练习SQL代码

---------PS:但凡有聚合函数（where），必然后Group by，Group by后面跟having 
---------面试题

SELECT name,sum(fen) as s
from score
GROUP BY NAME
ORDER BY s;

PS:一般聚合查询后面跟having

SELECT title, COUNT(DISTINCT emp_no) AS t FROM titles  --内部再次过滤
GROUP BY title HAVING t >= 2

分页查询

查询第1条到第10条的数据的sql是：select * from table limit 0,10;   ->对应我们的需求就是查询第一页的数据：select * from table limit (1-1)*10,10;

查询第10条到第20条的数据的sql是：select * from table limit 10,20;  ->对应我们的需求就是查询第二页的数据：select * from table limit (2-1)*10,10;

查询第20条到第30条的数据的sql是：select * from table limit 20,30;  ->对应我们的需求就是查询第三页的数据：select * from table limit (3-1)*10,10;

------------------题目---------------------
--1.查找最晚入职员工的所有信息
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`)); 
--答案：
--子查询
SELECT * FROM employees WHERE hire_date = (SELECT MAX(hire_date) FROM employees);
--排序
SELECT * FROM employees ORDER BY hire_date DESC LIMIT 0,1;
--2.查找入职员工时间排名倒数第三的员工所有信息
SELECT * 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 `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`));
--Answer:
select
salaries.emp_no,salaries.salary,salaries.from_date,salaries.to_date,dept_manager.dept_no
from salaries inner join dept_manager
on dept_manager.emp_no = salaries.emp_no 
where dept_manager.to_date = '9999-01-01'   --没想起来
and salaries.to_date = '9999-01-01';

--4.查找所有已经分配部门的员工的last_name和first_name
-- 内链接推荐
SELECT e.last_name,e.first_name,d.dept_no
FROM dept_emp d INNER JOIN employees e
ON d.emp_no = e.emp_no


--left join(左连接)：join左表中所有记录和右表中满足连接条件的记录信息
--right join(右连接)：join右表中所有记录和左表中满足连接条件的记录信息
--5.查找所有员工的last_name和first_name以及对应部门编号dept_no
SELECT e.last_name,e.first_name,d.dept_no
FROM employees e  LEFT JOIN dept_emp d
ON d.emp_no = e.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
SELECT e.emp_no,s.salary
FROM employees e  INNER JOIN salaries s
ON s.emp_no = e.emp_no AND e.hire_date = s.from_date  --少考虑这一种情况
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
-- 本题主要考聚合查询
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
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'
-- PS: on已经是条件的判断了，不能再用where,              INNER JOIN 左右两表交换 没有 错，oj有问题
SELECT d.dept_no,d.emp_no,s.salary
FROM dept_manager d  INNER JOIN salaries s
ON s.emp_no = d.emp_no 
AND d.to_date = '9999-01-01'
AND s.to_date = '9999-01-01'
---WHERE to_date='9999-01-01';
---注：有人反映将连接语句改成FROM dept_manager AS d INNER JOIN salaries AS s后，结果通不过。
--INNER JOIN对于左右两表并无顺序要求，此为本题OJ系统Bug所致。

--10.获取所有非manager的员工emp_no  !!!!!!
--  -PS:考察NOT IN
--方法一：使用NOT IN选出在employees但不在dept_manager中的emp_no记录 
SELECT emp_no FROM employees
WHERE emp_no NOT IN (SELECT emp_no FROM dept_manager)

-- 方法二：先使用LEFT JOIN连接两张表，再从此表中选出dept_no值为NULL对应的emp_no记录 
SELECT emp_no?FROM?(SELECT * FROM employees LEFT JOIN dept_manager 
ON employees.emp_no = dept_manager.emp_no)
WHERE 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。

SELECT de.emp_no, dm.emp_no AS manager_no?
FROM dept_emp AS de INNER JOIN dept_manager AS dm
ON de.dept_no = dm.dept_no?
WHERE dm.to_date = '9999-01-01' AND de.emp_no <> dm.emp_no 

--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
SELECT de.dept_no,de.emp_no,MAX(sa.salary) AS salary
FROM dept_emp AS de INNER JOIN salaries AS sa
ON de.emp_no = sa.emp_no
WHERE de.to_date = '9999-01-01' AND sa.to_date = '9999-01-01'
GROUP BY de.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。
--PS:正确的过滤是在having以后过滤
SELECT title,COUNT(title) AS t
FROM titles
WHERE t>=2  --错误
GROUP BY title 
-- 正确的答案          
SELECT title, COUNT(title) AS t FROM titles
GROUP BY title HAVING t >= 2; --- 聚合就聚合到这里了

--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。
--PS:对COUNT中进行分解
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。
--PS:不知道为什么不行
--SELECT t.title AVG(s.salary);
--FROM salaries s INNER JOIN titles t
--ON s.emp_no = t.emp_no AND t.to_date='9999-01-01' AND s.to_date='9999-01-01'
--GROUP BY title

select t.title,avg(s.salary) from titles t, salaries s where
s.emp_no = t.emp_no  
and s.to_date =?'9999-01-01'  
and t.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
SELECT  emp_no,salary
FROM salaries
WHERE to_date='9999-01-01'
ORDER BY salary DESC  --注意要降序
LIMIT 1,1

--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
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 e.emp_no = s.emp_no
AND s.to_date = '9999-01-01'
AND s.salary NOT IN (
       SELECT MAX(salary) FROM salaries
       WHERE salaries.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
--毁三观！！！！！！

--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 e 
left join dept_emp de on e.emp_no =de.emp_no 
left join departments d on d.dept_no=de.dept_no;


--20.查找员工编号emp_now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'
--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。
--PS:不知道为什么不行
--SELECT t.title AVG(s.salary);
--FROM salaries s INNER JOIN titles t
--ON s.emp_no = t.emp_no AND t.to_date='9999-01-01' AND s.to_date='9999-01-01'
--GROUP BY title

select t.title,avg(s.salary) from titles t, salaries s where
s.emp_no = t.emp_no  
and s.to_date =?'9999-01-01'  
and t.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
SELECT  emp_no,salary
FROM salaries
WHERE to_date='9999-01-01'
ORDER BY salary DESC  --注意要降序
LIMIT 1,1

--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
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 e.emp_no = s.emp_no
AND s.to_date = '9999-01-01'
AND s.salary NOT IN (
       SELECT MAX(salary) FROM salaries
       WHERE salaries.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
--毁三观！！！！！！

--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 e 
left join dept_emp de on e.emp_no =de.emp_no 
left join departments d on d.dept_no=de.dept_no;


--20.查找员工编号emp_now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'

--21.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_noy以及其对应的薪水涨幅growth，并按照growth进行升序
select t1.emp_no, t1.salary - t2.salary as growth
from    
(select e.emp_no,s.salary from salaries s,employees e where e.emp_no=s.emp_no and s.to_date='9999-01-01' )as t1,  
(select e.emp_no,s.salary from salaries s,employees e where e.emp_no=s.emp_no and s.from_date=e.hire_date)as t2 
where t1.emp_no=t2.emp_no 
order by growth;

--22.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
-- 多个表查询，可以多次链接
SELECT de.dept_no, dp.dept_name, COUNT(s.salary) AS sum?
FROM (dept_emp AS de INNER JOIN salaries AS s ON de.emp_no = s.emp_no)?
INNER JOIN departments AS dp ON de.dept_no = dp.dept_no?
GROUP BY de.dept_no

--23.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
SELECT emp_no,salary, AS rank
FROM salaries
WHERE to_date='9999-01-01'
ORDER BY emp_no

--24.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
select s1.emp_no , s1.salary ,count(distinct s2.salary) as rank 
from salaries s1, salaries s2 
where s1.to_date = '9999-01-01' and s2.to_date = '9999-01-01' and
s1.salary<=s2.salary  --最大的数只小于等于自己，第二大的数只小于等于两个数，以此类推。。。为他们的rank 
group by s1.emp_no 
order by rank ;

--25.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'

SELECT de.dept_no, s.emp_no, s.salary?
FROM (employees AS e INNER JOIN salaries AS s ON s.emp_no = e.emp_no AND s.to_date = '9999-01-01')
INNER JOIN dept_emp AS de ON e.emp_no = de.emp_no
WHERE de.emp_no NOT IN (SELECT emp_no FROM dept_manager)

--25.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',结果第一列给出员工的emp_no，第二列给出其manager的manager_no，第三列给出该员工当前的薪水emp_salary,第四列给该员工对应的manager当前的薪水manager_salary
--SELECT 
--FROM dept_emp e INNER JOIN dept_manager m
--ON e.emp_no = m.emp_no

--本题主要思想是创建两张表（一张记录当前所有员工的工资，另一张只记录部门经理的工资）进行比较，具体思路如下： 
-- 1、先用INNER JOIN连接salaries和demp_emp，建立当前所有员工的工资记录sem 
-- 2、再用INNER JOIN连接salaries和demp_manager，建立当前所有员工的工资记录sdm 
-- 3、最后用限制条件sem.dept_no = sdm.dept_no AND sem.salary >
-- sdm.salary找出同一部门中工资比经理高的员工，并根据题意依次输出emp_no、manager_no、emp_salary、manager_salary

SELECT sem.emp_no AS emp_no, sdm.emp_no AS manager_no, sem.salary AS emp_salary, sdm.salary AS manager_salary
FROM (SELECT s.salary, s.emp_no, de.dept_no FROM salaries s INNER JOIN dept_emp de 
ON s.emp_no = de.emp_no AND s.to_date = '9999-01-01' ) AS sem, 
(SELECT s.salary, s.emp_no, dm.dept_no FROM salaries s INNER JOIN dept_manager dm 
ON s.emp_no = dm.emp_no AND s.to_date = '9999-01-01' ) AS sdm
WHERE sem.dept_no = sdm.dept_no AND sem.salary > sdm.salary

--26.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
--PS：对一个表进行连接以后，可以再连接两一个表

本题的关键在于用 GROUP BY 同时对 de.dept_no 和 t.title
进行分组，具体思路如下： 
1、先用 INNER JOIN 连接 dept_emp 与 salaries，根据测试数据添加限定条件 de.to_date =
'9999-01-01' AND t.to_date = '9999-01-01'，即当前员工的当前头衔 
2、再用 INNER JOIN 连接departments，限定条件为 de.dept_no =
dp.dept_no，即部门编号相同 
3、最后用 GROUP BY 同时对 de.dept_no 和 t.title   进行分组，用 COUNT(t.title)
统计相同部门下相同头衔的员工个数 

SELECT de.dept_no, dp.dept_name, t.title, COUNT(t.title) AS count
FROM titles AS t INNER JOIN dept_emp AS de 
ON t.emp_no = de.emp_no AND de.to_date = '9999-01-01' AND t.to_date = '9999-01-01'
INNER JOIN departments AS dp 
ON de.dept_no = dp.dept_no
GROUP BY de.dept_no, t.title

--27.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)

--SELECT emp_no,from_date,  AS salary_growth
--FROM salaries

SELECT s2.emp_no, s2.from_date, (s2.salary - s1.salary) AS salary_growth
FROM salaries AS s1, salaries AS 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

--28.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

SELECT c.name AS name, COUNT(c.name) AS amount
FROM (film AS f INNER JOIN film_category AS fc ON f.film_id = fc.film_id )
INNER JOIN category AS c
ON fc.category_id = c.category_id
WHERE f.description LIKE '%robot%'
GROUP BY c.name HAVING amount >= 2

--29.使用join查询方式找出没有分类的电影id以及名称

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

--30.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, film_category fc, category c
WHERE f.film_id = fc.film_id 
AND fc.category_id = c.category_id 
AND c.name = 'Action'

--32.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

--不同数据库连接字符串的方法不完全相同，MySQL、SQL
--Server、Oracle等数据库支持CONCAT方法，而本题所用的SQLite数据库只支持用连接符号"||"来连接字符串 

SELECT last_name||" "||first_name AS Name FROM employees

--33.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'))
) 

--34.批量插入数据

INSERT INTO actor 
VALUES (1, 'PENELOPE', 'GUINESS', '2006-02-15 12:34:33'),
(2, 'NICK', 'WAHLBERG', '2006-02-15 12:34:33')

--35.批量插入数据
INSERT OR IGNORE INTO actor VALUES (3, 'ED', 'CHASE', '2006-02-15 12:34:33')

--36.创建一个actor_name表，将actor表中的所有first_name以及last_name导入改表。 actor_name表结构如下：

CREATE TABLE actor_name
(
    first_name varchar(45) NOT NULL,
    last_name varchar(45) NOT NULL
);
INSERT INTO actor_name SELECT first_name, last_name FROM actor;

--37.对first_name创建唯一索引uniq_idx_firstname，对last_name创建普通索引idx_lastname

CREATE UNIQUE INDEX uniq_idx_firstname ON actor(first_name);
CREATE INDEX idx_lastname ON actor(last_name);

--38.针对actor表创建视图actor_name_view，只包含first_name以及last_name两列，并对这两列重新命名，fist_name为first_name_v，last_name修改为last_name_v：
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')))
-------------------------------
--视图是用来干什么的？视图时虚表，仅存储建立视图的sql语句，查询的时候跟普通的表一样,
CREATE VIEW actor_name_view AS
SELECT first_name AS fist_name_v, last_name AS last_name_v
FROM actor 

--39.针对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 INDEXED BY idx_emp_no WHERE emp_no='10005';

--40.现在在last_update后面新增加一列名字为create_date, 类型为datetime, NOT NULL，默认值为'0000 00:00:00'
--ADD COLUMN create_date 
ALTER TABLE actor ADD COLUMN create_date datetime NOT NULL DEFAULT '0000-00-00 00:00:00';

--41.构造一个触发器audit_log，在向employees表中插入一条数据的时候，触发插入相关的数据到audit中。
CREATE TRIGGER audit_log AFTER INSERT ON employees_test
BEGIN
    INSERT INTO audit VALUES (NEW.ID, NEW.NAME);
END;

--42.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);

--43.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 FROM titles_test SET to_date =NULL, from_date='2001-01-01'
WHERE to_date='9999-01-01';

--44.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。
update titles_test set emp_no=replace(emp_no,'10001','10005') where id='5'

--45.将titles_test表名修改为titles_2017。
-- RENAME TABLE titles_test AS titles_2017;
ALTER TABLE titles_test RENAME TO titles_2017;

--46.在audit表上创建外键约束，其emp_no对应employees_test表的主键id。
--CREATE FOREIGN KEY for   REFERENCES ON  

--47.存在如下的视图：create view emp_v as select * from employees where emp_no >10005;如何获取emp_v和employees有相同的数据？
--PS ：使用INTERSECT取交集
SELECT * FROM employees INTERSECT SELECT * FROM emp_v

--48.将所有获取奖金的员工当前的薪水增加10%。
update salaries
set salary=1.1*salary
where emp_no in (select emp_no from emp_bonus)

--50.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;

--51.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",",","")))/length(",") AS cnt

--52.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
SELECT first_name
FROM employees
ORDER BY substr(first_name,-2) ;

--53.按照dept_no进行汇总，属于同一个部门的emp_no按照逗号进行连接，结果给出dept_no以及连接出的结果employees
--SELECT dept_no,(SELECT ) AS employees
--FROM dept_emp

--本题要用到SQLite的聚合函数group_concat(X,Y)，其中X是要连接的字段，Y是连接时用的符号，可省略，默认为逗号。此函数必须与 GROUP BY 配合使用。此题以 dept_no 作为分组，将每个分组中不同的emp_no用逗号连接起来（即可省略Y）。可参考：       http://www.sqlite.org/lang_aggfunc.html#groupconcat         http://blog.csdn.net/langzxz/article/details/16807859    1

SELECT dept_no, group_concat(emp_no) AS employees
FROM dept_emp GROUP BY dept_no

--54.查找排除当前最大、最小salary之后的员工的平均工资avg_salary。
--PS：不会选择最大和最小
--SELECT AVG(salary) AS avg_salary
--FROM salaries

SELECT AVG(salary) AS avg_salary FROM salaries 
WHERE to_date = '9999-01-01' 
AND salary NOT IN (SELECT MAX(salary) FROM salaries WHERE to_date = '9999-01-01')
AND salary NOT IN (SELECT MIN(salary) FROM salaries WHERE to_date = '9999-01-01')

--55.分页查询employees表，每5行一页，返回第2页的数据
SELECT * FROM employees LIMIT 5,5

--56.获取所有员工的emp_no、部门编号dept_no以及对应的bonus类型btype和recevied，没有分配具体的员工不显示

SELECT em.emp_no, de.dept_no, eb.btype, eb.recevied
FROM employees AS em INNER JOIN dept_emp AS de
ON em.emp_no = de.emp_no
LEFT JOIN emp_bonus AS eb 
ON de.emp_no = eb.emp_no

--57.使用含有关键字exists查找未分配具体部门的员工的所有信息。

SELECT * FROM employees WHERE NOT EXISTS 
(SELECT emp_no FROM dept_emp WHERE emp_no = employees.emp_no)

SELECT * FROM employees WHERE emp_no NOT IN (SELECT emp_no FROM dept_emp)

--58.获取employees中的行数据，且这些行也存在于emp_v中。注意不能使用intersect关键字。
--根据题意，不能使用 INTERSECT 关键字，但由于视图 emp_v 的记录是从 employees 中导出的，因此要判断两者中相等的数据，只需要判断emp_no相等即可。 
--   方法一：用 WHERE 选取二者 emp_no 相等的记录  1
SELECT em.* FROM employees AS em, emp_v AS ev WHERE em.emp_no = ev.emp_no

--59.获取有奖金的员工相关信息.
--给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况salary以及奖金金额bonus。 
--bonus类型btype为1其奖金为薪水salary的10%，btype为2其奖金为薪水的20%，其他类型均为薪水的30%。 当前薪水表示to_date='9999-01-01'
SELECT e.emp_no, e.first_name, e.last_name, b.btype, s.salary, 
(CASE b.btype 
 WHEN 1 THEN s.salary * 0.1
 WHEN 2 THEN s.salary * 0.2
 ELSE s.salary * 0.3 END) AS bonus
FROM employees AS e INNER JOIN emp_bonus AS b ON e.emp_no = b.emp_no
INNER JOIN salaries AS s ON e.emp_no = s.emp_no AND s.to_date = '9999-01-01'

--60.按照salary的累计和running_total，其中running_total为前两个员工的salary累计和，其他以此类推。 具体结果如下Demo展示。
--SELECT emp_no,salary, ADD(salary) AS running_total
--FROM salaries

SELECT s1.emp_no, s1.salary, 
(SELECT SUM(s2.salary) FROM salaries AS s2 
 WHERE s2.emp_no <= s1.emp_no AND s2.to_date = '9999-01-01') AS running_total 
FROM salaries AS s1 WHERE s1.to_date = '9999-01-01'

--61.对于employees表中，给出奇数行的first_name
--SELECT first_name
--FROM employees
--WHERE 

SELECT e.first_name FROM employees AS e
WHERE e.first_name IN(
    SELECT e1.first_name
    FROM employees AS e1 INNER JOIN employees AS e2
    ON e1.first_name >= e2.first_name
    GROUP BY e1.first_name HAVING count(e2.first_name) % 2 = 1);

 乔虹姐帮忙修改

SELECT * FROM PTE_ProfessionalTitleApplication


SELECT t.* FROM EEP_Base.dbo.Users t

SELECT t.* FROM EEP_Base.dbo.db_tmp t

SELECT * from EEP_Base.dbo.Department

SELECT t.userType FROM EEP_Base.dbo.db_tmp t

SELECT t.*,u.ParentId FROM EEP_Base.dbo.db_tmp t
inner join EEP_Base.dbo.Users u
on u.WorkerNO=t.workId


SELECT  FROM EEP_Base.dbo.db_tmp t
inner join EEP_Base.dbo.Users u
on u.WorkerNO=t.workId


SELECT u.ParentID FROM  EEP_Base.dbo.Users u
inner join EEP_Base.dbo.Department d
on u.ParentID=d.ID
inner join EEP_Base.dbo.db_tmp t
on u.WorkerNO=t.workId


SELECT t.* FROM EEP_Base.dbo.db_tmp t 
WHERE t.workId in (SELECT u.*,u.Number FROM  EEP_Base.dbo.Users u
inner join EEP_Base.dbo.Department d
on u.ParentID=d.ID)


SELECT u.name,u.Number FROM EEP_Base.dbo.Users u
WHERE u.WorkerNO in (SELECT t.workId FROM EEP_Base.dbo.db_tmp t)
 

insert into db_tmp(number) SELECT u.Number FROM EEP_Base.dbo.db_tmp t
inner join EEP_Base.dbo.Users u
on u.WorkerNO=t.workId


UPDATE db_tmp  SET(number) =(SELECT u.Number FROM  EEP_Base.dbo.Users u
WHERE u.WorkerNO=db_tmp.workId)


UPDATE EEP_Base.dbo.db_tmp  SET deptName =d.name FROM  EEP_Base.dbo.Users u, EEP_Base.dbo.db_tmp t,EEP_Base.dbo.Department d
WHERE u.WorkerNO=t.workId
AND u.ParentID=d.ID

 帮李姐整理的数据（没有用户ID）

需求：

只有红色的两列，找出人员编号和部门编号；
步骤：
1. 查找不重复的数据，把人员编号补充进去

SELECT t.ID,u.Number,u.name, t.Username FROM EEP_Base.dbo.tb_temp t
LEFT join EEP_Base.dbo.Users u
on u.Name=t.Username 
WHERE u.name !='刘x'
and u.name !='刘x'
and u.name !='张建x'

ORDER BY t.ID

2、查找重复人名 的部门的人，补充部门

SELECT u.Name,d.Name,u.Number FROM EEP_Base.dbo.Users u
inner join EEP_Base.dbo.Department d
on u.ParentID=d.ID
WHERE u.Name='沈中华'

