mysql - 如何减去作为子查询结果的 2 列

Question

我有这个查询，它按预期工作正常，即它显示所有列以及每个子查询的 2 列...

SELECT 
    a.group_id,
    a.code,
    a.description_en,
    c.size_code,
    (
        SELECT
            SUM(b.qty)
        FROM
            receiving_details AS b
        WHERE
            b.code = c.size_code
    ) AS in_qty,
    (
        SELECT
            SUM(d.qty)
        FROM
            requisition_details AS d
        WHERE
            d. matl_code = c.size_code
    ) AS out_qty
FROM products AS a
INNER JOIN products_sizes AS c ON c.prod_code = a.code
ORDER BY a.group_id ASC, a.code ASC, c.size_code ASC

但是，当我尝试添加这一行时...

(in_qty - out_qty) AS balance,

就在“FROM”语句之前，我收到一个错误:Unknown column 'in_qty' in 'field list'。

我做错了什么？

编辑:

根据接受的答案，我又做了一些修复并得到了我想要的结果。

SELECT *, (e.in_qty - e.out_qty) AS balance FROM
(SELECT 
    a.group_id,
    a.code,
    a.description_en,
    c.size_code,
    (
        SELECT
            IFNULL(SUM(b.qty),0)
        FROM
            receiving_details AS b
        WHERE
            b.code = c.size_code
    ) AS in_qty,
    (
        SELECT
            IFNULL(SUM(d.qty),0)
        FROM
            requisition_details AS d
        WHERE
            d. matl_code = c.size_code
    ) AS out_qty
FROM products AS a
INNER JOIN products_sizes AS c ON c.prod_code = a.code) AS e
ORDER BY e.group_id ASC, e.code ASC, e.size_code ASC

Answer 1

您不能在自己的 select 语句中引用同一字段。一种选择是将结果移动到另一个子查询中，然后执行计算:

select *, (in_qty - out_qty) AS balance
from (
    SELECT 
        a.group_id,
        a.code,
        a.description_en,
        c.size_code,
        (
            SELECT
                SUM(b.qty)
            FROM
                receiving_details AS b
            WHERE
                b.code = c.size_code
        ) AS in_qty,
        (
            SELECT
                SUM(d.qty)
            FROM
                requisition_details AS d
            WHERE
                d. matl_code = c.size_code
        ) AS out_qty
    FROM products AS a
    INNER JOIN products_sizes AS c ON c.prod_code = a.code
) t
ORDER BY group_id, code, size_code