Question

DBMS：Derby

语言：Java

外部查询从内部查询中获取结果它在某个日期内获得了某种子纹理的销售但有时候某种类型的订单细节没有 subtexture。所以没什么可加入的，我想看一个“0” 对于这种情况。抱歉，错误的命名和缩进。谢谢千斤顶

SELECT s1.TextureName AS TYPE,
       T.month AS MONTH,
       NULLIF(sum(S.Sales),0) AS NS,
       NULLIF(T.Total,0) AS total,
       NULLIF(NULLIF(cast(S.Sales AS Float),0.0) / NULLIF(Cast(T.Total AS float),0.0) * 100.0,0.0) AS percent
FROM subtexture s1
LEFT JOIN
  (SELECT s.SubTextureID,
          sum(od.NetSales) AS Sales,
          MONTH(o.PURCHASEDATETIME) AS MONTH
   FROM subtexture s
   LEFT JOIN products p ON (s.SubTextureID = p.SubTextureID)
   LEFT JOIN order_details od ON (p.ProductCode = od.ProductCode)
   LEFT JOIN orders o ON (od.OrderCode = o.OrderCode)
   WHERE TRIM(CAST(CAST(YEAR(o.PurchaseDateTime) AS CHAR(4)) AS VARCHAR(4))) || '-' || TRIM(CAST(CAST(MONTH(o.PurchaseDateTime) AS CHAR(2)) AS VARCHAR(2))) = '2013-11'
   GROUP BY s.SubTextureID,
            MONTH(o.PurchaseDateTime)) AS S ON (S.SubTextureID = s1.SubTextureID) ,
  (SELECT sum(od.NetSales) AS Total,
          MONTH(o.PurchaseDateTime) AS MONTH
   FROM orders o
   LEFT JOIN order_details od ON (o.OrderCode = od.OrderCode)
   WHERE TRIM(CAST(CAST(YEAR(o.PurchaseDateTime) AS CHAR(4)) AS VARCHAR(4))) || '-' ||TRIM(CAST(CAST(MONTH(o.PurchaseDateTime) AS CHAR(2)) AS VARCHAR(2))) = '2013-11'
   GROUP BY MONTH(o.PURCHASEDATETIME)) AS T
WHERE T.MONTH = S.MONTH
GROUP BY s1.TextureName,
         T.Total,
         T.MONTH,
         S.Sales

Answer 1

NULLIF(sum(S.Sales),0) AS NS,
...
NULLIF(NULLIF(cast(S.Sales AS Float),0.0) / NULLIF(Cast(T.Total AS float),0.0) * 100.0,0.0) AS percent

这永远无法奏效。稍微简化一下，你这样做：

SELECT sum(S.Sales),
       S.Sales

您如何设想将销售总额和个人销售同时归入同一个退货记录？由于您按销售分组，因此SUM（Sales）在任何时候都等于Sales，这让我相信您可能正在尝试完成查询中未反映的内容。

Answer 2

我认为代码存在一些问题。不幸的是我没有表格和测试数据。所以请记住，我的结果或多或少都是出于我的想法。仔细测试。

为了便于阅读，我将代码分为两个步骤。第一步获得销售总额，第二步计算所有纹理的总销售额并构建百分比。对于您的测试，将第一个代码复制两次到第二个代码中，替换SALES_ONE_TEXTURE的两个出现。我只是不想在必要时淹没答案。

以下是一个纹理（以及月份和年份）的销售总额的第一部分

-- Use this as subselect. And you're right: you need it twice
(   select  s.TextureName, s.SubTextureId, o.year, o.month, sum(COALESCE(d.NetSales, 0)) NetSales
    from    (select SubTextureId, TextureName from subtexture) as s
    join
            (select SubTextureId, ProductCode from products) as p
        on  (p.SubTextureId = s.SubTextureId)
    left outer join
            (select ProductCode, OrderCode, NetSales from order_details) as d
        on  (d.ProductCode = p.ProductCode)
    join
            (   select  YEAR(o.PurchaseDateTime) year,
                        MONTH(o.PurchaseDateTime) month,
                        OrderCode
                from    orders
                where   o.PurchaseDateTime between '2013-11-01 00:00:00' and '2013-11-30 23:59:59' -- make use of an index if one exists
            ) as o
        on  (o.orderCode = d.orderCode)
    group by 1,2,3,4
)   as SALES_ONE_TEXTURE;

这里是将数字组合在一起的外部代码：

-- The outer query
select  O.TextureName,
        A.year, A.month,
        A.NetSales, O.NetSales,
        cast(O.NetSales as float) / cast(A.NetSales as float) * 100.0 as percent
from    SALES_ONE_TEXTURE O -- replace here
join
        (   select  year, month, sum(NetSales) as NetSales
            from    SALES_ONE_TEXTURE -- replace here
            group by 1, 2 
        ) as SALES_ALL_TEXTURES A
    on  (A.year = O.year and A.month = O.month)

说明

COALESCE返回第一个非空值，通常用于将空值替换为默认值。 NULLIF执行相反的操作：将值替换为null。
替换来自外部联接的null不适用于所有RDBMS上的所有方言。你必须尝试，如果它在这里工作。如果没有，您需要将整个查询放在子选择中并在外部选择中执行coalesce替换
我过了一年又一个月，你过了一个月。每个月都会重复一个月，所以只有这种组合才是独一无二的。我也改变了获得月份的方式。我的方式可以使用索引（如果存在）。

如何从结果为null的左连接返回一种子纹理的“0”销售额

2 个答案: