php - 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，我已经使用这两个表构建选择查询，两个查询输出都是 inventory_transfer_details 表中数量的总和。我需要从“total_from”中减去“total_to”。请检查下面是 mysql 表。

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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'

第二个查询:

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 中减去 transfer_from

你有任何指南可以将这个最终查询转换为 Laravel 查询吗？

Answer 1

您可以组合这些查询并获得差异

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，这个查询可能看起来像

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()