Question

我有一个包含多个字段的产品表，其中包含各种属性的用户评估：

product  | attr_1_eval   | attr_2_eval   | attr_3_eval
ABC      | Correct       | Incorrect     | Null
DEF      | Incorrect     | Null          | Null
XYZ      | Undetermined  | Null          | Incorrect
123      | Null          | Undetermined  | Correct
456      | Incorrect     | Correct       | Correct

我需要编写一个查询，在所有产品中总计这些属性评估（其中not null）：

evaluation  | correct   | incorrect   | undetermined
attr_1      | 1         | 2           | 1
attr_2      | 1         | 1           | 1
attr_3      | 2         | 1           | 0

这个SQL让我分道扬：

SELECT 
SUM(CASE WHEN attr_1_eval = 'Correct' then 1 else 0 END) AS correct,
SUM(CASE WHEN attr_1_eval = 'Incorrect' then 1 else 0 END) AS incorrect,
SUM(CASE WHEN attr_1_eval = 'Undetermined' then 1 else 0 END) AS undetermined,
SUM(CASE WHEN attr_2_eval = 'Correct' then 1 else 0 END) AS correct,
...
FROM product

但它没有按列中的错误计数行对attr_1, attr_2..进行分组（如上面所需的结果集）。我正在使用Postgres，但欢迎任何SQL的帮助。

Answer 1

你能做3个工会吗？

SELECT 
  'attr_1' AS evaluation,
  SUM(CASE WHEN attr_1_eval = 'Correct' then 1 else 0 END) AS correct,
  SUM(CASE WHEN attr_1_eval = 'Incorrect' then 1 else 0 END) AS incorrect,
  SUM(CASE WHEN attr_1_eval = 'Undetermined' then 1 else 0 END) AS undetermined
FROM product
UNION
SELECT 
  'attr_2' AS evaluation,
  SUM(CASE WHEN attr_2_eval = 'Correct' then 1 else 0 END) AS correct,
  SUM(CASE WHEN attr_2_eval = 'Incorrect' then 1 else 0 END) AS incorrect,
  SUM(CASE WHEN attr_2_eval = 'Undetermined' then 1 else 0 END) AS undetermined
FROM product
UNION
SELECT 
  'attr_3' AS evaluation,
  SUM(CASE WHEN attr_3_eval = 'Correct' then 1 else 0 END) AS correct,
  SUM(CASE WHEN attr_3_eval = 'Incorrect' then 1 else 0 END) AS incorrect,
  SUM(CASE WHEN attr_3_eval = 'Undetermined' then 1 else 0 END) AS undetermined
FROM product

它可能不是最优雅/最有效的解决方案，但它应该得到你想要的

Answer 2

这是一种蛮力，我讨厌它扫描桌子三次，但这看起来似乎得到了所需的输出。对不起，我不知道PostGres，但这应该适用于Oracle：

select
  Attribute_name,
  Sum (correct) as Correct,
  sum (incorrect) as Incorrect,
  sum (undetermined) as Undetermined
from
  (
  select
    'attr_1' as Attribute_Name,
    decode (attr_1_eval, 'Correct', 1, 0) as correct,
    decode (attr_1_eval, 'Incorrect', 1, 0) as incorrect,
    decode (attr_1_eval, 'Undetermined', 1, 0) as undetermined
  from product
  union all
  select
    'attr_2',
    decode (attr_2_eval, 'Correct', 1, 0),
    decode (attr_2_eval, 'Incorrect', 1, 0),
    decode (attr_2_eval, 'Undetermined', 1, 0)
  from product
  union all
  select
    'attr_3',
    decode (attr_3_eval, 'Correct', 1, 0),
    decode (attr_3_eval, 'Incorrect', 1, 0),
    decode (attr_3_eval, 'Undetermined', 1, 0)
  from product
)
group by Attribute_Name

在单个表中查询字段值的计数

2 个答案: