基于 table1_id MySQL 中的 SUM 排序？

Question

我有一些 table 即 Table_A、Table_B、Table_C 和 Table_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

我想为每个 table_a.id_a 获取“SUM(table_c.value) as tot1 AND SUM(table_d.bill) as tot2”，如下所示：

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

我使用了脚本，如下所示：

$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，我得到了错误的结果，如下所示：

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

有人帮帮我吗？谢谢

Answer 1

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

这应该可以解决问题，您需要分组依据才能获得 1 个以上的结果。

Answer 2

我没有测试，但是应该可以:

$sqlQuery = "
SELECT *, SUM(`bill`) AS `total_bill` 
FROM `table1` 
LEFT JOIN `table2` 
ON table1.table1_id = table2.table2_id
GROUP BY `table1_id` 
ORDER BY `total_bill` DESC
";

$result = mysql_query($sqlQuery);
while($row = mysql_fetch_array($result))
    echo "{$row['table1_id']} - {$row['total_bill']}<br>";

Answer 3

尝试如下。它应该工作。如果您在特定列上使用聚合函数，则应将该列添加到 group by 子句中。

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;

基于 table1_id MySQL 中的 SUM 排序？

ORDER BY SUM in MySQL based table1_id?

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