题1

https://www.nowcoder.com/practice/2ed07ff8f67a474d90523b88402e401b?tpId=240&tags=&title=&difficulty=0&judgeStatus=0&rp=0

select uid,nick_name,achievement
FROM    user_info left join exam_record using(uid) # user_info作主表left join practice_record as p_r using(uid)
where achievement between 1200 and 2500and nick_name like "牛客%号"and (start_time like "_____09%" or month(p_r.submit_time)=9 )
group by uid

select uid,nick_name,achievement
from user_info
where nick_name like '牛客%号'
and achievement between 1200 and 2500
and (uid in(select uidfrom exam_recordwhere date_format(submit_time, '%Y%m') = '202109')or uid in(select uidfrom practice_recordwhere  date_format(submit_time, '%Y%m') = '202109')
)

题2

https://www.nowcoder.com/practice/1c5075503ccf4de1882976b2fff2c072?tpId=240&tags=&title=&difficulty=0&judgeStatus=0&rp=0

select uid,exam_id,round(avg(score), 0) as avg_score
from user_info left join exam_record using(uid)left join examination_info  using(exam_id)
where (nick_name like "牛客%号" or nick_name rlike "^[0-9]+$" )and tag rlike "^(c|C).*"and score is not null
group BY    uid,exam_id
order by uid,avg_score

select uid, exam_id, round(avg(score), 0) as avg_score
from exam_record
where uid in (select uid from user_info# 正则表达式where nick_name rlike '^牛客[0-9]+号$'or nick_name rlike '^[0-9]+$')and exam_id in (select exam_idfrom examination_info# 通配符where tag like 'C%'or tag like 'c%')and score IS NOT NULL
group by uid, exam_id
order by uid, avg_score;

select uid, exam_id, round(avg(score), 0) as avg_score
from exam_record
group by uid, exam_id
having uid in (select uid from user_info# 正则表达式where nick_name rlike '^牛客[0-9]+号$'or nick_name rlike '^[0-9]+$')and exam_id in (select exam_idfrom examination_info# 通配符where tag like 'C%'or tag like 'c%')and avg_score IS NOT NULL #  注意这里若是score is not null，则报"Unknown column 'score' in 'having clause'"
order by uid, avg_score;

题3

https://www.nowcoder.com/practice/f72d3fc27dc14f3aae76ee9823ccca6b?tpId=240&tags=&title=&difficulty=0&judgeStatus=0&rp=0

with…as创建临时表方式

WITH t_tag_count as (SELECT uid, `level`,COUNT(start_time) - COUNT(submit_time) as incomplete_cnt, -- 未完成数ROUND(IFNULL(1 - COUNT(submit_time) / COUNT(start_time), 0), 3) as incomplete_rate, -- 此人未完成率COUNT(start_time) as total_cnt -- 总作答数
#     FROM exam_record RIGHT JOIN user_info USING(uid) FROM user_info LEFT JOIN exam_record USING(uid) #user_info 作主表GROUP BY uid
)SELECT uid, incomplete_cnt, incomplete_rate
FROM t_tag_count
WHERE EXISTS ( SELECT uid FROM t_tag_count WHERE `level` = 0 AND incomplete_cnt > 2 ) AND `level` = 0 UNION ALLSELECT uid, incomplete_cnt, incomplete_rate
FROM t_tag_count
WHERE NOT EXISTS (SELECT uid FROM t_tag_count WHERE `level` = 0 AND incomplete_cnt > 2 ) AND total_cnt > 0ORDER BY incomplete_rate;

子表方式：

SELECT uid, incomplete_cnt, incomplete_rate
FROM (select ui.uid uid, level,     ###一定要是ui.uid 才能和level一一对应count(start_time)-count(submit_time)  incomplete_cnt,round(ifnull((count(start_time)-count(submit_time))/count(start_time),0),3)  incomplete_rate,COUNT(start_time) as total_cntfrom exam_record erright join user_info ui on er.uid=ui.uidgroup by uid) as a
WHERE EXISTS (           #判断为 有任意一个0级用户未完成试卷数大于2SELECT uid FROM (select ui.uid uid, level,     ###一定要是ui.uid 才能和level一一对应count(start_time)-count(submit_time)  incomplete_cnt,round(ifnull((count(start_time)-count(submit_time))/count(start_time),0),3)  incomplete_rate,COUNT(start_time) as total_cntfrom exam_record erright join user_info ui on er.uid=ui.uidgroup by uid) as aWHERE `level` = 0 AND incomplete_cnt > 2 )
AND level = 0    #输出0级用户的未完成数和未完成率union all        ##两种情况用union all 连接SELECT uid, incomplete_cnt, incomplete_rate
FROM (select ui.uid uid, level,     ###一定要是ui.uid 才能和level一一对应count(start_time)-count(submit_time)  incomplete_cnt,round(ifnull((count(start_time)-count(submit_time))/count(start_time),0),3)  incomplete_rate,COUNT(start_time) as total_cntfrom exam_record erright join user_info ui on er.uid=ui.uidgroup by uid) a
WHERE not EXISTS (       #判断为 没有任意一个0级用户未完成试卷数大于2SELECT uid FROM (select ui.uid uid, level,     ###一定要是ui.uid 才能和level一一对应count(start_time)-count(submit_time)  incomplete_cnt,round(ifnull((count(start_time)-count(submit_time))/count(start_time),0),3)  incomplete_rate,COUNT(start_time) as total_cntfrom exam_record erright join user_info ui on er.uid=ui.uidgroup by uid) aWHERE `level` = 0 AND incomplete_cnt > 2 )
AND total_cnt >0                      #筛选有作答记录的用户 即总作答数大于0即可
order by incomplete_rate asc

其它：

WITH target_user AS (SELECT user_info.uid, COUNT(1) AS incomplete_cntFROM exam_record LEFT JOIN user_info ON exam_record.uid = user_info.uidWHERE user_info.level = 0 AND submit_time IS NULLGROUP BY user_info.uidHAVING incomplete_cnt > 2
), target_user_exist AS (
SELECT COUNT(1) AS `exist` FROM target_user)
, total_summary AS (SELECTuser_info.uid,MAX(user_info.level) AS level,SUM(IF(submit_time IS NULL AND start_time IS NOT NULL, 1, 0)) AS incomplete_cnt,SUM(IF(submit_time IS NULL AND start_time IS NOT NULL, 1, 0)) / COUNT(1) AS incomplete_rate,SUM(IF(start_time IS NOT NULL, 1, 0)) AS has_submitFROM user_info LEFT JOIN exam_recordON user_info.uid = exam_record.uidGROUP BY user_info.uid
)SELECTuid,incomplete_cnt,ROUND(incomplete_rate, 3)
FROM total_summary LEFT JOIN target_user_exist ON 1=1
WHERE (exist=0 AND has_submit>0) OR (exist=1 AND level=0)
ORDER BY incomplete_rate ASC

题4

https://www.nowcoder.com/practice/ebff819fd38c46db8a42dfe43ca7b33a?tpId=240&tags=&title=&difficulty=0&judgeStatus=0&rp=0
本题关于case…when…then…[else]…end有多种写法：

(case when score<60 then '差'when score<75 then '中'when score<90 then '良'else '优' end) as score_grade(case when score >= 90 then '优'when score >= 75 then '良'when score >= 60 then '中'else '差' end) as score_grade(case when score>=90 then "优"when score between 75 and 89 then "良"when score between 60 and 74 then "中"else "差" end) as score_grade      (case when score>=90 then '优'when score>=75 and score<90 then '良'when score>=60 and score<75 then '中'else '差' end) as score_grade

法1：

selectlevel, score_grade, round(count(uid) / total, 3) as ratio
from (select u_i.uid,exam_id, score, level,(case when score >= 90 then '优'when score >= 75 then '良'when score >= 60 then '中'else '差' end) as score_grade,count(*) over(partition by level) as totalfrom user_info u_i join exam_record e_r on u_i.uid = e_r.uidwhere score is not null ) as  user_grade_table
group by level, score_grade
order by level desc, ratio desc

法2：

with t1 as(select level,case when score<60 then '差'when score<75 then '中'when score<90 then '良'else '优' end as score_gradefrom exam_record join user_info on exam_record.uid=user_info.uidwhere score is not null)select t1.level, score_grade, round(count(score_grade) / ct,3) as cnt
from t1 join (select level,count(level) ctfrom t1group by level) as t2
on t1.level=t2.level
group by t1.level,score_grade
order by t1.level desc,cnt desc;

上面方法的改写：

with t1 as(select level,case when score<60 then '差'when score<75 then '中'when score<90 then '良'else '优' end as score_gradefrom exam_record join user_info on exam_record.uid=user_info.uidwhere score is not null),
t2 as (select level,count(level) as ctfrom t1group by level)select t1.level, score_grade, round(count(score_grade) / ct,3) as cnt
from t1 join t2
on t1.level=t2.level
group by t1.level,score_grade
order by t1.level desc,cnt desc;

法3：

with t as (select u_i.uid,exam_id,score,level,case when score>=90 then '优' when score>=75 and score<90 then '良'when score>=60 and score<75 then '中'else '差' end as score_grade,count(*) over (partition by level) as totalfrom user_info u_i join exam_record  using(uid)where score is not NULL)select level,score_grade,round(count(*) / total,3) as ratio
from t
group by level, score_grade
order by level desc, ratio desc

sql常用操作之高级条件语句相关推荐

MYSQL 中 SQL 常用操作
SQL常用操作大全 1.SQL简单的操作 -- 常见SQL-- 查询 SELECT id,username,password FROM admin;-- 插入 INSERT INTO admin(us ...
MySQL中级优化教程（一）——SQL常用优化工具及explain语句的使用
序言: 说来惭愧,java学了两年,期间虽在博客上记了一些东西,可也不曾写过什么系统的教程,前一段时间开始学习MySQL数据库优化相关的知识,就想着趁着这个机会好好整理一份电子档出来,即方便自己之后回 ...
Python 炫技操作(01)：条件语句的七种写法
首发于微信公众号:Python编程时光系列导读 Python 炫技操作:条件语句的七种写法 Python 炫技操作:合并字典的七种方法 Python 炫技操作:连接列表的八种方法有的人说 Pyth ...
mysql删除所有男生_My SQL常用操作汇总
写这篇随笔的目的是我发现了在上一篇关于My SQL的随笔中存在一些不严谨的代码问题,在这里再次简单的总结一下并加以改进,以代码为主. # !每行命令必须以分号(;)结尾先通过命令行进入数据库客户端 ...
sql常用操作（含指定位置添加字段、修改到指定位置后等）
1) 创建用户表(user) 要求:字段类型长度uid intuname varchar 20password varchar 20birthday date create table user( ...
[FineReport]高级条件分组、斜线、自动查询、控件编辑属性、条件属性
通常我们在做按某一条件分组汇总数据的时候,都是透过SQL去处理.如果您不熟悉SQL怎么办?FR高级条件分组可以帮您实现. 一.新建数据集 select A0188 AS ID, CAST(FLOOR( ...
MySQL基础学习（三）————SQL语句的常用操作
文章目录 1.库 1.1库的创建 1.2 库的删除 1.3 库的修改 1.4 库的查找 2.表 2.1 表的创建 2.2 表的删除 2.3 表的修改 2.4 表的查找 3.数据或者记录 3.1 数据的 ...
【数据库】第一章数据库的分类、SQL、数据库、表和表记录的常用操作
[数据库]第一章数据库的分类.SQL.数据库与表的常用操作文章目录 [数据库]第一章数据库的分类.SQL.数据库与表的常用操作一.数据库的分类 1.关系型数据库 2.非关系型数据库 3.MyS ...
数据库常用操作语句总结
数据库常用操作语句总结一.基础 1.select 语句 2.select distinct 语句 3.where 子句 4.and 和 or 运算符 5.order by 语句 6.insert i ...

sql常用操作之高级条件语句

文章目录

题1

题2

题3

with…as创建临时表方式

子表方式：

题4

sql常用操作之高级条件语句相关推荐

最新文章

热门文章