3、找到部门编号填充

SELECT Name,Description FROM EEP_Base.dbo.Department

4、校验数据的正确性

SELECT t.ID,u.Name,d.Name,t.DeptName,u.Number,t.UserNumber FROM EEP_Base.dbo.Users u
inner join EEP_Base.dbo.Department d
on u.ParentID=d.ID
inner join EEP_Base.dbo.tb_temp t
ON u.Number=t.UserNumber
--WHERE u.Number!=t.UserNumber
ORDER BY t.ID

 李姐布置的sql查询

2、各部门本月里程碑节点完成率、年度目标完成率（未完成）
SELECT aa.OwnerDepartmentID,d.name,aa.sum1 AS '里程碑个数',aa.sum2 AS '已完成里程碑个数',ltrim(Convert(numeric(9,2),sum2*100.0/sum1))+'%' As '百分比' FROM
 
(SELECT OwnerDepartmentID,SUM(t1.cc) AS sum1,ISNULL(SUM(t2.cc), 0) as sum2  FROM ProjectArchive.dbo.Project AS p
INNER JOIN (SELECT COUNT(t.ID) as cc ,t.ProjectID as ProjectID FROM ProjectArchive.dbo.ISOForm io
INNER JOIN ProjectArchive.dbo.PlanFormProjectTask t
ON t.ProjectID=io.ProjectID  and io.RowGuid=t.FormRowGuid
WHERE io.FlowState>=4 and io.IsAvailability=1
AND t.IsEventTask=1
AND t.TaskState<>3
AND t.EndDate>='2020-10-01 00:00:00'
AND t.EndDate<'2020-11-01 00:00:00'
GROUP BY t.ProjectID) as t1 
ON t1.ProjectID=p.ID
left JOIN (SELECT COUNT(t.ID) as cc ,t.ProjectID as ProjectID FROM ProjectArchive.dbo.ISOForm io
INNER JOIN ProjectArchive.dbo.PlanFormProjectTask t
ON t.ProjectID=io.ProjectID  and io.RowGuid=t.FormRowGuid
WHERE io.FlowState>=4 and io.IsAvailability=1
AND t.IsEventTask=1
AND t.TaskState=2
AND t.EndDate>='2020-10-01 00:00:00'
AND t.EndDate<='2020-11-01 00:00:00'
GROUP BY t.ProjectID) as t2 
ON t2.ProjectID=p.ID
GROUP BY p.OwnerDepartmentID) AS aa

INNER JOIN EEP_Base.dbo.Department d
ON aa.OwnerDepartmentID=d.Id

3、计划变更项目清单

SELECT p.ID,p.Name,sp.Number,sp.Name,aa.date1 as '第一次时间',bb.date2 as '最新时间'
FROM ProjectArchive.dbo.Project p
INNER JOIN ProjectArchive.dbo.SubProject sp 
ON p.ID=sp.ProjectID
 left join(SELECT FormContent.query('PlanFormContent/PlanReleaseDateStr' ).value('(//PlanReleaseDateStr)[1]', 'varchar(50)') as date1,subprojectid FROM ProjectArchive.dbo.ISOForm io
WHERE io.FlowState>=4 and io.VersionNumber=1) as aa on sp.id=aa.SubProjectid 
left join(SELECT FormContent.query('PlanFormContent/PlanReleaseDateStr' ).value('(//PlanReleaseDateStr)[1]', 'varchar(50)') as date2,subprojectid FROM ProjectArchive.dbo.ISOForm io
WHERE io.FlowState>=4 and io.IsAvailability=1) as bb on sp.id=bb.SubProjectid 


4、未及时发布计划项目清单

SELECT p.ID,p.Name FROM ProjectArchive.dbo.Project p
LEFT  JOIN  ProjectArchive.dbo.ISOForm io
ON io.ProjectId=p.Id

WHERE io.FlowState>=4 and io.VersionNumber=1
AND DATEDIFF(day,ValidDate,FormContent.query('PlanFormContent/PlanReleaseDateStr' ).value('(//PlanReleaseDateStr)[1]', 'varchar(50)'))>20 

UNION
SELECT p.id,p.Name
from ProjectArchive.dbo.Project as p
where p.ProjectState=20 and DATEDIFF(day,ValidDate,GETDATE())>20  and (SELECT Count(id) from ProjectArchive.dbo.ISOForm where Projectid=p.id)=0