MySQL从另一个选择查询值中减去一个选择查询

Question

我有两个mysql表，并且已经建立了使用这两个表选择查询的方法，两个查询输出都是来自stock_transfer_details表的数量的总和。我需要从“ total_from”中减去“ total_to”。请检查以下是mysql表。

库存转移

inventory_transfer_details

下面是我的两个查询。

第一个查询：

select sum(b.transfer_quantity) as total_to 
from inventory_transfers as a 
join inventory_transfer_details as b on a.id = b.inventory_transfer_id 
where a.status="approved" and b.inventory_or_composite_id = '1' and a.to_warehouse_id = '2'

第二次查询：

select sum(b.transfer_quantity) as total_from 
from inventory_transfers as a 
join inventory_transfer_details as b on a.id = b.inventory_transfer_id 
where a.status="approved" and b.inventory_or_composite_id = '1' and a.from_warehouse_id = '2'

我需要从查询transfer_to减去to transfer_from

您是否有任何指南将最终查询转换为laravel查询？

Answer 1

您可以结合使用这些查询并获得不同之处

SELECT
  sum(
    IF(a.to_warehouse_id = '2', b.transfer_quantity, 0)
  ) - sum(
    IF(a.from_warehouse_id = '2', b.transfer_quantity, 0)
  ) as total
FROM
  inventory_transfers AS a
  JOIN inventory_transfer_details AS b ON a.id = b.inventory_transfer_id
WHERE
  a.status = "approved"
  AND b.inventory_or_composite_id = '1'

在laravel查询构建器中，该查询可能类似于

DB::table('inventory_transfers as a')
    ->select(DB::raw('sum(IF(a.to_warehouse_id = '2', b.transfer_quantity, 0)) - sum(IF(a.from_warehouse_id = '2', b.transfer_quantity, 0)) as total'))
    ->join('inventory_transfer_details as b', DB::raw('a.id'), '=', DB::raw('b.inventory_transfer_id'))
    ->where([
        ['a.status', '=', 'approved']
        ['b.inventory_or_composite_id', '=', 1]
    ])
    ->get()

Answer 2

我想为这个问题提供一个更通用的答案：从任何其他SUM()中减去任何SUM()。你可以简单地做...

SELECT Total1, Total2, Total1 - Total2
FROM
    (SELECT SUM(id) Total1 FROM TableA) AS a
INNER JOIN
    (SELECT SUM(id) Total2 FROM TableB) AS b;

在您的情况下，两个a和b子查询将被替换为当前用于计算总和的两个查询。