Question

我需要帮助将行从mysql数据库传递到div弹出窗口？

我如何将product id传递给我通过<a href="#shirt" class="view_btn">View</a>调用的弹出式div？

例如我想要这样

<a href="#shirt?id=<?php $row['id']" class="view_btn">View</a></li>

这里是我的代码

 <div class="latest_products_sec">
 <div class="latest_products">
 <div class="title_box">

 </div>
 <!--Latest Products Slider -->
 <div class="latest_pro_box">
 <h4><a href="detail.html">Latest Shirt's Collection</a></h4>
 <div class="latest_products_slider">

 <?php 
 $queryshirt=mysql_query("select image,id from products 
 where cid='1' LIMIT 5") or die ('die shirt query');
 while($rowshirt=mysql_fetch_array($queryshirt))
 {
 echo '<ul>';
 echo '<li><img src="admin/'.$rowshirt['image'].'" 
 width="225" height="300" alt="" />

 <a href="#shirt" class="view_btn">View</a></li>';

 echo '</ul>';?>
 }
 ?>

#shirt Div弹出窗口

 <div id="shirt" class="proquickview">
 <span class="close_btn" onclick="parent.jQuery.fancybox.close();">Close</span>
 <h2 class="title">quick view</h2>
 <div class="quickviewinfo">

 <?php
 // I Need To Get ID Here 
 echo $id=$_REQUEST['id'];
 ?>

 <div class="quickviewinforight">
 <div class="titlerow">
 <h2>Latest Shirt's Collection</h2>
 <div class="start_ratings">
 <div class="start_rating">
 </div>
 </div>
 <div>
 <div class="quick_review">
 <h3>Quick Discription</h3>
 <p>TEST.</p>
 </div>

<div class="qty_row">
<div class="qty_field">
<label>Select Quantity</label>
<span>
<select name="S">
<option>5</option>
</select>
</span>
</div>
<br class="clear" />
<span class="total_price">Price <strong>$88.00</strong> 
<del>$102.00</del></span>
<a href="cart.html" class="add_cart_btn">ADD TO CART</a>
</div>

<div class="total_price_row">
</div>
</div>
</div>
</div>

用于弹出窗口的Javascript

$(document).ready(function() {
$(".view_btn").fancybox({
'titlePosition'     : 'inside',
'transitionIn'      : 'none',
'transitionOut'     : 'none'
 });
 });

 $(document).ready(function(){
 // Cufon Functions //  
 Cufon.replace ('.latest_products_slider ul li a.view_btn',{hover:true});
 });

Answer 1

您可以使用新的'echo'

echo '<a href="#shirt?id='.$row['id'].'" class="view_btn">View</a></li>';

而不是使用此

echo '<li><img src="admin/'.$rowshirt['image'].'" width="225" height="300" alt="" />

<a href="#shirt" class="view_btn">View</a></li>';

组合'echo'语句用于图像和链接到View

Answer 2

使用jQuery .load函数。然后在动作上创建一个隐藏的div，tgen，使用load函数加载隐藏的div并将css选择更改为可见。非常简单，也可以添加效果。

如何将行从mysql数据库传递到div弹出窗口

2 个答案: