如何從不同的表中添加2個總數？

如何從收到的總金額中扣除總費用並將其顯示在總收入中？如何從不同的表中添加2個總數？

的數據來自不同的表從我的數據庫..可惜我不能上傳更清晰的視圖PIC ..

TOTAL AMOUNT RECEIVED || 15610 

TOTAL EXPENSES || 11300 

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

這裏是我的代碼：

<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>

來源

2013-02-02 Anna Mae

在第一循環中，添加'$行[ 'sum_payment']''到$ total_received'。在第二個循環中添加'$ res ['tprice']'到'$ total_expenses'。最後，計算'$ total_revenues = $ total_received - $ total_expenses'。 – Barmar

我首先要找出與結構/風格無關的一般答案，但本質上你想存儲第一個兩個查詢的返回值，然後做一些數學計算。讓我們先簡單化（KISS - 保持簡單，愚蠢），並從樣式中抽象出邏輯。

<?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/>'; 
?>

您所使用的$_GET['id']的注意事項 - 如果您計劃在東西最終將進入一個SQL查詢帶來，你應該逃避它：他們mysql_庫，使用mysql_real_escape_string http://php.net/manual/en/function.mysql-real-escape-string.php但是，在一般情況下，你應該切換到使用MySQLi庫 - http://www.php.net/manual/en/book.mysqli.php - 因爲它更安全，並且PHP最終不會支持標準的mysql函數。

來源

2013-02-02 08:34:44 phatskat

它不起作用，它顯示11300等於11300，並在另一個11300負號後打破。 –

每個while循環後$ sum_payment和$費用的值是多少？ – phatskat

如果有兩個表之間的關係，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'

來源

2013-02-02 08:51:36

nope他們不是。 –

其中r.id ='$ id'和r.id ='$ id' – 2013-02-02 11:42:28

它只顯示11300總費用:( –

如何從不同的表中添加2個總數？

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