Question

我希望能够总结页面中显示的所有收入，并且每次我将其他数据添加到收入列时它会自动汇总：

以下是我的代码：

<?php 
   require_once('Connections/connect.php');
   $id_customer = mysql_real_escape_string($_GET['id_customer']);                                   
   $sql_PK = "SELECT * FROM tbl_delivery_details WHERE tbl_customer_id_customer =    {$id_customer}";
   $PK = mysql_query($sql_PK, $connect);
   if ( mysql_error() ) {
      die ( mysql_error());
   }
   $row_PK = mysql_fetch_assoc($PK);            
   $customer_name = $row_PK['tbl_customer_id_customer'];
   $customer_name = mysql_real_escape_string($customer_name);                       

   $sql = "SELECT tbl_customer.customer_name, 
       tbl_delivery_details.delivery_details_route, 
       tbl_delivery_details.delivery_details_destination, 
       tbl_delivery_details.delivery_details_van_no, 
       tbl_delivery_details.delivery_details_waybill_no, 
       tbl_delivery_details.delivery_details_charge_invoice,
       tbl_delivery_details.delivery_details_revenue,
       tbl_delivery_details.delivery_details_strip_stuff,
       tbl_delivery_details.delivery_details_date           
       FROM tbl_customer, tbl_delivery_details          
       WHERE tbl_customer.id_customer = tbl_delivery_details.tbl_customer_id_customer 
       AND tbl_customer.id_customer = '{$customer_name}'";

       $res = mysql_query($sql) or die(mysql_error());
       $row = mysql_fetch_array($res);
       $sum = 0;

?>


    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/x   html">
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Customer Revenue</title>
    <link rel="stylesheet" type="text/css" href="qcc.css"/>
    </head>             
    <body>              
    <table border="1">
      <tr>
         <th>Reveneu</th>                                                 
      </tr>
      <?php do { ?>
          <tr>                          
              <td><?php echo $row_PK['delivery_details_revenue'];?></td>                                                       
      </tr>
      <?php } while ($row_PK = mysql_fetch_assoc($PK));?>
      <?php { ?>
        <?php  $sum+=$row_PK['delivery_details_revenue'] ?>
      <?php } ?>

      </table>

      <?php echo $sum; ?>
    </body>
  </html>

当我加载页面时，echo $ sum始终为零如何正确地总结我所做的列，如果我向其添加另一个数据，它将自动求和：

Answer 1

为什么不让MySQL在查询中为你做这个呢？

而不是在PHP中添加收益值

$sql = "SELECT SUM(tbl_delivery_details.delivery_details_revenue) as revenue, 
    tbl_customer.customer_name, 
    tbl_delivery_details.delivery_details_route, 
    tbl_delivery_details.delivery_details_destination, 
    tbl_delivery_details.delivery_details_van_no, 
    tbl_delivery_details.delivery_details_waybill_no, 
    tbl_delivery_details.delivery_details_charge_invoice,
    tbl_delivery_details.delivery_details_revenue,
    tbl_delivery_details.delivery_details_strip_stuff,
    tbl_delivery_details.delivery_details_date

FROM tbl_customer, tbl_delivery_details 

WHERE tbl_customer.id_customer = tbl_delivery_details.tbl_customer_id_customer 
AND tbl_customer.id_customer = '{$customer_name}'";

然后在你看来，只需回应SUM数字......

echo $row_PK['revenue'];

Answer 2

好吧，我脑子里没有PHP解释器来运行你的代码。所以，我可以发现一些事情

首先，在第一个查询中有一个SQL注入。将变量转换为整数

$id_customer = intval($_GET['id_customer']);

或将其视为查询中的字符串

$sql_PK = "SELECT * FROM tbl_delivery_details WHERE tbl_customer_id_customer = '$id_customer'";

或 - 更好 - 使用一些数据库包装器，允许您使用占位符来表示查询中的实际数据。

接下来，您的查询难以理解如果您的字段名称没有干扰，则没有理由使用table.field表示法如果你想要表格中的大多数字段，也可以使用短地别名并考虑使用*：

$sql = "SELECT SUM(delivery_details_revenue) as revenue, 
               customer_name, tbl_delivery_details.* 
        FROM tbl_customer, tbl_delivery_details 
        WHERE id_customer = tbl_customer_id_customer 
              AND id_customer = '$customer_name'";

顺便说一句，在编辑查询时，我注意到命名不一致：id_customer = '$customer_name'。不要将自己与错误的变量名混淆。如果是id，则将其称为“id”，而不是“name”

如果id_customer等于tbl_customer_id_customer，我在第一个查询中也没有任何意义。我认为你需要简化你的代码 - 我相信它的复杂性是你没有得到你的结果的主要原因。

从非常简单的查询开始，如

$sql = "SELECT SUM(delivery_details_revenue) as revenue, 
        FROM tbl_delivery_details
        WHERE tbl_customer_id_customer = '$id_customer'";

并查看是否返回任何内容如果是这样 - 开始添加更多数据来获取如果不是 - 检查您的数据和整体数据结构是否正常。

Answer 3

如果我正确地阅读了这个，那么你总结了你的while循环之外的值。那不行。

我认为你正在混合正常的while循环和'do while'循环。

请参阅此代码：

<?php do { ?> <tr> <td><?php echo $row_PK['delivery_details_revenue'];?></td> </tr> <?php } while ($row_PK = mysql_fetch_assoc($PK));?> <?php { ?> <?php $sum+=$row_PK['delivery_details_revenue'] ?> <?php } ?>

应该更多地沿着这些方向：

<?php do { ?> <tr> <td><?php echo $row_PK['delivery_details_revenue']; $sum+=$row_PK['delivery_details_revenue'] ?> </td></tr> <?php } while ($row_PK = mysql_fetch_assoc($PK));?>
如果你能更清楚地编写代码，那么这不会发生;尽量避免交错html和php：

<?php do { $revenue = $row_PK['delivery_details_revenue']; $sum += revenue; println("<tr><td>$revenue</td></tr>"); } while ($row_PK = mysql_fetch_assoc($PK)); ?>

如果你问我，这会更清楚。

它不会总结，我的代码有什么问题？

3 个答案: