左外部连接问题与mysql

Question

我有下面的代码，与我从表中选择第一个所有ID的用户，然后我选择并从属于这个用户的优惠券表中找到点的总和，然后我也选择所有从零售商表中属于此用户的点。然后我做这些和之间的区别。

但是出了点问题，我得到了完全不同的观点。

$query4 = 'SELECT u.*, sum(c.points) as total_sum1, sum(r.basket_value) as total_sum 
      FROM users u 
      left outer join coupon c on u.user_id=c.user_id 
      left outer join retailer r on u.user_id=r.user_id 
      group by user_id'; 
$result4 = mysql_query($query4) or die(mysql_error()); 

$total1=0; 
$total=0; 
$total2=0; 
while($row = mysql_fetch_array($result4)) { 
    $total1 += $row['total_sum1']; 
    $total += $row['total_sum']; 
    echo "<table>"; 
    echo "<tr>"; 
    echo "<td>"; 
    echo $total2=$total-$total1; 
    echo "</td>"; 
    echo "<td>"; 

    echo "</td>"; 
    echo "</tr>"; 
    echo "</table>"; 
} 

样品输出：

total points remaining | user_id 
0        9839467227 
0        9853125067 
0        9937770769 
0        9974837329 
222060      A101 
0        A102 
0        A103 
0        A104 

Answer 1

我猜你的问题是：

$total1 += $row['total_sum1']; 
    $total += $row['total_sum']; 

$total1和$total被追加为所有用户的所有记录。我想，那一定是：

$total1 = $row['total_sum1']; 
    $total = $row['total_sum']; 

试试这个：

SELECT u.*, SUM(c.ts) AS total_sum1, SUM(r.bv) AS total_sum 
FROM users u 
LEFT JOIN 
(SELECT user_id ,SUM(points) AS ts FROM coupon GROUP BY user_id) c 
ON u.user_id=c.user_id 
LEFT JOIN 
(SELECT user_id ,SUM(basket_value) AS bv FROM retailer GROUP BY user_id) r 
ON u.user_id=r.user_id 
GROUP BY u.user_id;