页面加载后显示弹出窗口

Question

我按照在线教程创建了一个jQuery弹出窗口.

我想在页面加载后显示此弹出窗口/页面加载后出现+如何编码它就像它在页面加载5秒后出现一样.

由于我对jQuery的了解不足,我无法按照我的要求使其工作.

知道怎么做吗？

Javascript成为

<script type="text/javascript">
function PopUp(){
        document.getElementById('ac-wrapper').style.display="none"; 
}
</script>

CSS-

#ac-wrapper {
position: fixed;
top: 0;
left: 0;
width: 100%;
height: 100%;
background: rgba(255,255,255,.6);
z-index: 1001;
}
#popup{
width: 555px;
height: 375px;
background: #FFFFFF;
border: 5px solid #000;
border-radius: 25px;
-moz-border-radius: 25px;
-webkit-border-radius: 25px;
box-shadow: #64686e 0px 0px 3px 3px;
-moz-box-shadow: #64686e 0px 0px 3px 3px;
-webkit-box-shadow: #64686e 0px 0px 3px 3px;
position: relative;
top: 150px; left: 375px;
}

HTML的

 <div id="ac-wrapper">
  <div id="popup">
  <center>
    <h2>Popup Content Here</h2>
      <input type="submit" name="submit" value="Submit" onClick="PopUp()" />
  </center>
  </div>
</div>

<p>Page Content Here......</p>

Answer 1

尝试这样的事情

<script type="text/javascript">
function PopUp(hideOrshow) {
    if (hideOrshow == 'hide') document.getElementById('ac-wrapper').style.display = "none";
    else document.getElementById('ac-wrapper').removeAttribute('style');
}
window.onload = function () {
    setTimeout(function () {
        PopUp('show');
    }, 5000);
}
</script>

和你的 html

<div id="ac-wrapper" style='display:none'>
    <div id="popup">
        <center>
             <h2>Popup Content Here</h2>    
            <input type="submit" name="submit" value="Submit" onClick="PopUp('hide')" />
        </center>
    </div>
</div>

演示 JsFiddle

Answer 2

当DOM完成加载时,您可以在$(document).ready()函数中添加代码.

从这里删除onclick:

<input type="submit" name="submit" value="Submit" onClick="PopUp()" />

试试这个:

$(document).ready(function(){
   setTimeout(function(){
      PopUp();
   },5000); // 5000 to load it after 5 seconds from page load
});