【问题标题】:Bootstrap with JS, result of a function in popover带有 JS 的引导程序,popover 中函数的结果
【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&nbsp;&nbsp;&nbsp;"+"<img src= \"img/Tlogo2.svg\"/>";
                    //document.getElementById("temp").innerHTML = Centigrade+"° C , Cool&nbsp;&nbsp;&nbsp;"+"<img src= \"img/Tlogo2.svg\"/>";
                }
                else if (Centigrade > 25) {
                    var temp = document.getElementById("temp");
                    temp.style.fontSize = "20px";
                    temp.innerHTML = Centigrade+"° C , Cool&nbsp;&nbsp;&nbsp;"+"<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。

  1. 设置图像而不是红色按钮,然后设置温度数据
  2. 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】:

标签: javascript html twitter-bootstrap


【解决方案1】:

也许你可以在悬停时弹出,这里是示例

  $(function() {
      $('[title]').attr("data-rel", "tooltip");
      $("[data-rel='tooltip']")
          .attr("data-placement", "top")
          .attr("data-content", function() {
              return $(this).attr("title")
          })
          .removeAttr('title');


      var showPopover = function() {
          $(this).popover('show');
      };
      var hidePopover = function() {
          $(this).popover('hide');
      };
      $("[data-rel='tooltip']").popover({
          trigger: 'manual'
      }).click(showPopover).hover(showPopover, hidePopover);

  });

这样使用

 <a href="#" tabindex="0" class="btn btn-lg btn-danger" title="40c">Temperature</a>

【讨论】:

  • 与我的解决方案无关。
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