jQuery打开弹出式按钮单击以进行引导 [英] Jquery open popup on button click for bootstrap

问题描述

当我单击下面的按钮

   <button class="btn btn-primary btn-lg" data-toggle="modal" data-target="#myModal">
        Launch demo modal
    </button>

我在弹出式窗口(模式)下打开

I open below popup(modal)

<!-- Modal -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
    <div class="modal-dialog">
        <div class="modal-content">
            <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>
                <h4 class="modal-title" id="myModalLabel">Modal title</h4>
            </div>
            <div class="modal-body">
                ...
            </div>
            <div class="modal-footer">
                <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
                <button type="button" class="btn btn-primary">Save changes</button>
            </div>
        </div>
    </div>
</div>

我参考以下示例

http://getbootstrap.com/javascript/

我的问题:

如何使用javascript/jquery在按钮单击上打开#myModal?

How can i open #myModal on button click by using javascript/jquery ?

任何帮助将不胜感激.

谢谢.

推荐答案

提供一个ID以唯一标识按钮，例如说myBtn

Give an ID to uniquely identify the button, lets say myBtn

// when DOM is ready
$(document).ready(function () {

     // Attach Button click event listener 
    $("#myBtn").click(function(){

         // show Modal
         $('#myModal').modal('show');
    });
});

JSFIDDLE

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