获取多列的总和并计算每行的比率

时间:2013-02-06 02:34:43

标签: mysql sql sum pivot

下面是我的brand_of_items表的架构。为简单起见,此处显示了两列:id(主要和AI),符号(varchar 50,唯一)

Table - brand_of_items
id   symbol
0    a
1    b
2    c
..   ..
10   j

下面是我的items_of_brand的架构。

Table - mainIndexQuantity
id  brand_of_items_id   vol  item_type  salefinalizeddate
0         1              5      0       2005-5-11
1         1              6      0       2004-5-11
2         1              7      0       2011-5-11
3         1              8      0       2011-5-12
4         1              9      0       2011-5-12
5         1              10     0       2011-5-11
6         1              5      1       2012-5-11
7         1              6      1       2012-5-11
8         1              7      1       2011-5-11
9         1              8      1       2010-5-12
10        1              9      1       2012-5-12
11        1              10     1       2005-5-12

mainIndexQuantity表brand_of_items_id列是指向brand_of_items(id)的外键。 mainIndexQuantity表item_type列不是外键,它应该是。

这两个项目类型是:0 =零售,1 =批发

我想计算每个brand_brand_of_items表条目的项目类型(零售与批发)的比率。目标是看一个品牌商品是在零售还是批发销售更多。


** 增加复杂性: 我想在mainIndexQuantity表中添加一个日期列,并希望找出RetailVolume和WholesaleVolume之和的差异,并按salefinalizeddate字段对结果进行分组。

这是为了帮助确定哪些季节中销售的商品更多以及零售价格和&的总和(delta)差异。 WholeSaleVolume将帮助选择最受关注的项目。

1 个答案:

答案 0 :(得分:5)

试试这个:

SELECT 
  b.id,
  b.symbol,
  IFNULL(SUM(m.item_type = 1), 0) / (COUNT(*) * 1.0) AS wholesaleRatio,
  IFNULL(SUM(m.item_type = 0), 0) / (COUNT(*) * 1.0) AS RetailRatio
FROM brand_of_items b
LEFT JOIN mainIndexQuantity m ON b.id = m.brand_of_items_id
GROUP BY b.id, 
         b.symbol;

SQL Fiddle Demo

这会给你:

| ID | SYMBOL | WHOLESALERATIO | RETAILRATIO |
----------------------------------------------
|  0 |      a |              0 |           0 |
|  1 |      b |            0.5 |         0.5 |
|  2 |      c |              0 |           0 |
| 10 |      j |              0 |           0 |

假设:

  • wholesaleRatio是整个销售类型的项目计数到所有项目的计数。
  • RetailRatio是所有项目计数类型retail的项目数。

如果这个比例用于vol列与总vol的总和,则可以改为:

SELECT 
  b.id,
  b.symbol,
  SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) / SUM(m.vol) AS wholesaleRatio,
  SUM(CASE WHEN m.item_type = 0 THEN m.vol ELSE 0 END) / SUM(m.vol) AS RetailRatio
FROM brand_of_items b
LEFT JOIN mainIndexQuantity m ON b.id = m.brand_of_items_id
GROUP BY b.id, 
         b.symbol;

请注意:

  • 我使用LEFT JOIN,因此您在结果集中获得了不匹配的行,即那些没有MainIndexQuantity表条目的品牌商品。如果您不想包含它们,请改用INNER JOIN
  • 1.0相乘得到小数位数,如@JW所述。

更新1

要包含Total VolumeRetail Volume SumWholesale Volume sum,请尝试以下操作:

SELECT 
  b.id,
  b.symbol,
  IFNULL(SUM(m.item_type = 1), 0) * 1.0 / COUNT(*)     AS wholesaleRatio,
  IFNULL(SUM(m.item_type = 0), 0) * 1.0 / COUNT(*)     AS RetailRatio,
  IFNULL(SUM(m.vol), 0)                                AS 'Total Volume',
  SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS 'Retail Volume sum',
  SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS 'Wholesale Volume sum'
FROM brand_of_items b
LEFT JOIN mainIndexQuantity m ON b.id = m.brand_of_items_id
GROUP BY b.id, 
         b.symbol;

