【问题标题】:How to do a subtract of 2 columns that were the results of a sub-query如何减去作为子查询结果的 2 列
【发布时间】:2026-02-02 10:25:01
【问题描述】:

我有这个查询,它按预期工作正常,即它显示了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 'field list' 中的未知列 'in_qty' 的错误

我做错了什么?

编辑:

根据接受的答案,我又做了一些修复,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

【问题讨论】:

  • 从 stmt 得到什么,你有 3 个
  • @Drew,对不起...在此之前,'FROM products AS'
  • 查看@sgeddes 所说的内容,否则另一个包装器可以使该列别名起作用。这意味着另一个使用派生表的外部包装器可以使用它们
  • @Drew,是的,我刚刚阅读了 sgeddes 的回复并接受了它作为答案。也感谢您的帮助。

标签: mysql subquery subtraction


【解决方案1】:

您不能在 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

【讨论】:

  • @sgeddes...太棒了!经过几次修复后,我确实设法得到了我想要的结果。您的回答提供了方法。非常感谢!