在MySQL和PHP上使用左连接对其他表中的字段求和

我有四张桌子：A桌、B桌、C桌和D桌

TABLE_A
id_a | Name
A1   | ASD
A2   | ZXC

TABLE_B
id_b | id_a
B1   | A1
B2   | A2

TABLE_C
id_c | id_b | Value
C1   | B1   | 1
C2   | B1   | 1
C3   | B2   | 1
C4   | B2   | 1
C5   | B2   | 1

TABLE_D
id_d | id_a | Bill
D1   | A1   | 5
D2   | A2   | 10

我想得到每个表的总和（表c.值）为tot1，总和（表d.账单）为tot2，如下所示：

id_a  | SUM_VALUE of table_c | SUM_BILL of table_d
A1    | 2                    | 5
A2    | 3                    | 10

我正在使用以下脚本：

$sql=" SELECT *, SUM(table_c.value) as tot1, SUM(table_d.bill) as tot2
FROM table_a
LEFT JOIN table_b ON table_b.id_a=table_a.id_a
LEFT JOIN table_c ON table_c.id_b=table_b.id_b
GROUP BY id_a ";

但是得到了错误的结果：

id_a  | SUM_VALUE of table_c | SUM_BILL of table_d
A1    | 2                    | 10
A2    | 3                    | 30

---

有什么想法吗？

首先找到表c的

和值

，然后

将结果与表d的
左连接
，找到表d的
和值
，这样可以避免在
表中求和重复值。试试这个

SELECT F.id_a,
       F.SUM_VALUE_of_table_c,
       Sum(d.Bill) AS SUM_BILL_of_table_d
FROM   (SELECT a.id_a,
               Sum(c.Value) AS SUM_VALUE_of_table_c
        FROM   table_a A
               LEFT JOIN table_b B
                      ON B.id_a = A.id_a
               LEFT JOIN table_c C
                      ON c.id_b = B.id_b
        GROUP  BY a.id_a) F
       LEFT JOIN TABLE_D D
              ON d.id_a = F.id_a 
Group by F.id_a 

您可以在加入之前执行总和计算：

SELECT a.id_a, b.sum_b, c.sum_c
FROM   table_a a
JOIN   table_b b ON a.id_a = b.id_b
JOIN   (SELECT   id_b, SUM(value)
        FROM     table_c
        GROUP BY id_a) c ON b.id_b = c.id_b
JOIN   (SELECT   id_a, SUM(value)
        FROM     table_d
        GROUP BY id_a) c ON a.id_a = d.id_b       

我得到了答案：）

这是不可能的。表b的总账单值重复：（@FedriAndhika-更新，错过
分组人
。添加了演示牙链主控器，请注意：）
SELECT *,
       Sum(bill) AS sum_of_tableD
FROM   table_d,
       table_a
       LEFT JOIN (SELECT *,
                         Sum(table_c.value) AS sum_of_tableC
                  FROM   table_c,
                         table_b
                  WHERE  table_c.id_b = table_b.id_b
                  GROUP  BY id_b) x
              ON x.id_a = USER.id_a
WHERE  table_d.id_a = table_a.id_a