Updated SQL Fiddle Demo

这会给你:

| ID | SYMBOL | WHOLESALERATIO | RETAILRATIO | TOTAL VOLUME | RETAIL VOLUME SUM | WHOLESALE VOLUME SUM |
--------------------------------------------------------------------------------------------------------
|  0 |      a |              0 |           0 |            0 |                 0 |                    0 |
|  1 |      b |            0.5 |         0.5 |           90 |                45 |                   45 |
|  2 |      c |              0 |           0 |            0 |                 0 |                    0 |
| 10 |      j |              0 |           0 |            0 |                 0 |                    0 |

如果要按这些总和和总和对结果集进行排序,请将此查询放在子查询中,然后执行此操作:

SELECT *
FROM
(
   SELECT 
      b.id,
      b.symbol,
      IFNULL(SUM(m.item_type = 1), 0) * 1.0 / COUNT(*)     AS wholesaleRatio,
      IFNULL(SUM(m.item_type = 0), 0) * 1.0 / COUNT(*)     AS RetailRatio,
      IFNULL(SUM(m.vol), 0)                                AS TotalVolume,
      SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS RetailVolumeSum,
      SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS WholesaleVolumeSum
    FROM brand_of_items b
    LEFT JOIN mainIndexQuantity m ON b.id = m.brand_of_items_id
    GROUP BY b.id, 
             b.symbol
) AS sub
ORDER BY RetailVolumeSum    DESC, 
         WholesaleVolumeSum DESC;

但是你的最后一个要求并不清楚,你是否正在寻找那些具有最高retio / wholesale ratis和volumns或者选择最高值的物品?

对于后者:

SELECT *
FROM
(
   SELECT 
      b.id,
      b.symbol,
      IFNULL(SUM(m.item_type = 1), 0) * 1.0 / COUNT(*)     AS wholesaleRatio,
      IFNULL(SUM(m.item_type = 0), 0) * 1.0 / COUNT(*)     AS RetailRatio,
      IFNULL(SUM(m.vol), 0)                                AS TotalVolume,
      SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS RetailVolumeSum,
      SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS WholesaleVolumeSum
    FROM brand_of_items b
    LEFT JOIN mainIndexQuantity m ON b.id = m.brand_of_items_id
    GROUP BY b.id, 
             b.symbol
) AS sub
ORDER BY RetailVolumeSum    DESC, 
         WholesaleVolumeSum DESC,
         TotalVolume        DESC
LIMIT 1;

更新2

要获得总体积最高的品牌,您可以这样做:

SELECT 
  b.id,
  b.symbol,
  IFNULL(SUM(m.item_type = 1), 0) * 1.0 / COUNT(*)     AS wholesaleRatio,
  IFNULL(SUM(m.item_type = 0), 0) * 1.0 / COUNT(*)     AS RetailRatio,
  IFNULL(SUM(m.vol), 0)                                AS TotalVolume,
  SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS RetailVolumeSum,
  SUM(CASE WHEN m.item_type = 1 THEN m.vol ELSE 0 END) AS WholesaleVolumeSum
FROM brand_of_items b
LEFT JOIN mainIndexQuantity m ON b.id = m.brand_of_items_id
GROUP BY b.id, 
         b.symbol
HAVING SUM(m.vol) = (SELECT MAX(TotalVolume)
                     FROM
                     (
                       SELECT brand_of_items_id, SUM(vol) AS TotalVolume
                       FROM mainIndexQuantity
                       GROUP BY brand_of_items_id
                     ) t);

Like this

请注意:

  • 这将为您提供总体积最高的品牌,如果您正在寻找具有最高比率的品牌,您必须更换having子句以获得比率的最大值而不是最大值总量。

  • 这将为您提供总体积最高的项目,因此如果有多个项目的总体积最高,您可能会有超过项目的内容,例如this updated fiddle demo。在这种情况下,要只获得一个,您必须使用LIMIT才能返回一个。