用户打开网站时在 Magento 主页上打开一个弹出窗口

Question

我想在用户每次访问网站时在主页中打开一个弹出窗口。 现在的问题是每次重新加载主页时都会出现弹出窗口。我只想显示此弹出窗口一次。

 <?php $currentUrl = Mage::helper('core/url')->getCurrentUrl();
 if($currentUrl=='http://website.com/')
 {
 ?>
<form action="<?php echo $this->getUrl('subscribe/index/subscriptionsave/'); ?>" id="subscribeForm" method="post">
<div id="boxes">
  <div id="dialog" class="window">
    <div id="popupfoot">  <a class="agree" href="#">close</a> </div>
    <div class="popupcontentbox">
        <div class="box_content">
            <h2>Stay in the loop!!</h2>
            <p>Be the first one to know about new arrivals,
fresh offers &  fashion updates.</p>
         <div class="book-column-left">
        <?php /*?> <label class="f">Email Address<span>*</span></label><?php */?>
         <input type="text" class="input-text required-entry validate-email" placeholder="Email Address *"  name="emailid" id="emailid">
          </div>
           <input type="submit" name="sub" value="Submit" class="btn">
           <div class="offer_text">Get flat 20% Off on your First purchase. Use coupon "INDULGE20".</div>
        </div>
    </div>
  
  </div>
  <div id="mask"></div>
</div>
</form>
 <?php } ?>
<script type="text/javascript">
//<![CDATA[
    var subscribeForm = new VarienForm('subscribeForm', true);
//]]>
    </script>



<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.js"></script> 
    <?php */?><script>
    jQuery(document).ready(function() { 
    var id = '#dialog'; 
    //Get the screen height and width
    var maskHeight = jQuery(window).height();
    var maskWidth = jQuery(window).width();
        
    //Set heigth and width to mask to fill up the whole screen
    jQuery('#mask').css({'width':maskWidth,'height':maskHeight});
     
    //transition effect
    jQuery('#mask').fadeIn(500);    
    jQuery('#mask').fadeTo("slow",0.9); 
        
    //Get the window height and width
    var winH = jQuery(window).height();
    var winW = jQuery(window).width();
                  
    //Set the popup window to center
    jQuery(id).css('top',  winH/2-jQuery(id).height()/2);
    jQuery(id).css('left', winW/2-jQuery(id).width()/2);
        
    //transition effect
    jQuery(id).fadeIn(2000);    
        
    //if close button is clicked
    jQuery('.window .agree').click(function (e) {
    //Cancel the link behavior
    e.preventDefault();
    
    jQuery('#mask').hide();
    jQuery('.window').hide();
    });
    
    //if mask is clicked
    jQuery('#mask').click(function () {
    jQuery(this).hide();
    jQuery('.window').hide();
    });
        
         
    jQuery(window).scroll(function() {    
            
    var scroll = jQuery(window).scrollTop();
            if(scroll > 300){
                jQuery(".gotop").addClass("active");
            }else if(scroll < 300){
                jQuery(".gotop").removeClass("active");
            }
    
            
        });
        
        jQuery('.gotop').click(function(){
        jQuery("html, body").animate({ scrollTop: 0 }, 700);
            return false;
         });
    });
    </script>

Answer 1

保存cookies 值，例如OPENED = 1，然后在下一页打开php 一边检查此类cookie 是否已经存在，然后不要添加您的popup 代码或在javascript 一边检查并执行不显示popup

JS 实现示例

1) 将 set 和 get cookie 函数添加到您的 js - 请参阅此处有关 setCookie 和 getCookie 的手册 - http://www.w3schools.com/js/js_cookies.asp

<script type="text/javascript">

function setCookie(cname, cvalue, exdays) {
    var d = new Date();
    d.setTime(d.getTime() + (exdays*24*60*60*1000));
    var expires = "expires="+d.toUTCString();
    document.cookie = cname + "=" + cvalue + "; " + expires;
} 
function getCookie(cname) {
    var name = cname + "=";
    var ca = document.cookie.split(';');
    for(var i=0; i<ca.length; i++) {
        var c = ca[i];
        while (c.charAt(0)==' ') c = c.substring(1);
        if (c.indexOf(name) == 0) return c.substring(name.length,c.length);
    }
    return "";
} 

</script>

2) 只需用if contion 代码括起来即可显示您的弹出窗口，正如我从您凌乱的代码中了解到的那样，它是jQuery(id).fadeIn(2000);

if( getCookie('popup_opened') == "" ){
     jQuery(id).fadeIn(2000);

     // creates popup_opened cookie and will keep it till browser close
     setCookie('popup_opened', '1', 0)
}

注意：通过此解决方案，您的访问者至少应该在每个使用过的浏览器上看到该弹出窗口，因为 cookie 分别存储在每个浏览器中