选择查询中的总和和最后一条记录

Question

这是计算 2 个日期期间的剩余货物的代码：

剩余=首次开始数量+当期收入-当期销售额。

是否可以更改 dt1 派生表，使其获得最后收入的收入价格？不是总金额而是上次收入的价格？

我想要的输出是获取包含这些列的结果表： 商品 ID、商品、最后收入价格、已售数量、其余

我正在使用 Firebird 3。

数据库表：

• 商品：goods_id，商品...
• 收入：收入_id、收入_dt、qnt、商品_id...
• 销售：sale_id，收入_id，sale_dt，qnt，goods_id...

SELECT
    g.goods ,
    g.goods_id,
    coalesce(dt2.startqnt,0) as startqnt,
    coalesce(dt2.income,0)as income_qnt,
    coalesce(dt2.sales,0)as sales_qnt,
    coalesce(dt2.rest,0) as rest

FROM goods g
LEFT JOIN
   (select                                                                          --dt2
      goods_id,
      coalesce(sum(startincome),0) as startincome,
      coalesce(sum(startsale),0) as startsale,
      sum(coalesce(startincome,0)-coalesce(startsale,0)) as startqnt,
   
      coalesce(sum(income_qnt),0) as income_qnt,
      coalesce(sum(sale_qnt),0) as sales_qnt,
      sum(coalesce(startincome,0)-coalesce(startsale,0)+coalesce(income_qnt,0)-coalesce(sale_qnt,0)) as rest

    from(
         select                                                                     --dt1
           i.goods_id,
           sum(case when i.income_dt<:d1 then coalesce(i.qnt,0)  end) as startincome,
           0 as startsale,
           sum(case when (cast(i.income_dt as date)>=:d1) and (cast(i.income_dt as date)<=:d2) then coalesce(i.qnt,0) end) as income_qnt,
           0 as sale
        from income i
       where i.income_dt<=:d2
        group by goods_id
    union all
         select
           s.goods_id,
           0,
           sum(case when (cast(s.sale_dt as date) <:d1) then coalesce(s.qnt,0) end) as startsale,
           0,
           sum(case when (cast(s.sale_dt as date) >=:d1) and (cast(s.sale_dt as date)<=:d2) then coalesce(s.qnt,0) end) as sale_qnt
        from sale s , income i
        where
          s.sale_dt<=:d2 and
          s.income_id=i.income_id
        group by goods_id
    ) dt1
   group by goods_id )dt2
on g.goods_id=dt2.goods_id
order by goods

Answer 1

要获取每种产品的最后收入价格，您需要添加逻辑来获取每种产品的最新收入日期，然后将其与收入表连接起来以获得相关价格。

下面是 SQL 的修改版本，应该可以达到预期的结果：

SELECT
    g.goods,
    g.goods_id,
    coalesce(dt2.startqnt,0) as startqnt,
    coalesce(dt2.income_qnt,0) as income_qnt,
    coalesce(dt2.sales_qnt,0) as sales_qnt,
    coalesce(dt2.rest,0) as rest,
    coalesce(dt2.last_income_price,0) as last_income_price

FROM goods g
LEFT JOIN
   (select
      goods_id,
      coalesce(sum(startincome),0) as startincome,
      coalesce(sum(startsale),0) as startsale,
      sum(coalesce(startincome,0)-coalesce(startsale,0)) as startqnt,
      coalesce(sum(income_qnt),0) as income_qnt,
      coalesce(sum(sale_qnt),0) as sales_qnt,
      sum(coalesce(startincome,0)-coalesce(startsale,0)+coalesce(income_qnt,0)-coalesce(sale_qnt,0)) as rest,
      li.price as last_income_price

    from(
         select
           i.goods_id,
           sum(case when i.income_dt<:d1 then coalesce(i.qnt,0) end) as startincome,
           0 as startsale,
           sum(case when (cast(i.income_dt as date)>=:d1) and (cast(i.income_dt as date)<=:d2) then coalesce(i.qnt,0) end) as income_qnt,
           0 as sale_qnt
        from income i
        where i.income_dt<=:d2
        group by goods_id

    union all

         select
           s.goods_id,
           0,
           sum(case when (cast(s.sale_dt as date) <:d1) then coalesce(s.qnt,0) end) as startsale,
           0,
           sum(case when (cast(s.sale_dt as date) >=:d1) and (cast(s.sale_dt as date)<=:d2) then coalesce(s.qnt,0) end) as sale_qnt
        from sale s , income i
        where s.sale_dt<=:d2 and s.income_id=i.income_id
        group by goods_id
    ) dt1

    -- Get the last income price for each product
    LEFT JOIN (
      SELECT
          i.goods_id,
          i.price
      FROM income i
      WHERE i.income_dt = (
          SELECT MAX(i2.income_dt)
          FROM income i2
          WHERE i2.goods_id = i.goods_id
      )
    ) li on dt1.goods_id = li.goods_id

   group by goods_id, li.price
) dt2
on g.goods_id = dt2.goods_id
order by g.goods