【发布时间】:2014-11-12 06:08:11
【问题描述】:
这是从天气 API 获取伦敦温度的代码。它工作正常(图像是本地的,因此不会显示):
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<meta name="format-detection" content="telephone=no" />
<meta name="viewport" content="user-scalable=no, initial-scale=1, maximum-scale=1, minimum-scale=1, width=device-width, height=device-height, target-densitydpi=device-dpi" />
<link rel="stylesheet" type="text/css" href="css/body.css" />
<meta name="msapplication-tap-highlight" content="no" />
</head>
<body>
<script src="http://code.jquery.com/jquery-2.0.0.js"></script>
<script language="javascript" type="text/javascript">
<!--
function foo(callback) {
$.ajax({
url: "http://api.openweathermap.org/data/2.5/weather?q=London",
dataType: 'JSON',
success: callback
});
}
function myCallback(result) {
var temp = JSON.stringify(JSON.parse(result.main.temp));
var Kelvin = 272;
var Centigrade = Math.round(temp-Kelvin);
if (Centigrade <= 25) {
//alert("Temperature : "+Math.round(Centigrade)+" C");
var temp = document.getElementById("temp");
temp.style.fontSize = "20px";
temp.innerHTML = Centigrade+"° C , Cool "+"<img src= \"img/Tlogo2.svg\"/>";
//document.getElementById("temp").innerHTML = Centigrade+"° C , Cool "+"<img src= \"img/Tlogo2.svg\"/>";
}
else if (Centigrade > 25) {
var temp = document.getElementById("temp");
temp.style.fontSize = "20px";
temp.innerHTML = Centigrade+"° C , Cool "+"<img src= \"img/Tlogo3.svg\"/>";
//document.getElementById("temp").innerHTML = Centigrade+"° C , It's Hot !!! "+"<img src= \"img/Tlogo3.svg\"/>";
}
}
</script>
<div style="position: absolute; left: 30px; top: 75px;">
<img src="img/temlogo.svg" width="35" height="35" onclick="foo(myCallback);"/>
</div>
<p id="temp"></p>
</body>
</html>
从教程点和 Bootstrap 网站我尝试使用可忽略的弹出框。它也可以正常工作:
<!DOCTYPE html>
<html>
<body>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="http://code.jquery.com/jquery-2.0.0.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap.min.css">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap-theme.min.css">
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/js/bootstrap.min.js"></script>
<script language="javascript" type="text/javascript">
$(function() {
$("[data-toggle='popover']").popover();
});
</script>
</body>
<a href="#" tabindex="0" class="btn btn-lg btn-danger" role="button" data-toggle="popover" data-trigger="focus" title="Temperature" data-content="40C">Temperature</a>
</html>
现在我正在尝试将温度作为弹出元素。 IE。如果我点击图像按钮,它应该触发温度获取功能,然后在弹出框中显示温度和与之相关的图像。所以这是我面临的两个挑战。
- 设置图像而不是红色按钮,然后设置温度数据
- 列表项和图像,即。
Tlogo2.svg将出现在弹出框中。
那么任何人都可以建议如何设置它?
编辑:我已经尝试过这个来达到我所说的。但什么也没发生。代码在这里:
<!DOCTYPE html>
<html>
<body>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="http://code.jquery.com/jquery-2.0.0.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap.min.css">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap-theme.min.css">
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/js/bootstrap.min.js"></script>
<script language="javascript" type="text/javascript">
//Function
function foo(callback) {
$.ajax({
url: "http://api.openweathermap.org/data/2.5/weather?q=London",
dataType: 'JSON',
success: callback
});
}
function myCallback(result) {
var temp = JSON.stringify(JSON.parse(result.main.temp));
var Kelvin = 272;
var Centigrade = temp-Kelvin;
alert("Temperature : "+Math.round(Centigrade)+" C");
//document.getElementById("temp").innerHTML = "Temperature : "+Math.round(Centigrade)+" C";
}
$(function() {
$("[data-toggle='popover']").popover(myCallback(result));
});
</script>
</body>
<a href="#" tabindex="0" class="btn btn-lg btn-danger" role="button" data-toggle="popover" data-trigger="focus" title="Temperature" data-content="40C">Temperature</a>
</html>
我正在添加一些内容。这样人们就不会感到困惑,看看我真正想要什么。我想要函数的结果,即该弹出元素的温度为 23 C 代码:
<!DOCTYPE html>
<html>
<body>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="http://code.jquery.com/jquery-2.0.0.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap.min.css">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/css/bootstrap-theme.min.css">
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.0/js/bootstrap.min.js"></script>
<script language="javascript" type="text/javascript">
//Function
function foo(callback) {
$.ajax({
url: "http://api.openweathermap.org/data/2.5/weather?q=London",
dataType: 'JSON',
success: callback
});
}
function myCallback(result) {
var temp = JSON.stringify(JSON.parse(result.main.temp));
var Kelvin = 272;
var Centigrade = temp-Kelvin;
alert("Temperature : "+Math.round(Centigrade)+" C");
//document.getElementById("temp").innerHTML = "Temperature : "+Math.round(Centigrade)+" C";
}
$(function() {
$("[data-toggle='popover']").popover(myCallback);
});
</script>
</body>
<a href="#" tabindex="0" class="btn btn-lg btn-danger" role="button" data-toggle="popover" data-trigger="focus" title="Temperature" data-content= "myCallback(result);" >Temperature</a>
</html>
所以让我知道我需要更改的地方。
【问题讨论】:
-
检查这可能对你有帮助stackoverflow.com/questions/8130069/…
标签: javascript html twitter-bootstrap