如何正确使用聚合条件组

Question

我希望能够计算每个大陆的苹果总类型（仅限有机），按国家/地区分类；如果它们混合在一起，包括总数。

例如，食品 B1 是来自美国的有机金苹果。因此应该有一个计数“1”golden_bag 和“1”有机。现在，A1 也是来自阿根廷的有机产品 - 但是，它既有美味的奶奶苹果，也有美味的红苹果 - 因此它被计为“1”mixed_bag 和“1”用于 granny_bag 和“1”用于 red_bag。

最后，E1和F1都是老挝的富士苹果，一个是有机的，一个不是；所以总计数是 2 fuji_bag，它的总计数应该是 Organic_fd。

Table X:
food_item | food_area | food_loc   | food_exp
A1          lxgs        argentina   1/1/20
B1          iyan        usa         5/31/21
C1          lxgs        peru        4/1/20
D1          wa8e        norway      10/1/19
E1          894a        laos        5/1/19
F1          894a        laos        9/17/19


Table Y:
food_item | organic
A1          Y
B1          Y
C1          N
D1          N
E1          Y
F1          N

Table Z:
food_item | food_type
A1          189
A1          190
B1          191
C1          189
D1          192
E1          193
F1          193

SELECT continent, country,
      SUM(organic)  AS organic_fd, SUM(Granny) AS granny_bag,
      SUM(Red_delc) AS red_bag,    SUM(Golden) AS golden_bag,
      SUM(Gala)     AS gala_bag,   SUM(Fuji)   AS fuji_bag,
      SUM(CASE WHEN Granny + Red_delc + Golden + Gala + Fuji > 1 THEN 1  ELSE 0 END) AS mixed_bag     
FROM (SELECT (CASE SUBSTR (x.food_area, 4, 1)
              WHEN 's' THEN 'SA' WHEN 'n' THEN 'NA'
              WHEN 'e' THEN 'EU' WHEN 'a' THEN 'AS' ELSE NULL END) continent,
          x.food_loc country, COUNT(y.organic) AS Organic
          COUNT(CASE WHEN z.food_type = '189' THEN 1 END) AS Granny,
          COUNT(CASE WHEN z.food_type = '190' THEN 1 END) AS Red_delc,
          COUNT(CASE WHEN z.food_type = '191' THEN 1 END) AS Golden,
          COUNT(CASE WHEN z.food_type = '192' THEN 1 END) AS Gala,
          COUNT(CASE WHEN z.food_type = '193' THEN 1 END) AS Fuji      
    FROM x LEFT JOIN z ON x.food_item = z.food_item
           LEFT JOIN y on x.food_item = y.food_item and y.organic = 'Y'    
               WHERE  x.exp_date > sysdate
    GROUP BY SUBSTR (x.food_area, 4, 1), x.food_loc, y.organic) h
GROUP BY h.continent, h.country, h.organic

我没有得到正确的输出，例如，老挝将显示 TWICE 来说明有机计数和非有机计数。因此它将显示1 organic_fd 和0 organic_fd 和1 fuji_bag，另一行将是另一行1 fuji_bag。我想要总数。 （另外，如果我添加更多的食物，我的混合包会显示每条记录/行的计数大多为“1”）。

以下是所需的输出：

| continent | country   |organic_fd | granny_bag| red_bag| golden_bag| gala_bag|fuji_bag | mixed_bag
| SA        | argentina |    1      | 1         |   1    | 0         | 0       | 0       | 1
| SA        | peru      |    0      | 1         |   0    | 0         | 0       | 0       | 0
| NA        | usa       |    1      | 0         |   0    | 1         | 0       | 0       | 0
| EU        | norway    |    0      | 0         |   0    | 0         | 1       | 0       | 0
| AS        | laos      |    1      | 0         |   0    | 0         | 0       | 2       | 0

所以，假设我想添加另一种食品，来自挪威的 G1，它有 3 种有机苹果：fuji, red, granny... 那么挪威现在将有 1 的以下列计数：@987654329 @、organic_fd、fuji_bag、red_bag、granny_bag（除了之前的1 gala_bag计数）。如果添加与 G1 完全相同的 H1，那么它现在将具有以下总数 2：mixed_bag、organic_fd、fuji_bag、red_bag、granny_bag

Answer 1

查询：

WITH
  t AS (
    SELECT
      CASE SUBSTR(X.food_area, LENGTH(X.food_area), 1)
        WHEN 's' THEN 'SA'
        WHEN 'n' THEN 'NA'
        WHEN 'e' THEN 'EU'
        WHEN 'a' THEN 'AS'
        ELSE NULL
      END AS continent,
      x.food_loc AS country,
      COUNT(DISTINCT CASE Y.organic WHEN 'Y' THEN X.food_item END) OVER (
        PARTITION BY x.food_loc
      ) AS organic_fd,
      CASE
        WHEN MIN(Z.food_type) OVER (
               PARTITION BY x.food_loc, X.food_item
             ) = Z.food_type AND
             MAX(Z.food_type) OVER (
               PARTITION BY x.food_loc, X.food_item
             ) > Z.food_type THEN 1 END AS mixed,
      Z.food_type
    FROM X
    JOIN Y ON X.food_item = Y.food_item
    JOIN Z ON Y.food_item = Z.food_item
  )
SELECT
  continent, country, organic_fd,
  COUNT(CASE WHEN food_type = '189' THEN 1 END) AS Granny,
  COUNT(CASE WHEN food_type = '190' THEN 1 END) AS Red_delc,
  COUNT(CASE WHEN food_type = '191' THEN 1 END) AS Golden,
  COUNT(CASE WHEN food_type = '192' THEN 1 END) AS Gala,
  COUNT(CASE WHEN food_type = '193' THEN 1 END) AS Fuji,
  COUNT(mixed) AS mixed_bag
FROM t
GROUP BY continent, country, organic_fd

您可以在此处尝试此查询：https://rextester.com/TSSH87409。

Answer 2

x 和 z 之间存在一对多关系，并且连接可能会为 x 中的每一行生成许多行，就像 A1 的情况一样。所以你首先必须对x 中的行进行编号，这是我的子查询t1 所做的，除了映射值。然后将它们分组为每个计数列（奶奶、有机等）采用max()，就像在子查询t2 中一样。最后求和。

^{dbfiddle demo}

with
  t1 as (
    select rn, food_item, food_area, food_loc country, food_exp, food_type,
           decode(substr(food_area, 4, 1), 's', 'SA', 'n', 'NA', 'e', 'EU', 'a', 'AS') continent,
           case organic when 'Y' then 1 else 0 end org,
           case when food_type = '189' then 1 else 0 end gra,
           case when food_type = '190' then 1 else 0 end red,
           case when food_type = '191' then 1 else 0 end gol,
           case when food_type = '192' then 1 else 0 end gal,
           case when food_type = '193' then 1 else 0 end fuj 
      from (select rownum rn, x.* from x) x join y using (food_item) join z using (food_item)
      where food_exp > sysdate),
  t2 as (
    select rn, country, continent, max(org) org, max(gra) gra, 
           max(red) red, max(gol) gol, max(gal) gal, max(fuj) fuj,
           case when max(gra) + max(red) + max(gol) + max(gal) + max(fuj) > 1 
                then 1 else 0 
            end mix
       from t1 group by rn, country, continent)
select continent, country, sum(org) organic_fd, sum(gra) granny, sum(red) red_delc, 
       sum(gol) golden_bag, sum(gal) gala_bag, sum(fuj) fuji_bag, sum(mix) mixed_bag 
  from t2 
  group by continent, country

以上查询给出了预期的输出，请对其进行测试并根据需要进行调整。我注意到你使用左连接。如果X 中的某些行可能在Y 或Z 中没有数据，您可能需要在计算中添加nvl()s。也许您还应该将映射的硬编码值放入表中。硬编码它们不是好的做法。希望这会有所帮助:)