Question

这是我的情况：

我想计算给定日期的给定项目的数量和价格。

价格是使用总商品数量和单价计算得出的，因此价格会随商品数量而变化。

warehouse_1指出该物品是从该仓库发货的，warehouse_2指出该物品已发送到该仓库。

这是我的逻辑：

获取每个项目的交货并汇总其数量。（第一CTE）

分别查找两个仓库中的数量总和。（第二CTE）

计算最终数量，然后乘以单价。

显示结果，其中包含商品ID，数量和价格。

我编写了一个查询，该查询可以正确执行计算，但是当数据数量增加时，它会以指数形式变慢。（在具有6k行的数据库上花费5秒，几乎将数据库锁定在具有21k行的同事数据库上）

如何优化此查询？我正在对来自第一CTE的每一行进行第二CTE的累积计算，我认为需要进行返工。

我可以使用 LAG() 此用例的功能？我用

之类的方法尝试过

LAG(a.deliveryTotal) over(order by a.updated desc rows between unbounded preceding and current row)

而不是第二CTE中的CASE块，但是我似乎无法弄清楚如何使用filter()或在LAG（）语句中放置条件。

这是我的查询：

`

with deliveriesCTE as (
select
    row_number() over(partition by it.id
order by
    dd.updated asc) as rn,
    sum(dd.quantity) as deliveryTotal,
    dd.updated as updated,
    it.id as item_id,
    d.warehouse_1 as outWH,
    d.warehouse_2 as inWH,
    d.company_code as company
from
    deliveries d
join deliveries_detail dd on
    dd.deliveries_id = d.id
join items it on
    it.id = dd.item_id
where
    ...
group by
    dd.updated,
    it.id,
    d.warehouse_1,
    d.warehouse_2,
    d.company_code
order by
    dd.updated asc),
cumulativeTotalsByUnit as (
select
    distinct on
    (a.item_id) a.rn,
    a.deliveryTotal,
    a.updated,
    a.item_id,
    a.outWH,
    a.inWH,
    a.company,
    case
        when a.rn = 1
        and a.outWH is not null then coalesce(a.deliveryTotal,
        0)
        else (
        select
            coalesce(sum(b.deliveryTotal) filter(
            where b.outWH is not null),
            0)
        from
            deliveriesCTE b
        where
            a.item_id = b.item_id
            and b.rn <= a.rn)
    end as outWHTotal,
    case
        when a.rn = 1
        and a.inWH is not null then coalesce(a.deliveryTotal,
        0)
        else (
        select
            coalesce(sum(b.deliveryTotal) filter(
            where b.inWH is not null),
            0)
        from
            deliveriesCTE b
        where
            a.item_id = b.item_id
            and b.rn <= a.rn)
    end as inWHTotal
from
    deliveriesCTE a
order by
    a.item_id,
    a.updated desc)
select
    resultView.item_id,
    resultView.quantity,
    resultView.price
from
    (
    select
        cumTotals.item_id,
        cumTotals.inWHTotal - cumTotals.outWHTotal as quantity,
        p.price * (cumTotals.inWHTotal - cumTotals.outWHTotal) as price
    from
        prices p
    join cumulativeTotalsByUnit cumTotals on
        cumTotals.item_id = p.item_id ) resultView
where
    resultView.rn = 1;

`

Answer 1

很难说不使用MCV，但我对您要执行的操作的猜测是执行Windowed SUM（）计算而不是LAG（）。有文档Here。

查询cumulativeTotalsByUnit并不是必需的，并且可以二次执行复杂的自引用联接。

您的交付CTE应该如下：

select
    sum(dd.quantity) over (partition by it.id ORDER BY dd.updated asc) as deliveryTotal,
    dd.updated as updated,
    it.id as item_id,
    d.warehouse_1 as outWH,
    d.warehouse_2 as inWH,
    d.company_code as company
from
    deliveries d
join deliveries_detail dd on
    dd.deliveries_id = d.id
join items it on
    it.id = dd.item_id
where
    ...
group by
    dd.updated,
    it.id,
    d.warehouse_1,
    d.warehouse_2,
    d.company_code
order by
    dd.updated asc

查找条件和分组依据的累积总数

1 个答案: