聚合函数调用不能嵌套？

Question

在PostgreSQL 数据库中，我有一个名为answers 的表。此表存储有关用户如何回答问题的信息。表中只有 4 个问题。同时，回答问题的用户数量可以是动态的，用户只能回答部分问题。

表answers:

| EMPLOYEE | QUESTION_ID | QUESTION_TEXT          | OPTION_ID | OPTION_TEXT  |
|----------|-------------|------------------------|-----------|--------------|
| Bob      | 1           | Do you like soup?      | 1         | Yes          |
| Alex     | 1           | Do you like soup?      | 2         | No           |
| Kate     | 1           | Do you like soup?      | 3         | I don't know |
| Bob      | 2           | Do you like ice cream? | 1         | Yes          |
| Alex     | 2           | Do you like ice cream? | 3         | I don't know |
| Oliver   | 2           | Do you like ice cream? | 1         | Yes          |
| Bob      | 3           | Do you like summer?    | 2         | No           |
| Alex     | 3           | Do you like summer?    | 1         | Yes          | 
| Jack     | 3           | Do you like summer?    | 2         | No           |
| Bob      | 4           | Do you like winter?    | 3         | I don't know |
| Alex     | 4           | Do you like winter?    | 1         | Yes          |
| Oliver   | 4           | Do you like winter?    | 3         | I don't know |

例如，使用下一个代码，我可以找到回答这些问题的每个人的问题 1 和 2 的平均答案。

select
    employee,
    avg(
        case when question_id in (1, 2) then option_id else null end
    ) as average_score
from
    answers
group by
    employee

结果：

| EMPLOYEE | AVERAGE_SCORE |
|----------|---------------|
| Bob      | 2             |
| Alex     | 2,5           |
| Kate     | 3             |
| Oliver   | 1             |

现在，我想知道问题 1 和 2 的答案平均值 >= 大于 2 的用户数量。我尝试了下一个代码，但它引发了错误：

select
    count(
        avg(
            case when question_id in (1, 2) then option_id else null end
        )
    ) as average_score
from
    answers
where
    average_score >= 2
group by
    answers.employee

错误：

SQL Error [42803]: ERROR: aggregate function calls cannot be nested

Answer 1

您需要在聚合之后进行过滤。这使用了having 子句。在 Postgres 中，也可以使用filter：

select employee,
       avg(option_id) filter (where question_id in (1, 2)) as average_score
from answers
group by employee
having avg(option_id) filter (where question_id in (1, 2)) > 2;

如果您想要计数，请将其用作子查询：select count(*) from <the above query>。

奇怪的是，您将“option_id”等同于“score”，但这就是您的问题的措辞。

Answer 2

你必须使用having子句..它可以简单地完成

select employee, [Average Score] = avg(case when question_id in (1, 2) 
                                            then option_id else null 
                                            end
                                       ) 
from answers group by employee having average_score > 2;

更新： 它现在必须工作......

select employee, average_score = avg(case when question_id in (1, 2) 
                                            then option_id else null 
                                            end
                                       ) 
from answers group by employee having average_score > 2;