Question

我想知道如何仅在特定功能中显示特定div。我想显示以下div，称为＆＃34; cd-popup＆＃34;在我的javascript函数中调用＆＃34; validateForm（）＆＃34;。

<script type="text/javascript">
   function validateForm() {
      var x = document.getElementById("email-id").value;
      if ( x == null || x == "" ) {
         $('.cd-popup').addClass('is-visible');
      }
   }
</script>
<div class="cd-popup" role="alert">
   <div class="cd-popup-container" style="margin-left: 10%;">
      <p style="color: #5b5b5b;">Are you sure you want to delete this element?</p>
      <a href="#0" class="cd-popup-close img-replace">Close</a>
      <!-- <img id="close"  style="position: absolute;right: -14px;top: -14px; cursor: pointer;" onclick ="div_hide()"> -->
   </div>
   <!-- cd-popup-container -->
</div>
<!-- cd-popup -->

此外，我已进入电子邮件div，我通过＆＃34; onblur＆＃34;

打电话给该功能

<!-- EMAIL -->
<div class="col-lg-6 col-xs-6">
   <div class="form-group">
       <input type="email" class="form form-control email requiretop" placeholder="Your email" name="email" id="email-id" onblur="return validateForm()">
   </div>
</div>

我只想在调用validateForm（）方法时显示div cd-popup

Answer 1

这是一个简单的show-hide div现象，jQuery .show（）和.hide（）元素是这个的完美解决方案。只需按以下方式修改脚本：

<script type="text/javascript">
function validateForm() {
    var x = document.getElementById("email-id").value;
    if ( x == null || x == "" ) {
        $('.cd-popup').show();
    }
    else{
        $('.cd-popup').hide(); 
    }
}

</script>

在javascript函数中显示特定的div

1 个答案: