MySQL COUNT意外结果

Question

我的数据库结构如下所示:

https://docs.google.com/open?id=0B9ExyO6ktYcOenZ1WlBwdlY2R3c

某些表的说明:

1. answer_chk_results - 检查答案表.因此,如果此表中不存在某些答案,则表示未经检查
2. lesson_questions - 课程< - >问题关联(通过id)表

当我像这样查询数据库:

SELECT
    q.id,
    q.content,
    q.type
FROM
    `questions_and_exercises` q,
    `lesson_questions` lq
WHERE
    q.id = lq.qid
AND lq.lid = 1

我收到了所有问题清单.但我想要的是每个问题的答案和检查答案.当我使用此查询时:

SELECT
    q.id,
    q.content,
    q.type,
    COUNT(DISTINCT a.ID) answer_count,
    COUNT(DISTINCT acr.id) checked_count
FROM
    `questions_and_exercises` q,
    `lesson_questions` lq
LEFT JOIN answers a ON a.qid = lq.qid,
LEFT JOIN `answer_chk_results` acr ON acr.aid = a.id
WHERE
    q.id = lq.qid
AND lq.lid = 1

结果只有一个问题,错误的答案数.(我的数据库中有大约9-10个问题)我错过了什么？

Answer 1

A group by会有所帮助,例如:

select  q.id
,       q.content
,       q.type
,       count(distinct a.id) as answer_count
,       count(acr.checked) as checked_count
from    questions_and_exercises q
join    lesson_questions lq
on      q.id = lq.id
left join
        answers a 
on      a.qid = lq.qid
left join
        answer_chk_results acr 
on      acr.aid = a.id
group by
        q.id
,       q.content
,       q.type

请注意,count(acr.checked)返回answer_chk_results表中有一行的答案数.并count(distinct acr.checked)返回已检查列的不同值的数量.如果checked是布尔值,则始终为1或0.