我正在一个项目上,我从数据库中合并图像src并保持隐藏状态,直到单击相对于图像位置的按钮。但是,图像绝对定位。以下是我的代码段。

的HTML

<div class="table-responsive">
    <table class="table table-bordered table-hover table-striped">
      <thead>
        <tr>
             <th>S/N</th>
             <th>Name</th>
             <th>Phone</th>
             <th>Proof of Payment</th>
             <th>Activate User</th>
        </tr>
    </thead>
    <tbody>

        <tr>
            <td>1</td>
            <td>Okolo Michael</td>
            <td>08062970094</td>
            <td><button class="btn btn-warning view_pop">View POP</button></td>
            <td><button class="btn btn-primary">Activate User</button></td>
            <div class="pop_view"><img src="pop/59a178206fade2.43644948.jpg" alt="File not Found"> <span class="close">X</span></div>
           </tr><tr>
            <td>2</td>
            <td>Okeke Chidimma</td>
            <td>08044323123</td>
            <td><button class="btn btn-warning view_pop">View POP</button></td>
            <td><button class="btn btn-primary">Activate User</button></td>
            <div class="pop_view"><img src="pop/59a178206fade2.43644948.jpg" alt="File not Found"> <span class="close">X</span></div>
          </tr><tr>
            <td>3</td>
            <td>Anibueze Chigozie</td>
            <td>08162657108</td>
            <td><button class="btn btn-warning view_pop">View POP</button></td>
            <td><button class="btn btn-primary">Activate User</button></td>
           <div class="pop_view"><img src="pop/59a178206fade2.43644948.jpg" alt="File not Found"> <span class="close">X</span></div>
          </tr> </tbody>
    </table>
</div>


jQuery查询

$('.view_pop').click(function() {
    $(this).parent().siblings('.pop_view').css('display', 'block');
    $(this).html('Viewed');
});

$('.close').click(function() {
    $(this).parent('.pop_view').css('display', 'none');
});


的CSS

.pop_view{
    display: none;
    position: absolute;
    z-index: 5;
    top: 25%;
    left: 25%;

}

.close{
    position: relative;
    top: 0;
    right: 0;
}


单击该按钮时,不会显示图像。如何基于单击的按钮唯一地选择这些图像?

最佳答案

实际上,<div>中的<tr>是无效的结构,浏览器将其呈现在顶部(整个表本身之外),这就是代码失败的原因(因为它不再是child/siblingstr/td)。

解决方案:-将<div>转换为<td>

工作示例:-



$('.view_pop').click(function() {
    $(this).parent().siblings('.pop_view').css('display', 'block');
    $(this).html('Viewed');
});

$('.close').click(function() {
    $(this).parent('.pop_view').css('display', 'none');
});

.pop_view{
display:none;
float:left;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="table-responsive">
<table class="table table-bordered table-hover table-striped">
<thead>
  <tr>
       <th>S/N</th>
       <th>Name</th>
       <th>Phone</th>
       <th>Proof of Payment</th>
       <th>Activate User</th>
  </tr>
</thead>
<tbody>

  <tr>
      <td>1</td>
      <td>Okolo Michael</td>
      <td>08062970094</td>
      <td><button class="btn btn-warning view_pop">View POP</button></td>
      <td><button class="btn btn-primary">Activate User</button></td>
      <td class="pop_view"><img src="pop/59a178206fade2.43644948.jpg" alt="File not Found"> <span class="close">X</span></td>
     </tr><tr>
      <td>2</td>
      <td>Okeke Chidimma</td>
      <td>08044323123</td>
      <td><button class="btn btn-warning view_pop">View POP</button></td>
      <td><button class="btn btn-primary">Activate User</button></td>
      <td class="pop_view"><img src="pop/59a178206fade2.43644948.jpg" alt="File not Found"> <span class="close">X</span></td>
    </tr><tr>
      <td>3</td>
      <td>Anibueze Chigozie</td>
      <td>08162657108</td>
      <td><button class="btn btn-warning view_pop">View POP</button></td>
      <td><button class="btn btn-primary">Activate User</button></td>
     <td class="pop_view"><img src="pop/59a178206fade2.43644948.jpg" alt="File not Found"> <span class="close">X</span></td>
    </tr> </tbody>
</table>
</div>

关于jquery - 我无法使用jquery唯一定位目标元素,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/45901016/

10-12 12:40
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