请查看下表(称为响应)。它显示了受访者对问答的回应。
questionid answerid respondentid
1 10 1
1 11 2
1 11 4
1 12 3
1 12 5
2 20 1
2 20 2
2 21 2
2 22 1
2 22 4
2 23 1
2 23 3
2 24 4
3 30 2
3 30 3
3 30 4
3 31 1
我们可以运行以下SQL:
select questionid, answerid, count(respondentid) as noOfRespondentsToQuestionAndAnswer
from response
group by questionid, answerid
...这将告诉我们有多少受访者回答了问题+答案的每个组合。
我们也可以这样做:
select questionid, count(distinct respondentid) as noOfRespondentsToQuestion
from response
group by questionid
...这将告诉我们有多少不同的受访者回答了每个问题。
我想将两个选项合并为一个,并且让不同的答复者的数量在每个问题的多行上表示(这是必要的,因为它仅基于问题而不是答案)。 / p>
所以,我想得到如下结果:
questionid,answerid,noOfRespondentsToQuestionAndAnswer,noOfRespondentsToQuestion
1 10 1 5
1 11 2 5
1 12 2 5
2 20 2 4
2 21 1 4
2 22 2 4
2 23 2 4
2 24 1 4
3 30 3 4
3 31 1 4
只用一个查询就可以实现这个目的吗?
答案 0 :(得分:5)
select one.questionid, answerid,
noOfRespondentsToQuestionAndAnswer,
noOfRespondentsToQuestion
FROM (
select questionid, answerid,
count(respondentid) as noOfRespondentsToQuestionAndAnswer
from response
group by questionid, answerid) AS one
JOIN (
select questionid, count(distinct respondentid) as noOfRespondentsToQuestion
from response
group by questionid) AS two
WHERE one.questionid = two.questionid;
答案 1 :(得分:0)
你没有指定哪种类型的数据库,这会简化这一点,但是从纯粹的sql想法,不使用任何分析,它可以完成,但你会失去效率。
select questionid, answerid,
(select a.count(*) FROM datatable a WHERE a.questionid=questionid AND a.answerid=answerid),
(select b.count(*) FROM datatable b WHERE b.questionid=questionid)
FROM datatable ORDER BY questionid, answerid;