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我试图运行下面的代码:麻烦一些openweathermap API示例代码
var mainWeather = {
init: function() {
$("#submitWeather").click(function() {
return mainWeather.getWeather();
});
},
getWeather: function() {
$.get('http://api.openweathermap.org/data/2.5/weather?q=' + $("#cityInput").val() + "," + $("#countryInput").val() + "&APPID=myweatherkey_removed", function(data) {
var json = {
json: JSON.stringify(data),
delay: 1
};
echo(json);
});
},
// Prints result from the weatherapi, receiving as param an object
createWeatherWidg: function(data) {
return "<div class='pressure'> <p>Temperature: " + (data.main.temp - 273.15).toFixed(2) + " C</p></div>" +
"<div class='description'> <p>Title: " + data.weather[0].main + "</p></div>" +
"<div class='description'> <p>Description: " + data.weather[0].description + "</p></div>" +
"<div class='wind'> <p>Wind Speed: " + data.wind.speed + "</p></div>" +
"<div class='humidity'> <p>Humidity: " + data.main.humidity + "%</p></div>" +
"<div class='pressure'> <p>Pressure: " + data.main.pressure + " hpa</p></div>";
}
};
var echo = function(dataPass) {
$.ajax({
type: "POST",
url: "/echo/json/",
data: dataPass,
cache: false,
success: function(json) {
var wrapper = $("#weatherWrapper");
wrapper.empty();
wrapper.append("<div class='city'> <p>Place: " + json.name + ", " + json.sys.country + "</p></div>");
wrapper.append(mainWeather.createWeatherWidg(json));
}
});
};
mainWeather.init();
和
<!DOCTYPE html>
<html>
<head>
<title>Open Weather API</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
</head>
<body>
<div id="userArea" class="container">
<div id="stateWrapper">
<input type="text" id="cityInput" placeholder="Enter a State" />
</div>
<br/>
<div id="countryWrapper">
<input type="text" id="countryInput" placeholder="Enter a Country" />
</div>
<br/>
<div id="buttonArea">
<input type="submit" id="submitWeather" class="btn btn-primary"/>
<br/>
</div>
<!- USED TO SHOW RESULT -->
<div id="weatherWrapper">
</div>
</div>
<script type="text/javascript" src="jquery-3.1.1.slim.js"></script>
<script type="text/javascript" src="mainWeather.js"></script>
</body>
</html>
,但它只是不会做任何事情,当我点击提交。
Chrome调试器首先声明$ .get不是第9行中的函数,因此在搜索后,我将其更改为jquery.get。现在它说这不是一个功能。我不知道我在做什么。有人可以弄清楚这一点吗?
感谢您的回复。然而,使用正常的jquery运行它现在产生了这个错误:XMLHttpRequest无法加载file:/// C:/ echo/json /。协议方案仅支持交叉源请求:http,data,chrome,chrome-extension,https,chrome-extension-resource。 –
您正在使用文件协议而不是http协议。使用本地服务器,它会工作。搜索并安装XAMPP或WampServer。如果您阅读错误:交叉源请求仅支持协议方案:http ... –
我不认为您拥有所有代码。 – hamza765