查询单个表中字段值的计数答案

【问题标题】：Querying Counts on Field Values in a Single Table查询单个表中字段值的计数
【发布时间】：2012-04-12 04:23:27
【问题描述】：

我有一个包含多个字段的产品表，其中包含用户对各种属性的评价：

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来汇总所有产品的属性评估（不为空）：

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

这条 SQL 让我走到了这一步：

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

但它不会将 attr_1, attr_2.. 按列中包含错误计数的行分组（如上面所需的结果集）。我正在使用 Postgres，但欢迎任何形式的 SQL 提供帮助。

【问题讨论】：

标签： sql postgresql group-by sum case

【解决方案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

这可能不是最优雅/最有效的解决方案，但它应该得到你想要的

【讨论】：

我正要发布那个。我要做的唯一更改是将UNION 替换为UNION ALL，因为静态'attr_1','attr_2','attr_3' 确保记录不同
在 Postgres 中完美运行，包括 @Lamak 的 UNION All。谢谢！

【解决方案2】：

这有点蛮力，我讨厌它扫描表 3 次这一事实，但这似乎确实获得了所需的输出。很抱歉我不知道 PostGres，但这应该可以在 Oracle 中使用：

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

【讨论】：