牛客数据库MySQL

傷城~ 2022-04-24 19:52 206阅读 0赞

### 1.查找最晚入职员工的所有信息 ###

select * from employees
    where hire_date =
    (select max(hire_date) from employees);

### 2.查找入职员工时间排名倒数第三的员工所有信息 ###

\#limit 的使用方法  
LIMIT m,n : 表示从第m+1条开始，取n条数据；  
LIMIT n ： 表示从第0条开始，取n条数据，是limit(0,n)的缩写。

可能存在多个相同的日期，使用distinct区分

select * from employees
    where hire_date = 
    (select distinct hire_date from employees
    order by hire_date desc limit 2,1
    );

### 3.查找各个部门当前(to\_date='9999-01-01')领导当前薪水详情以及其对应部门编号dept\_no ###

CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

SELECT s.*, d.dept_no FROM salaries s ,  dept_manager d
    where s.to_date = '9999-01-01'
    and d.to_date = '9999-01-01'
    and s.emp_no = d.emp_no;

### 4.查找所有已经分配部门的员工的last\_name和first\_name ###

CREATE TABLE \`dept\_emp\` (  
\`emp\_no\` int(11) NOT NULL,  
\`dept\_no\` char(4) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`dept\_no\`));

select e.last_name, e.first_name, d.dept_no
    from employees as e 
    join dept_emp as d 
    on e.emp_no= d.emp_no;

### 5.查找所有员工的last\_name和first\_name以及对应部门编号dept\_no，也包括展示没有分配具体部门的员工 ###

CREATE TABLE \`dept\_emp\` (  
\`emp\_no\` int(11) NOT NULL,  
\`dept\_no\` char(4) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`dept\_no\`));

e 作为左连接的部分

#考察左连接，
    select e.last_name, e.first_name, d.dept_no 
    from employees as e 
    left join dept_emp as d
    on e.emp_no = d.emp_no;

### 6.查找所有员工入职时候的薪水情况，给出emp\_no以及salary， 并按照emp\_no进行逆序 ###

CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

确保入职的时间和工作人员的工号

select e.emp_no, s.salary
    from employees as e
    join salaries as s
    on e.hire_date = s.from_date
    and e.emp_no = s.emp_no
    order by e.emp_no desc;

### 7.查找薪水涨幅超过15次的员工号emp\_no以及其对应的涨幅次数t ###

CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

> GROUP BY语句在HAVING语句之后；SQL会在分组之后计算HAVING语句。

select emp_no, count(emp_no) as t from salaries 
    group by emp_no having t > 15;

### 8.找出所有员工当前(to\_date='9999-01-01')具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示 ###

CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

select distinct salary from salaries
    where to_date='9999-01-01'
    order by salary desc;

### 9.获取所有部门当前manager的当前薪水情况，给出dept\_no, emp\_no以及salary，当前表示to\_date='9999-01-01' ###

CREATE TABLE \`dept\_manager\` (  
\`dept\_no\` char(4) NOT NULL,  
\`emp\_no\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`dept\_no\`));  
CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

select d.dept_no, d.emp_no, s.salary
    from dept_manager as d 
    join salaries as s 
    on d.emp_no = s.emp_no
    and s.to_date='9999-01-01'
    and d.to_date='9999-01-01';

### 10.获取所有非manager的员工emp\_no ###

CREATE TABLE \`dept\_manager\` (  
\`dept\_no\` char(4) NOT NULL,  
\`emp\_no\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`dept\_no\`));  
CREATE TABLE \`employees\` (  
\`emp\_no\` int(11) NOT NULL,  
\`birth\_date\` date NOT NULL,  
\`first\_name\` varchar(14) NOT NULL,  
\`last\_name\` varchar(16) NOT NULL,  
\`gender\` char(1) NOT NULL,  
\`hire\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`));

**方法1 ，使用not in**

select emp_no from employees 
    where emp_no not in 
    (select emp_no from dept_manager);

**方法2 使用is null**

**最后一个where是选择左连接后的表**

select e.emp_no from employees as e 
    left join dept_manager as d 
    on e.emp_no = d.emp_no
    where d.dept_no is null;

### 11.获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to\_date='9999-01-01'。   结果第一列给出当前员工的emp\_no,第二列给出其manager对应的manager\_no。 ###

CREATE TABLE \`dept\_emp\` (  
\`emp\_no\` int(11) NOT NULL,  
\`dept\_no\` char(4) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`dept\_no\`));  
CREATE TABLE \`dept\_manager\` (  
\`dept\_no\` char(4) NOT NULL,  
\`emp\_no\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`dept\_no\`));

select e.emp_no, m.emp_no as manager_no
    from dept_emp as e
    join dept_manager as m 
    on e.dept_no = m.dept_no
    where e.emp_no != m.emp_no
    and e.to_date='9999-01-01'
    and m.to_date='9999-01-01';

### 12.获取所有部门中当前员工薪水最高的相关信息，给出dept\_no, emp\_no以及其对应的salary ###

CREATE TABLE \`dept\_emp\` (  
\`emp\_no\` int(11) NOT NULL,  
\`dept\_no\` char(4) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`dept\_no\`));  
CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

select e.dept_no, e.emp_no, max(s.salary) as salary
    from salaries as s 
    join dept_emp as e
    on s.emp_no = e.emp_no 
    where e.to_date = '9999-01-01'
    and s.to_date = '9999-01-01'
    group by e.dept_no;

### 13.从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。 ###

CREATE TABLE IF NOT EXISTS "titles" (  
\`emp\_no\` int(11) NOT NULL,  
\`title\` varchar(50) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date DEFAULT NULL);

select title, count(title) as t
    from titles
    group by title;

### 14.从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。   注意对于重复的emp\_no进行忽略。 ###

CREATE TABLE IF NOT EXISTS "titles" (  
\`emp\_no\` int(11) NOT NULL,  
\`title\` varchar(50) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date DEFAULT NULL);

select title, count(distinct emp_no) as t
    from titles
    group by title 
    having t >= 2;

### 15.查找employees表所有emp\_no为奇数，且last\_name不为Mary的员工信息，并按照hire\_date逆序排列 ###

select * from employees
    where emp_no % 2=1
    and last_name != 'Mary'
    order by hire_date desc;

### 16.统计出当前各个title类型对应的员工当前薪水对应的平均工资。结果给出title以及平均工资avg。 ###

CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));  
CREATE TABLE IF NOT EXISTS "titles" (  
\`emp\_no\` int(11) NOT NULL,  
\`title\` varchar(50) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date DEFAULT NULL);

select t.title, avg(s.salary) as avg
    from titles as t
    join salaries as s 
    on t.emp_no = s.emp_no
    where t.to_date = '9999-01-01'
    and s.to_date = '9999-01-01'
    group by title;

### 17.获取当前（to\_date='9999-01-01'）薪水第二多的员工的emp\_no以及其对应的薪水salary ###

CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

**使用distinct防止有多个相同的最大值**

select emp_no, salary from salaries 
    where salary = (
    select distinct salary from salaries 
    order by salary desc
    limit 1,1
    )
    and to_date='9999-01-01';

### 18.查找当前薪水(to\_date='9999-01-01')排名第二多的员工编号emp\_no、薪水salary、last\_name以及first\_name，不准使用order by ###

CREATE TABLE \`employees\` (  
\`emp\_no\` int(11) NOT NULL,  
\`birth\_date\` date NOT NULL,  
\`first\_name\` varchar(14) NOT NULL,  
\`last\_name\` varchar(16) NOT NULL,  
\`gender\` char(1) NOT NULL,  
\`hire\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`));  
CREATE TABLE \`salaries\` (  
\`emp\_no\` int(11) NOT NULL,  
\`salary\` int(11) NOT NULL,  
\`from\_date\` date NOT NULL,  
\`to\_date\` date NOT NULL,  
PRIMARY KEY (\`emp\_no\`,\`from\_date\`));

select e.emp_no, max(s.salary) as salary, e.last_name, e.first_name
    from employees as e inner join salaries as s
    on s.emp_no = e.emp_no
    where s.to_date='9999-01-01'
    // 不能使用  e.to_date = ...上述是新表
    and s.salary not in (select max(salary) from salaries where to_date='9999-01-01');

### 19.查找所有员工的last\_name和first\_name以及对应的dept\_name，也包括暂时没有分配部门的员工 ###

1、第一次LEFT JOIN连接employees表与dept\_emp表，得到所有员工的last\_name和first\_name以及对应的**dept\_no**，也包括暂时没有分配部门的员工

2、第二次LEFT JOIN连接上表与departments表，即**连接dept\_no与dept\_name**，得到所有员工的last\_name和first\_name以及对应的**dept\_name**，也包括暂时没有分配部门的员工

select e.last_name, e.first_name, d.dept_name
    from (employees as e left join dept_emp on e.emp_no = dept_emp.emp_no)
    left join departments as d on d.dept_no = dept_emp.dept_no;

### 查找员工编号emp\_no为10001其自入职以来的薪水salary涨幅值growth ###

select max(salary)-min(salary) as growth 
    from salaries
    where emp_no = 10001;

### 查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp\_no以及其对应的薪水涨幅growth，并按照growth进行升序 ###

可能存在降薪的情况

链接：[https://www.nowcoder.com/questionTerminal/fc7344ece7294b9e98401826b94c6ea5?f=discussion][https_www.nowcoder.com_questionTerminal_fc7344ece7294b9e98401826b94c6ea5_f_discussion]  
来源：牛客网  
  
先分别用两次LEFT JOIN左连接employees与salaries，建立两张表，分别存放员工当前工资（sCurrent）与员工入职时的工资（sStart），再用INNER JOIN连接sCurrent与sStart，最后限定在同一员工下用当前工资减去入职工资。

使用当前的工资减去入职的工资

select e.emp_no, (s.salary - sa.salary) as growth
    from
        (employees as e inner join salaries as s
        on e.emp_no = s.emp_no and s.to_date = '9999-01-01')
        
        inner join salaries as sa 
        on e.emp_no = sa.emp_no and e.hire_date = sa.from_date
    order by growth asc;

第二种方法：

select sCur.emp_no, (sCur.salary - sStart.salary) as growth
    from (select e.emp_no, s.salary from employees e left join salaries s on e.emp_no=s.emp_no where s.to_date='9999-01-01') as sCur
    inner join (select e.emp_no, s.salary from employees e left join salaries s on e.emp_no=s.emp_no where s.from_date=e.hire_date) as sStart
    on sCur.emp_no = sStart.emp_no
    order by growth asc;

### 统计各个部门对应员工涨幅的次数总和，给出部门编码dept\_no、部门名称dept\_name以及次数sum ###

select de.dept_no, dp.dept_name, count(sa.salary) as sum
    from (dept_emp as de inner join salaries as sa on de.emp_no = sa.emp_no)
    inner join departments as dp on de.dept_no = dp.dept_no
    group by de.dept_no;

### 对所有员工的当前(to\_date='9999-01-01')薪水按照salary进行按照1-N的排名，相同salary并列且按照emp\_no升序排列 ###

链接：[https://www.nowcoder.com/questionTerminal/b9068bfe5df74276bd015b9729eec4bf?f=discussion][https_www.nowcoder.com_questionTerminal_b9068bfe5df74276bd015b9729eec4bf_f_discussion]  
来源：牛客网

本题的主要思想是**复用salaries表进行比较排名**，具体思路如下：

1、从两张相同的salaries表（分别为s1与s2）进行对比分析，先将两表限定条件设为to\_date = '9999-01-01'，挑选出当前所有员工的薪水情况。

2、本题的精髓在于 **s1.salary <= s2.salary，意思是在输出s1.salary的情况下，有多少个s2.salary大于等于s1.salary**，比如当s1.salary=94409时，有3个s2.salary（分别为94692,94409,94409）大于等于它，但由于94409重复，利用COUNT(DISTINCT s2.salary)去重可得工资为94409的rank等于2。其余排名以此类推。

3、千万不要忘了GROUP BY s1.emp\_no，否则输出的记录只有一条（可能是第一条或者最后一条，根据不同的数据库而定），因为用了合计函数COUNT()

4、最后先以 s1.salary 逆序排列，再以 s1.emp\_no 顺序排列输出结果

select s1.emp_no, s1.salary, count(distinct s2.salary)
    from salaries s1, salaries s2
    where s1.to_date='9999-01-01'
    and s2.to_date='9999-01-01'
    and s1.salary <= s2.salary
    group by s1.emp_no
    order by s1.salary desc, s1.emp_no asc;

### 获取所有非manager员工当前的薪水情况，给出dept\_no、emp\_no以及salary ，当前表示to\_date='9999-01-01' ###

1、先用INNER JOIN连接employees和salaries，找出当前所有员工的工资情况

2、再用INNER JOIN连接dept\_emp表，找到所有员工所在的部门

3、最后用限制条件de.emp\_no NOT IN (SELECT emp\_no FROM dept\_manager WHERE

select de.dept_no, e.emp_no, s.salary 
    from (employees e inner join salaries s on e.emp_no = s.emp_no and s.to_date='9999-01-01')
    inner join dept_emp de on de.emp_no = s.emp_no
    where e.emp_no 
    not in (select emp_no from dept_manager where to_date='9999-01-01');

### 获取员工其当前的薪水比其manager当前薪水还高的相关信息，当前表示to\_date='9999-01-01', ###

select de.emp_no, dm.emp_no as manager_no, s1.salary as emp_salary, s2.salary as manager_salary
    from dept_emp de, dept_manager dm, salaries s1, salaries s2
    where de.dept_no = dm.dept_no and de.to_date='9999-01-01' and dm.to_date='9999-01-01'
    and de.emp_no = s1.emp_no and s1.to_date='9999-01-01'
    and dm.emp_no = s2.emp_no and s2.to_date='9999-01-01'
    and s1.salary > s2.salary;

### 汇总各个部门当前员工的title类型的分配数目，结果给出部门编号dept\_no、dept\_name、其当前员工所有的title以及该类型title对应的数目count ###

select de.dept_no, de.dept_name, t.title, count(t.title) as cnt
    from titles as t inner join dept_emp as d
    on t.emp_no = d.emp_no and t.to_date = '9999-01-01' and d.to_date = '9999-01-01'
    inner join departments as de on de.dept_no = d.dept_no
    group by de.dept_no, t.title;

### 给出每个员工每年薪水涨幅超过5000的员工编号emp\_no、薪水变更开始日期from\_date以及薪水涨幅值salary\_growth，并按照salary\_growth逆序排列。 ###

提示：在sqlite中获取datetime时间对应的年份函数为strftime('%Y', to\_date)

链接：[https://www.nowcoder.com/questionTerminal/eb9b13e5257744db8265aa73de04fd44?f=discussion][https_www.nowcoder.com_questionTerminal_eb9b13e5257744db8265aa73de04fd44_f_discussion]  
来源：牛客网

本题的难点在于**如何理解** **每年薪水涨幅**，以及**复用salaries表求出每年薪水涨幅超过5000的员工**，具体思路如下：

1、假设s1是涨薪水前的表，s2是涨薪水后的表，因为每个员工涨薪水的时间不全固定，有可能一年涨两次，有可能两年涨一次，所以每年薪水的涨幅，应该理解为两条薪水记录的from\_date相同或to\_date相同。

/\*\* 如果只限定to\_date相同，则将第三条原始测试数据的52668改成62668时，就会少一条【62668-48584=14084】的记录

INSERT INTO salaries VALUES(10008,46671,'1998-03-11','1999-03-11');

INSERT INTO salaries VALUES(10008,48584,'1999-03-11','2000-03-10');

INSERT INTO salaries VALUES(10008, **62668** ,'2000-03-10','2000-07-31'); \*\*/

2、找到s1与s2符合要求的记录后，用s2的薪水减去s1的薪水，用salary\_growth表示，加上限定条件 s1.emp\_no = s2.emp\_no AND salary\_growth > 5000，即同一员工每年涨幅超过5000的记录

select s2.emp_no, s2.from_date, (s2.salary - s1.salary) as salary_growth  // 使用表  2 的列
    from salaries s1, salaries s2
    where s1.emp_no = s2.emp_no
    and salary_growth > 5000
    and (strftime('%Y', s2.to_date) - strftime('%Y', s1.to_date) = 1 or 
        strftime('%Y', s2.from_date) - strftime('%Y', s1.from_date) = 1
        )
    order by salary_growth desc;

### 查找描述信息中包括robot的电影对应的分类名称以及电影数目，而且还需要该分类对应电影数量>=5部 ###

链接：[https://www.nowcoder.com/questionTerminal/3a303a39cc40489b99a7e1867e6507c5?f=discussion][https_www.nowcoder.com_questionTerminal_3a303a39cc40489b99a7e1867e6507c5_f_discussion]  
来源：牛客网

1、找到**对应电影数量>=5的所有分类，建立成虚表cc**：(select category\_id, count(film\_id) as category\_num from film\_category group by category\_id having count(film\_id)>=5) as cc

2、设定**限制条件** f.description like '%robot%'

3、在表cc、f、fc、c中查找**包括robot的电影对应的分类名称和对应的电影数目**。

SELECT c.name, COUNT(fc.film_id) FROM
     (select category_id, COUNT(film_id) AS category_num FROM
         film_category  GROUP BY category_id HAVING count(film_id)>=5) AS cc,
     film AS f, film_category AS fc, category AS c
    WHERE  f.description LIKE '%robot%'
    AND f.film_id = fc.film_id
    AND c.category_id = fc.category_id
    AND c.category_id=cc.category_id

### 使用join查询方式找出没有分类的电影id以及名称 ###

使用左连接

select f.film_id, f.title
    from film f left join film_category fc 
    on f.film_id = fc.film_id
    where fc.category_id is null;

### 使用子查询的方式找出属于Action分类的所有电影对应的title,description ###

select f.title, f.description
    from film f inner join film_category fc
    on f.film_id = fc.film_id
    inner join
    (select category_id from category where name = 'Action') as ca
    on fc.category_id = ca.category_id;

### 获取select \* from employees对应的执行计划 ###

explain select * from employees;

### 将employees表的所有员工的last\_name和first\_name拼接起来作为Name，中间以一个空格区分 ###

select (last_name || ' ' || first_name) as Name from employees;
    //  使用 mysql时
    select contact(last_name, " ", first_name) as Name from employees;

### 创建一个actor表，包含如下列信息 ###

create table if not exists actor
    (
        actor_id smallint(5) NOT NULL PRIMARY KEY,
        first_name varchar(45) not null,
        last_name varchar(45) not null,
        last_update timestamp not null DEFAULT (datetime('now','localtime'))
    )

### 对于表actor批量插入如下数据 ###

insert into actor values(1, "PENELOPE", "GUINESS", "2006-02-15 12:34:33"), 
    (2, "NICK", "WAHLBERG", "2006-02-15 12:34:33");

### 对于表actor批量插入如下数据,如果数据已经存在，请忽略，不使用replace操作 ###

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

### 创建一个actor\_name表，将actor表中的所有first\_name以及last\_name导入改表。 ###

create table actor_name as select first_name, last_name from actor;

### 针对如下表actor结构创建索引 ###

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

### 针对actor表创建视图actor\_name\_view ###

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

### 针对salaries表emp\_no字段创建索引idx\_emp\_no ###

select * from salaries indexed by idx_emp_no where emp_no = 10005;

### 在last\_update后面新增加一列名字为create\_date ###

alter table actor add column create_date datetime not null default('0000-00-00 00:00:00');

### 构造一个触发器audit\_log ###

create trigger audit_log after insert on employees_test begin insert into audit values(new.id, new.name);
    end;

### 删除emp\_no重复的记录，只保留最小的id对应的记录。 ###

delete from titles_test  where id not in (select min(id) from titles_test  group by emp_no);

### 将所有to\_date为9999-01-01的全部更新为NULL,且 from\_date更新为2001-01-01。 ###

update titles_test  set to_date = null, from_date = '2001-01-01' where to_date = '9999-01-01';

### 将 id=5 以及 emp\_no=10001 的行数据替换成id=5以及emp\_no=10005 ###

update titles_test set emp_no = replace(emp_no, 10001, 10005) where id = 5;

### 将titles\_test表名修改为titles\_2017。 ###

alter table titles_test rename to titles_2017;

### 在audit表上创建外键约束，其emp\_no对应employees\_test表的主键id。 ###

drop table audit;
    create table audit(
        EMP_no int not null,
        create_date datetime not null,
        foreign key(EMP_no) references employees_test(ID));

### 如何获取emp\_v和employees有相同的数据？ ###

select * from emp_v  intersect select * from employees; -- 求交集

### 将所有获取奖金的员工当前的薪水增加10%。 ###

update salaries set salary = salary *1.1 where emp_no in
    (select s.emp_no from emp_bonus e, salaries s where 
    s.emp_no = e.emp_no and s.to_date = '9999-01-01');

### 针对库中的所有表生成select count(\*)对应的SQL语句 ###

select "select count(*) from " || name || ";" as cnts
    from sqlite_master where type = 'table';

### 将employees表中的所有员工的last\_name和first\_name通过(')连接起来。 ###

select last_name || "'" || first_name as name from employees;

### 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', ',', '')) as cnt;

### 获取Employees中的first\_name，查询按照first\_name最后两个字母，按照升序进行排列 ###

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