0].main.temp);
document.write('<br>');
document.write(data.list[0].main[0].dt_txt);
document.write('<br>');
});
</script>
</body>
</html>
用于提取JSON数据的Html页面
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.10.0.min.js"></script>
<!-- Javascript -->
<script type="text/javascript">
function loadUrl(newLocation){
window.location = newLocation;
return false;
}
</script>
<script type="text/javascript">
$(document).ready(function (){
$("#btn382").click(function(){
/* set no cache */
$.ajaxSetup({ cache: false });
$.getJSON("weather.json", function(data){
var html = [];
/* loop through array */
$.each(data, function(index, d){
html.push("Team : ", d.Teams, ", ",
"Long : ", d.Long, ", ",
"Lat : ", d.Lat, ", ",
"Weather : ", d.Weather, "<br>");
});
$("#div381").html(html.join('')).css("background-color", "orange");
}).error(function(jqXHR, textStatus, errorThrown){ /* assign handler */
/* alert(jqXHR.responseText) */
alert("error occurred!");
});
});
});
</script>
<!-- HTML -->
<a name="#ajax-getjson-example"></a>
<div id="example-section38">
<div>Football weather</div>
<div id="div381"></div>
<button id="btn382" type="button">Team location</button>
</div>
weather.json
{
"Teams":"Wycombe Wanderers",
"Long":-0.800299,
"Lat":51.6306,
"Weather":" api.openweathermap.org/data/2.5/weather?lat=51.6306&lon=-0.800299&mode=html"
},
{
"Teams":"Livingston",
"Long":-3.52207,
"Lat":55.8864,
"Weather":" api.openweathermap.org/data/2.5/weather?lat=55.8864&lon=-3.52207&mode=html"
},
{
"Teams":"Brighton and Hove Albion",
"Long":-0.08014,
"Lat":50.8609,
"Weather":" api.openweathermap.org/data/2.5/weather?lat=50.8609&lon=-0.08014&mode=html"
},
我有基本的知识可以帮助你.这是你的两页的混搭.
首先,我修改了你的getWeather功能来调用天气而不是预测.它接受一个city参数,并在调用回调之前将该参数附加到数据.
function getWeather(city, callback) {
var url = 'http://api.openweathermap.org/data/2.5/weather';
$.ajax({
dataType: "jsonp",
url: url,
jsonCallback: 'jsonp',
data: { q: city },
cache: false,
success: function (data) {
data.city = city;
callback(data);
}
});
}
在这里,代替你的团队JSON,我以JS对象的形式创建了一个,阿森纳和利物浦及其相应的城市作为数据.该函数循环遍历对象,提取城市名称并将其传递给getWeather.返回数据并将其附加到div.
$(document).ready(function () {
$("#btn382").click(function () {
var teams = {
arsenal: { city: 'london' },
liverpool: { city: 'liverpool' }
};
for (var team in teams) {
var city = teams[team].city;
getWeather(city, function(data) {
var html = [];
html.push('<div>')
html.push('City: ', data.city, ', ');
html.push('Lat: ', data.coord.lat, ', ');
html.push('Lon: ', data.coord.lon, ', ');
html.push('Weather: ', data.weather[0].description);
html.push('</div>')
$("#div381").append(html.join('')).css("background-color", "orange");
});
}
});
});
希望这会给你一些关于如何解决这个项目的想法.