Php 在基于MySQL的table1\u id中按和排序？

我有一些桌子，那就是桌子A，桌子B，桌子C和桌子D

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

对于每个表a.id\U a，我希望将SUMtable_c.值设置为tot1，将SUMtable_d.bill设置为tot2，如下所示：

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

$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 ";

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

我使用了脚本，如下所示：

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

$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 ";

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

但是，我得到了一个错误的结果，如下所示：

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

$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 ";

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

有人帮我吗？ 谢谢

这应该可以做到，你需要分组来获得不止一个结果。

我没有测试它，但它应该可以工作：

---

尝试如下。它应该会起作用。如果对特定列使用聚合函数，则该列应为GROUPBY子句中的外接程序

SELECT SUM(bill) AS total_bill 
from table1, table2 
WHERE table1.table1_id=table2.table2_id 
AND table2_id= yourId
Group by table1_id
order by bill;

---

请添加样本数据和预期结果。