如何从不同的表中添加2个总计?

时间:2013-02-02 08:12:58

标签: php html-table sum

如何从收到的总金额中减去总金额并将其显示在总收入上?

数据来自我的数据库中的不同表格..很遗憾,我无法上传图片以获得更清晰的视图..

TOTAL AMOUNT RECEIVED || 15610

TOTAL EXPENSES || 11300

TOTAL REVENUES ||  (this must be equal to TOTAL AMOUNT RECEIVED - TOTAL EXPENSES)

这是我的代码:

<table width="383" border="1" bordercolor="#00CCFF">
<tr>
<td width="245" bgcolor="#0099FF">TOTAL AMOUNT RECIEVED</td>
<td width="128" bgcolor="#FFFFFF">
            <?php
            include("confstudents.php");
            $id = $_GET['id'];
            $query = "SELECT id, SUM(1stPayment + 2ndPayment + 3rdPayment + 4thPayment) um_payment FROM student_payments"; 
            $result = mysql_query($query) or die(mysql_error());
            // Print out result
            while($row = mysql_fetch_array($result)){
            echo "" . $row['sum_payment'];
            echo "<br/>";
            }
            ?>
</td>
</tr>
<tr>
<td bgcolor="#0099FF">TOTAL EXPENSES</td>
<td bgcolor="#FFFFFF">
            <?php
            include("confexpenses.php");
            $id = $_GET['id'];
            $query = 'SELECT SUM(piece * price) tprice FROM expenses'; 
            $result = mysql_query($query) or die(mysql_error());
            while($res = mysql_fetch_assoc($result)){
            echo " " . $res['tprice']; " ";
            }
            ?>
</td>
</tr>


<tr>
<td bgcolor="#0099FF">TOTAL REVENUES</td>
<td bgcolor="#FFFFFF">
            <?php
            include("totalrev.php");
            ?>
</td>
</tr>
</table>

2 个答案:

答案 0 :(得分:0)

我将找出独立于结构/样式的一般答案,但实际上你想要从frist两个查询存储返回值,然后做一些数学运算。让我们先简单一点(KISS - Keep It Simple,Stupid),并从样式中抽象出逻辑。

<?php
$query = "
SELECT
    id,
    SUM(1stPayment + 2ndPayment + 3rdPayment + 4thPayment) AS sum_payment
FROM
    student_payments"; 

$result = mysql_query($query) || die(mysql_error());

// Create a variable to store the sum of payments
$sum_payment = 0;

// Print out result
while($row = mysql_fetch_array($result))
{
    echo "" . $row['sum_payment'];
    echo "<br/>";

    $sum_payment += (int)$row['sum_payment'];
}

$query = 'SELECT SUM(piece * price) tprice FROM expenses'; 
$result = mysql_query($query) or die(mysql_error());

// Variable to store the expenses
$expenses = 0;

while($res = mysql_fetch_assoc($result))
{
    echo " " . $res['tprice']; " ";
    $expenses += $res['tprice'];
}

// Calculate the difference
$total_rev = $sum_payments - $expenses;

echo '<br/>', $total_rev, '<br/>';
?>

关于你使用$_GET['id']的说明 - 如果你计划引入一些最终会进入SQL查询的东西,你应该逃避它:使用mysql_ library,使用mysql_real_escape_string http://php.net/manual/en/function.mysql-real-escape-string.php但是,一般来说,你应该切换到使用MySQLi库 - http://www.php.net/manual/en/book.mysqli.php - 因为它更安全,最终PHP不支持标准的mysql函数。

答案 1 :(得分:0)

如果两个表之间存在关系,例如id:

SELECT 
SUM(r.stPayment) as RECIEVED, sum(e.piece * e.price) as EXPENSES
FROM
student_payments as r,
expenses as e
WHERE r.id = e.id and r.id = '$id'