如何在MVC 4按钮上显示jQuery模式弹出窗口

Question

I am doing one mvc project where I have one button. 我正在做一个只有一个按钮的mvc项目。 When I clicked on button one popup with textbox and button should appear. 当我单击按钮时，将出现一个带有文本框和按钮的弹出窗口。 I tried as below. 我尝试如下。 Instead of getting popup when the page first loads required textbox and button in page only(supposed to display when the button clicked) This is what I tried. 当页面第一次加载所需的文本框和仅在页面中的按钮时（而不是在单击按钮时显示），而不是弹出窗口。这是我尝试的。

<head>
 <link rel="stylesheet" href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css">

    <script src="//code.jquery.com/jquery-1.10.2.js"></script>

    <script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script>

    <link rel="stylesheet" href="/resources/demos/style.css">
</head>

Style information I copied from https://jqueryui.com/dialog/#modal-form(many lines of code so I am not posting here) This is my script and popup code. 我从https://jqueryui.com/dialog/#modal-form复制的样式信息（很多行代码，所以我不在这里发布）这是我的脚本和弹出代码。

<div id="dialog-form" title="Create new user">
    <form>
        <fieldset>

            <label for="name">Enter your Comments Here</label>

            <input type="text" name="name" id="name" value="Jane Smith" class="text ui-widget-content ui-corner-all">
            <!-- Allow form submission with keyboard without duplicating the dialog button -->

            <input type="submit" tabindex="-1" style="position:absolute; top:-1000px">

        </fieldset>

    </form>

</div>
<script>
    $("#Delete").button().on("click", function () {

        dialog.dialog("open");

    });


</script>

This is my mvc code. 这是我的MVC代码。

@using (Html.BeginForm("Index", "TestRendering", FormMethod.Post))
    {
   <td scope="col"><input type="button" id="Delete" class="btn btn-primary btn-cons red" value="Delete" /></td>

    }

Am I going wrong in this process? 在这个过程中我做错了吗？ Thanks. 谢谢。

Answer 1

Try this 尝试这个

<input type="button" id="Delete" class="btn btn-primary btn-cons red" value="Delete" />

<div id="dialog-form" title="Create new user">
  <form>
    <fieldset>
      <label for="name">Enter your Comments Here</label>
      <input type="text" name="name" id="name" value="Jane Smith" class="text ui-widget-content ui-corner-all">
      <!-- Allow form submission with keyboard without duplicating the dialog button -->

      <input type="submit" tabindex="-1" style="position:absolute; top:-1000px">
    </fieldset>
  </form>
</div>

JavaScript JavaScript的

$('#Delete').click(function() {
  $('#dialog-form').dialog('open');
});

jQuery(document).ready(function() {
  jQuery("#dialog-form").dialog({
    autoOpen: false,
    modal: true,
    open: function() {
      jQuery('.ui-widget-overlay').bind('click', function() {
        jQuery('#dialog').dialog('close');
      })
    }
  });
});

Updated Demo 更新的演示

如何在MVC 4按钮上显示jQuery模式弹出窗口

问题描述

1 个解决方案

解决方案1
0 已采纳 2016-03-10 05:40:34

如何在MVC 4按钮上显示jQuery模式弹出窗口

问题描述

1 个解决方案

解决方案1 0 已采纳 2016-03-10 05:40:34

解决方案1
0 已采纳 2016-03-10 05:40:34