Question

我正在尝试在下面的查询中显示四列：总回复计数，错误回复计数以及基于前两列的%错误。结果将按question_id分组。

SELECT 
    COUNT(correct) as total_resp,
    SUM(case when correct = 'f' then 1 else 0 end) as incor_resp,
    (incor_resp / total_resp) as percent_incor,
    question_id
FROM answers
WHERE student_id IN (
    SELECT id FROM students
    WHERE lang = 'es'
    LIMIT 50
)
GROUP BY question_id;

我的问题是，为什么上述percent_incor定义不起作用？我是否无法访问total_resp和incor_resp以便能够执行第3个字段定义的操作？如果没有，我怎么能在输出中包含这个字段？

谢谢！

Answer 1

不可能在同一个选择中通过别名引用其他字段。您需要再次重复表达式，或者您可以将select包装在另一个select中并在那里计算：

SELECT total_resp, incor_resp, incor_resp/total_resp as percent_incor
FROM (
    SELECT 
        COUNT(correct) as total_resp,
        SUM(case when correct = 'f' then 1 else 0 end) as incor_resp,
        (incor_resp / total_resp) as percent_incor,
        question_id
    FROM answers
    WHERE student_id IN (
        SELECT id FROM students
        WHERE lang = 'es'
        LIMIT 50
    )
    GROUP BY question_id
) t;

Answer 2

你不能以SELECT COUNT(correct) as total_resp, SUM(case when correct = 'f' then 1 else 0 end) as incor_resp, (SUM(case when correct = 'f' then 1 else 0 end) / COUNT(correct)) as percent_incor, question_id FROM answers WHERE student_id IN ( SELECT id FROM students WHERE lang = 'es' LIMIT 50 ) GROUP BY question_id;中的方式引用字段的别名。

试试这个：

=SUM(C4:N4)

Answer 3

此代码是您正在寻找的吗？

SELECT 
COUNT(correct) as total_resp,
SUM(case when correct = 'f' then 1 else 0 end) as incor_resp,
(SUM(case when correct = 'f' then 1 else 0 end) / COUNT(correct)) as percent_incor,
question_id
FROM answers
WHERE student_id IN (
SELECT id FROM students
WHERE lang = 'es'
LIMIT 50
GROUP BY question_id;

Answer 4

如果要使用别名，则必须访问子查询

SELECT total_resp, incor_resp, total_resp / incor_resp as percent_incor
FROM (
        SELECT 
                COUNT(correct) as total_resp,
                SUM(case when correct = 'f' then 1 else 0 end) as incor_resp,
                question_id
            FROM answers
            WHERE student_id IN (
                SELECT id FROM students
                WHERE lang = 'es'
                LIMIT 50
            )
            GROUP BY question_id
        ) T

Answer 5

Coudl是案件sintax的问题

SELECT 
   COUNT(correct) as total_resp,
   SUM(case correct when 'f' then 1 else 0 end) as incor_resp,
   (incor_resp / total_resp) as percent_incor,
    question_id
FROM answers
WHERE student_id IN (
   SELECT id FROM students
   WHERE lang = 'es'
   LIMIT 50
)
GROUP BY question_id;

在SQL中的2个聚合字段上执行操作

5 个答案: