牛客网mysql经典题目

特别注意：

mysql日期函数转化：MySQL DATE_FORMAT() 函数

不要使用DISTINCT关键字，要用group by替换。

MySQL_lead()函数_判断同一id同一列两行是否相等。lead()函数

mysql8新特性：

mysql之CTE通用表达式：地址

mysql之窗口函数：窗口函数详解

mysql查询易错题：

SQL23 对所有员工的薪水按照salary降序进行1-N的排名。

select emp_no,salary,
cast(@rownum := @rownum + (@pre <> (@pre := salary)) as signed) as t_rank
from salaries a,(select @rownum := 0) b,(select @pre := -1) c
order by salary desc

mysql关联关系图： 

SQL35 批量插入数据，不使用replace操作:

insert IGNORE into actor values(3,'ED','CHASE','2006-02-15 12:34:33');

SQL37 对first_name创建唯一索引uniq_idx_firstname:

对first_name创建唯一索引uniq_idx_firstname，
对last_name创建普通索引idx_lastname
create只能添加这两种索引;
创建普通索引  CREATE INDEX index_name ON table_name (column_list)
创建唯一索引  CREATE UNIQUE INDEX index_name ON table_name (column_list)
多个索引间用分号；连接

1. create (unique) index 索引名 on 表名（列名）

1 2	create unique index uniq_idx_firstname on actor(first_name); create index idx_lastname on actor(last_name);

2. alter table 表名 add (unique) index 索引名（列名）

1 2	alter table actor add unique index uniq_idx_firstname(first_name); alter table actor add index idx_lastname(last_name);

SQL38 针对actor表创建视图actor_name_view

create view actor_name_view (first_name_v,last_name_v) 
as select first_name,last_name from actor 

SQL39 针对上面的salaries表emp_no字段创建索引idx_emp_no，查询emp_no为10005，使用强制索引。

select * from salaries force index(idx_emp_no) where emp_no = 10005

SQL41 构造一个触发器audit_log

构造触发器时注意以下几点：

1、用 CREATE TRIGGER 语句构造触发器，用 BEFORE或AFTER 来指定在执行后面的SQL语句之前或之后来触发TRIGGER

2、触发器执行的内容写出 BEGIN与END 之间

3、可以使用 NEW与OLD 关键字访问触发后或触发前的employees_test表单记录

常用关键字解析：

• trigger_name：标识触发器名称，用户自行指定；
• trigger_time：标识触发时机，取值为 BEFORE 或 AFTER；
• trigger_event：标识触发事件，取值为 INSERT、UPDATE 或 DELETE；
• tbl_name：标识建立触发器的表名，即在哪张表上建立触发器；
• trigger_stmt：触发器程序体，可以是一句SQL语句，或者用 BEGIN 和 END 包含的多条语句，每条语句结束要分号结尾。

MySQL 中定义了 NEW 和 OLD，用来表示触发器的所在表中，触发了触发器的那一行数据。具体地：

1.在 INSERT 型触发器中，NEW 用来表示将要（BEFORE）或已经（AFTER）插入的新数据；

2.在 UPDATE 型触发器中，OLD 用来表示将要或已经被修改的原数据，NEW 用来表示将要或已经修改为的新数据；

3.在 DELETE 型触发器中，OLD 用来表示将要或已经被删除的原数据；使用方法： NEW.columnName （columnName 为相应数据表某一列名）

create trigger audit_log after insert on employees_test
for each row
begin 
    insert into audit values (NEW.ID,NEW.NAME);
end;

SQL42 删除emp_no重复的记录：

delete a from titles_test a 
join titles_test b 
on a.emp_no = b.emp_no 
where a.id > b.id

DELETE FROM titles_test 
WHERE id NOT IN( 
    SELECT a.id FROM (
        SELECT emp_no,count(*) total,min(id) id 
        FROM titles_test 
        GROUP BY emp_no 
    ) a
)

以下两种方法适用于mysql8.0之后版本，需用到窗口函数或者CTE语法：

select * from (
    select *,row_number() over(partition by emp_no order by id desc) as num 
    from titles_test
) a where a.num = 1

with dups as(
    select id,row_number() over(partition by emp_no order by emp_no) asrow_no
    from titles_test
)
DELETE FROM titles_test
WHERE id IN (
    SELECT id
    FROM dups
    WHERE dups.row_num > 1)

SQL44 将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005,其他数据保持不变，使用replace实现，直接使用update会报错。

REPLACE INTO titles_test VALUES(5,10005,...);

#另一种写法

#UPDATE titles_test SET emp_no = REPLACE(emp_no,10001,10005) WHERE id = 5;

SQL51 查找字符串 10,A,B 中逗号,出现的次数cnt :

select length("10,A,B") - length(replace("10,A,B",",",""))

SQL53 按照dept_no进行汇总(多行拼接成一行):

select dept_no,group_concat(DISTINCT emp_no order by emp_no separator ',')employees
from dept_emp group by dept_no

扩展（一行拆分成多行）：

tb_attendance_manager表：

temp表：

select a.cadreCategory,a.attendanceType, substring_index(substring_index(a.vacationType, ';', b.myId + 1), ';', -1) as vacationType from tb_attendance_manager a join tb_temp b on b.myId < (length(a.vacationType) - length(replace(a.vacationType, ';', '')) + 1) 

SQL60 统计salary的累计和running_total(重点 循环求和):

1.窗口函数
select emp_no,salary,sum(salary) over(order by emp_no) as running_total
from salaries
where to_date = '9999-01-01'
order by emp_no
2.子查询
select emp_no,salary,(
    select sum(b.salary) from salaries b 
    where b.emp_no <= a.emp_no
    and b.to_date = '9999-01-01'
    order by b.emp_no
)
from salaries a
where a.to_date = '9999-01-01'
GROUP BY a.emp_no
3.自连接1
select a.emp_no,a.salary,sum(b.salary) as running_total
from salaries a join salaries b on a.emp_no >= b.emp_no
where a.to_date = '9999-01-01'
and b.to_date = '9999-01-01'
group by a.emp_no
order by a.emp_No asc
4.自连接2
select a.emp_no,a.salary,sum(b.salary) as running_total
from salaries a ,salaries b
where a.to_date = '9999-01-01'
and b.tp_date = '9999-01-01'
and a.emp_no >= b.emp_no
group by a.emp_no
order by a.emp_no asc

SQL61 对于employees表中，给出奇数行的first_name(重点 计数排序求奇数行):

1.子查询，查找升序排列的奇数位名称，就可以理解为：比该名称还小的名字有偶数个

select a.first_name from employees a where (select count(*) from employees b where a.first_name >= b.first_name)%2 = 1

2.窗口函数：

select first_name,emp_no,rank() over(order by first_name) as ranking from employees

3.使用临时变量：

select first_name,@norank:=@norank+1 as ra from employees es,(select @norank:=0) t1 order by first_name

SQL63 刷题通过的题目排名（双条件排序）

select id,number,dense_rank() over (order by number desc) t_rank 
from passing_number
order by t_rank,id

SQL65 异常的邮件概率(关联条件做运算)

select date,round(sum(type='no_completed')*1.0/count(*),3) as p 
from (
    select a.date,a.type from email a 
    left join `user` b on a.send_id = b.id 
    left join `user` c on a.receive_id = c.id 
    and b.is_blacklist = 0 and c.is_blacklist
) c group by date

SQL68 牛客每个人最近的登录日期(自链接查询连续两天都出现的数据)

select round(count(c.user_id)/count(b.user_id),3) from(
    select a.user_id,min(a.date) date from login a group by user_id
) b left join login c on b.user_id = c.user_id
and b.date = DATE_SUB(c.date,interval 1 day)

SQL69 牛客每个人最近的登录日期(查询每天新用户数量)

#明确问题：登录的当前日期=该用户所有登录日期的最小值

窗口函数：

select a.date,sum(case when num=1 then 1 else 0 end) new from (
select date,row_number() over(partition by user_id order by date) as num 
from login 
)a group by date
计数神器——sum+case方法

1 2 3 4 5 6 7

select distinct date         ,sum(case when (user_id,date) in      (select user_id,min(date)from login group by user_id)     then 1 else 0 end) from login group by date order by date;

SQL70 牛客每个人最近的登录日期(五 难) 

简单方法：

select a.date,(case when round(count(b.old)/count(a.new),3) 
                    then round(count(b.old)/count(a.new),3)
                    else 0.000 end) 
from (

T1（首次登录标记为1，作为新用户标签，记作new_u）：
select user_id,date,(
    case when (user_id,date) in (
        select user_id,min(date) from login group by user_id
    ) then 1 else null end
) as new from login
)a left join (

T2（次日登录标记为2，作为次日留存标签，记作rlog）：
select user_id,date,2 as old 
from login 
where (user_id,date) 
in (select user_id,date_add(date,interval 1 day) from login) 
)b on a.user_id = b.user_id group by date

T1和T2进行左连接（T1为全表，进行左连接可将次日登录标签进行（2\None）标记)

对new_u和rlog进行计数，次日登录标签总数/新用户标签总数（rlog/new_u）即可完成此答题

备注：由于题目需要将不存在表示为0.000，这里使用case方法进行值分配（不为空则为计数值，为空则为0.000）

使用2个窗口函数：第一个窗口函数看用户是否次日登陆，第二个窗口函数看用户是否计入当日新用户。

select b.date,round(ifnull(su/tal,0),3) from (
select date ,count(*) su from(
    select user_id,date,lead(date,1) over(partition by user_id order by date) old,
    rank() over(partition by user_id order by date) myrank from login
) b where datediff(old,date) = 1
group by date
) a right join (
select sum(case when c.new = 1 then 1 else 0 end) tal,date from(
    select date,rank() over(partition by user_id order by date) new from login
)c group by date
) b on a.date =b.date 

SQL71 牛客每个人最近的登录日期(六)

使用窗口函数SUM，将日期作为“窗口”计算累加刷题数量:

select b.name,a.date,a.num from (
select user_id,date,sum(number) over(partition by user_id order by date) as num 
from passing_number ) a
left join user b on a.user_id = b.id order by a.date

使用自联结:

select b.name,a.date,a.num from (
    select a.user_id,a.date,sum(b.number) as num FROM passing_number a,
    passing_number b where a.date >= b.date and a.user_id = b.user_id
    group by a.date,a.user_id
)a left join user b on a.user_id = b.id group by a.user_id,a.date order by a.date,b.name

SQL75 考试分数(四)(求奇数 偶数方法)

求中位数，只需要知道每个岗位的总数，用聚合函数count；得出总数后，再分奇/偶讨论中位数；

偶数：start=count/2，end=count/2+1;

奇数：start=count/2，得到的会是x.5,用round取整；end同理

select a.job,
case when(a.c%2=0) then round((a.c/2),0) else (round(a.c/2,0)) end as start,
case when(a.c%2=0) then round((a.c/2+1),0) else (round(a.c/2,0)) end as end
from(
    select job,count(*) c
    from grade
    group by job
) a
order by a.job asc

SQL76 考试分数(五)(有点难，头疼)

select a.id,a.job,a.score,a.rank from (
    select id,score,row_number() over(partition by job order by score desc) as `rank`,job from grade
) a inner join (
    select b.job,
    case when(b.num%2=0) then round((b.num/2),0) else (round(b.num/2,0)) end as start,
    case when(b.num%2=0) then round((b.num/2+1),0) else (round(b.num/2,0)) end as end 
    from grade a right join (
        select count(job) num,job from grade group by job
    ) b on 1=1 group by b.job
) b on a.job = b.job and (a.rank = b.start or a.rank = b.end) order by a.id

扩展操作：

柏拉图求递增：

SELECT TP,TJ,SC,SR,SUM(SR) OVER(ORDER BY TJ DESC ROWS BETWEEN UNBOUNDED PRECEDING AND 0 PRECEDING) AS PS,TP TPNAME
  FROM (SELECT AA.REGION TP,COUNT(ID) TJ,SUM(COUNT(ID)) OVER(PARTITION BY AA.REGION) SC,ROUND(SUM(COUNT(ID)) OVER(PARTITION BY AA.REGION) / SUM(COUNT(ID)) OVER(),4) SR
        FROM (SELECT DOA.ID,DOA.REGION
              FROM T_QM_CHECKIN_INFO DOA
              LEFT JOIN T_KOF_CALENDARS TKC
              ON TO_CHAR(DOA.RETURN_TIME, 'YYYY-MM-DD') = TKC.DD
                ) AA
  WHERE AA.MON>='2021-01' and AA.MON<='2022-01'
  GROUP BY AA.REGION
  ORDER BY SC
) BB ORDER BY TJ DESC, TP, SC, SR