现在运营想要查看所有来自浙江大学的用户题目回答明细情况,请你取出相应数据。
select
qpd.device_id,
qpd.question_id,
qpd.result
from
question_practice_detail as qpd
inner join user_profile as up on up.device_id = qpd.device_id
and up.university = '浙江大学'
order by
question_id
select
device_id,
question_id,
result
from
question_practice_detail
where
device_id in (
select
device_id
from
user_profile
where
university = '浙江大学'
)
order by
question_id
运营想要了解每个学校答过题的用户平均答题数量情况,请你取出数据。
select
university,
count(question_id) / count(distinct qpd.device_id) as avg_answer_cnt
from
question_practice_detail as qpd
inner join user_profile as up on qpd.device_id = up.device_id
group by
university
运营想要计算一些参加了答题的不同学校、不同难度的用户平均答题量,请你写SQL取出相应数据。
//每个学校:按学校分组group by university
//不同难度:按难度分组group by difficult_level
//平均答题数:总答题数除以总人数count(qpd.question_id) / //count(distinct qpd.device_id)
//来自上面信息三个表,需要联表,up与qpd用device_id连接,
//qd与qpd用question_id连接。
//平均值精度:保留4位小数round(x, 4)
select
university,
difficult_level,
round(
count(qpd.question_id) / count(distinct qpd.device_id),
4
) as avg_answer_cnt
from
question_practice_detail as qpd
left join user_profile as up on up.device_id = qpd.device_id
left join question_detail as qd on qd.question_id = qpd.question_id
group by
university,
difficult_level
运营想要查看参加了答题的山东大学的用户在不同难度下的平均答题题目数,请取出相应数据。
SELECT
t1.university,
t3.difficult_level,
COUNT(t2.question_id) / COUNT(DISTINCT (t2.device_id)) as avg_answer_cnt
from
user_profile as t1,
question_practice_detail as t2,
question_detail as t3
WHERE
t1.university = '山东大学'
and t1.device_id = t2.device_id
and t2.question_id = t3.question_id
GROUP BY
t3.difficult_level;
现在运营想要分别查看学校为山东大学或者性别为男性的用户的device_id、gender、age和gpa数据,请取出相应结果,结果不去重。
select
device_id,
gender,
age,
gpa
from
user_profile
where
university = '山东大学'
union all
select
device_id,
gender,
age,
gpa
from
user_profile
where
gender = 'male'