SQL进阶day12——空值处理

1 统计有未完成状态的试卷的未完成数和未完成率

（复习if的用法）

我的思路：

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select exam_idm
count(if submit_time is NULL then 1 else 0 end) incomplete_cnt,
count(if submit_time is NULL then 1 else 0 end)/count(submit_time) complete_rate
from exam_record
group by id

报错好像是因为(if submit_time is NULL then 1 else 0 end)不对。

查询改正：

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select exam_id,
sum(if (submit_time is NULL,1,0)) incomplete_cnt,
count(if (submit_time is NULL,1,0))/count(submit_time) complete_rate
from exam_record
group by id

正确答案：

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select exam_id,
sum(if (submit_time is NULL,1,0)) incomplete_cnt,
round(avg(if (submit_time is NULL,1,0)),3) complete_rate
from exam_record
group by exam_id
having complete_rate != 0

复盘:

（1）【条件判断函数】------根据满足不同条件，执行相应流程（用法不要搞混啦！）
● if(expr,v1,v2)

如果表达式expr是true返回值v1，否则返回v2

例如：if(1<2,'Y','N')返回Y，if(1>2,'Y','N')返回N if后面要加括号！！！！
● case when

case expr when v1 then r1 $when v2 then r2$ ... $else rn$ end

例如：case 2 when 1 then 'one' when 2 then 'two' else 'more' end 返回two

case后面的值为2，与第二条分支语句when后面的值相等相等，因此返回two

（2）是按照exam_id进行分组，统计有未完成状态的试卷的，是基于试卷来的。

2 0级用户高难度试卷的平均用时和平均得分

我的代码：

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#2、再筛选出试卷------difficulty = hard
select uid,
round(avg((if(score is NULL,0,score))),0) avg_score,
avg(if(submit_time is NULL,ei.duration,timestampdiff(minute,submit_time,start_time))) avg_time_took
from exam_record er join examination_info ei
on er.exam_id = ei.exam_id
where uid in  #1、先筛选出用户------level=0
(select uid from user_info
where level = 0)
and ei.difficulty = 'hard'
group by er.uid,er.exam_id
# 未完成的默认试卷最大考试时长
# if(submit_time is NULL,ei.duration,timestampdiff(minute,submit_time,start_time))

# 未完成的默认0分处理
# if(score is NULL,0,score)

答案错误原因：

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# 未完成的默认试卷最大考试时长
# if(submit_time is NULL,ei.duration,timestampdiff(minute,submit_time,start_time))

改正：

时间差函数：timestampdiff，如计算差多少分钟

timestampdiff(minute,时间1,时间2)，是时间2-时间1。

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# 未完成的默认试卷最大考试时长
# if(submit_time is NULL,ei.duration,timestampdiff(minute,start_time,submit_time))

题目是要求：请输出每个0级用户所有的(..)试卷,所以是直接按照用户分组就好，

不要group by er.uid,er.exam_id

正确答案：

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#2、再筛选出试卷------difficulty = hard
select uid,
round(avg((if(score is NULL,0,score))),0) avg_score,
round(avg(if(submit_time is NULL,ei.duration,timestampdiff(minute,start_time,submit_time))),1) avg_time_took
from exam_record er join examination_info ei
on er.exam_id = ei.exam_id
where uid in  #1、先筛选出用户------level=0
(select uid from user_info
where level = 0)
and ei.difficulty = 'hard'
group by er.uid
# 未完成的默认试卷最大考试时长
# if(submit_time is NULL,ei.duration,timestampdiff(minute,start_time,submit_time))

# 未完成的默认0分处理
# if(score is NULL,0,score)