Sql practice

employee表 数据准备

use tempdb

go

   

if OBJECT_ID('employee') is not null

drop table employee

;with employee(id,name,salary,manager_id) as

(

select * from

(

values

(1,'John',300,3),

(2,'Mike',200,3),

(3,'Sally',550,4),

(4,'Jane',500,7),

(5,'Joe',600,7),

(6,'Dan',600,3),

(7,'Phil',550,NULL)

) as ve(id,name,salary,manager_id)

)

select * into employee from employee

 

--1.Give the names of employees, whose salaries are greater than their immediate managers':

SELECT e.name

FROM employee AS e

JOIN employee AS m

ON e.manager_id = m.id

WHERE e.salary > m.salary

 

--2.What is the average salary of employees without direct reports

--method1

SELECT Avg(e.salary) AS avgsalry

FROM employee AS e

LEFT JOIN employee AS m

ON m.manager_id = e.id

WHERE m.id IS NULL

--method2

SELECT Avg(e.salary) AS avgsalary

FROM employee AS e

WHERE NOT EXISTS (SELECT *

FROM employee AS m

WHERE m.manager_id = e.id)

 

/******************************************************************************************/

第二题的数据准备:student course courseSelection 三张表

if OBJECT_ID('student','u') is not null

drop table student

if OBJECT_ID('course','u') is not null

drop table course

if OBJECT_ID('courseSelection','u') is not null

drop table courseSelection

   

;with student(student_no,student_name) as

(

SELECT * FROM

(values

(1,'John'),

(2,'Mike'),

(3,'Sally'),

(4,'Jane'),

(5,'Joe'),

(6,'Dan'),

(7,'Phil')

) as vstudent(student_no,student_name)

)

select * into student from student

;with course (Course_no,Course_name,Course_teacher,Course_credit) as

(

SELECT * FROM (

VALUES

(1,'Java','Steve',12),

(2,'SQLServer','Bill',8),

(3,'Windows','Robert',16),

(4,'Art','Evan',7),

(5,'C#','Steve',9),

(6,'HTML','Robert',12),

(7,'Finance','Tom',9)

) as vcourse(Course_no,Course_name,Course_teacher,Course_credit)

)

select * into course from course

   

;with CourseSelection(student_no,Course_no,Grade) as

(

select * from

(values

(3,3,57),

(3,3,52),

(3,3,59),

(3,6,57),

(3,6,75),

(6,2,89),

(1,3,93),

(1,6,88),

(6,7,88),

(6,1,99)

) as vcs (student_no,Course_no,Grade)

)

select * into CourseSelection from CourseSelection

 

--1.Find the students name who pass both "Finance" and "SQLServer" and their average grade(pass means "grade" >= 60).

SELECT DISTINCT s.student_name,

cs.avggrade

FROM student AS s

JOIN (SELECT Avg(grade)

OVER(

partition BY student_no) AS avggrade,

*

FROM courseselection) AS cs

ON s.student_no = cs.student_no

JOIN course AS c

ON cs.course_no = c.course_no

WHERE c.course_name IN ( 'SQLServer', 'Finance' )

AND cs.grade >= 60

 

--2.Find the students name who failed one course more than 3 times and current still not passed.

--max(grade) <60 and group by course_no having count(*)>=3

SELECT s.student_name

FROM student AS s

JOIN (SELECT student_no

FROM courseselection AS cs

GROUP BY student_no,

course_no

HAVING Count(*) > 2

AND Max(grade) < 60) AS cs

ON cs.student_no = s.student_no

 

--3.Update teacher "Tom" 's grade for everyone, for those grade >= 90, deduct 10, for those grade between 65 and 89, deduct 5, the rest remain.

UPDATE cs

SET cs.grade = CASE

WHEN cs.grade > 90 THEN cs.grade - 10

WHEN cs.grade BETWEEN 65 AND 89 THEN cs.grade - 5

ELSE cs.grade

END

FROM course AS c

JOIN courseselection AS cs

ON c.course_no = cs.course_no

WHERE c.course_teacher = 'Tom'

 

   

--4.Find the average grade each teacher give to their students, sort the result by descending,

-- if one student attend one course more than once, only take the highest grade into account.

SELECT c.course_teacher,

Avg(cs.grade) AS avgGrade

FROM course AS c

JOIN (SELECT course_no,

Max(grade) AS grade

FROM courseselection

GROUP BY student_no,

course_no) AS cs

ON c.course_no = cs.course_no

GROUP BY c.course_teacher

ORDER BY avggrade DESC

 

--5.        Find the student names who is qualify to graduate with following conditions:

--a.        Total earn course_credit >= 50

--b.        Failed no more than 5 courses

--c.        The maximum of course_credit from one teacher is 20.(one teacher may have more than one courses)

 

SELECT s.student_name

FROM student AS s

JOIN (SELECT student_no,

Sum (CASE

WHEN totalcreditfromoneteacher > 20 THEN 20

ELSE totalcreditfromoneteacher

END) AS TotalCredit

FROM (SELECT student_no,

course_teacher,

Sum (CASE

WHEN cs.grade > 60 THEN c.course_credit

ELSE 0

END) AS TotalCreditFromOneTeacher

FROM course AS c

JOIN courseselection AS cs

ON cs.course_no = c.course_no

GROUP BY student_no,

course_teacher) AS A

GROUP BY student_no) AS cs

ON cs.student_no = s.student_no

JOIN (SELECT DISTINCT student_no

FROM courseselection cs

GROUP BY student_no,

course_no

HAVING Count(DISTINCT course_no) < 5) AS stuentfaillessthan5courses

ON s.student_no = stuentfaillessthan5courses.student_no

WHERE cs.TotalCredit>50