如何在php

时间:2015-10-18 06:47:26

标签: php jquery

我正在开发一个php的电子商务项目。除了一件事" cart.php"外,其余的事情已经完成。在这个"购物车"以下项目显示在表格中:产品图像,数量和价格。在数量列中有一个下拉列表,其中填充了1到10之间的数字。

在此,我想" 显示存储在表格中的产品数量,以显示为选定数量"下拉列表。我将如何实现这一目标?

以下是我的代码:

$pickrecs = "SELECT t_ordid, DATE_FORMAT(tord_dt,'%d-%m-%Y') as label_date, file_path, file_name, lbl_qty FROM ord where uid = '{$getid}'";
    $result = mysqli_query($connection, $pickrecs); ?>
    <table id='display'>
      <thead>
         <tr>
        <th>Product</th>
        <th>Creation&nbsp;Date</th>
        <th>Labels</th>
        <th>Total&nbsp;Price (&nbsp;&nbsp;)</th>
        <th></th>
         </tr>
      </thead>
      <?php while($row = mysqli_fetch_assoc($result)){ ?>
        <tbody>
          <tr>
        <td style="display:none;"><?php echo $row["t_ordid"] ?></td>
        <td><img src="<?php echo $row['file_path'] ?>" alt="" style="width:61px; height:54px" /></td>
        <td><?php echo $row["label_date"] ?></td>
        <td>
          <?php
                 $pickrecs = "SELECT r.rt_qty, r.rt_cost FROM rm as r join fy as f on r.fin_yr = f.fyear where f.status='A'";
                 $rtlist = mysqli_query($connection, $pickrecs); 
                 echo "<select id='rtcrt' name='rtcrt' value='' class='rdd'>";
                   while($rtlistarr = mysqli_fetch_assoc($rtlist)){ 
                 //echo "<option value = $rtlistarr[rt_qty]>$rtlistarr[rt_qty]</option>";
                 echo "<option value = $row[lbl_qty]>$rtlistarr[rt_qty]</option>";
                   }
                 echo "</select>";
              ?>
              <?php echo $_POST['rtcrt'] = $row['lbl_qty'] ?>
        </td>
        <td><input type="text" name="i-cost" value="5.95" readonly class="icost"></td>
          </tr>
        </tbody>
      <?php } ?>
    </table>

购物车 Snapshot of Shopping Cart

0 个答案:

没有答案