Relational Schema:
Students (**sid**, name, age, major)
Courses (**cid**, name)
Enrollment (**sid**, **cid**, year, term, grade)
Write a SQL query that returns the name of the students who took all courses.I'm not sure how I capture the concept of 'ALL' in a SQL query.
EDIT:
I want to be able write it without aggregation as I want to use the same logic for writing the query in relational algebra as well.
Thanks for the help!
One way of writing such queries is to count the number of course and number of courses each student took, and compare them:
SELECT s.*
FROM students s
JOIN (SELECT sid, COUNT(DISTINCT cid) AS student_courses
FROM enrollment
GROUP BY sid) e ON s.sid = e.sid
JOIN (SELECT COUNT(*) AS cnt
FROM courses) c ON cnt = student_cursed
This gives course combinations that are possible but haven't been taken...
SELECT s.sid, c.cid FROM students CROSS JOIN courses
EXCEPT
SELECT sid, cid FROM enrollment
So, you can then do the same with the student list...
SELECT sid FROM students
EXCEPT
(
SELECT DISTINCT
sid
FROM
(
SELECT s.sid, c.cid FROM students CROSS JOIN courses
EXCEPT
SELECT sid, cid FROM enrollment
)
AS not_enrolled
)
AS slacker_students
I don't like it, but it avoids aggregation...
SELECT *
FROM Students
WHERE NOT EXISTS (
SELECT 1 FROM Courses
LEFT OUTER JOIN Enrollment ON Courses.cid = Enrollment.cid
AND Enrollment.sid = Students.sid
WHERE Enrollment.sid IS NULL
)
btw. names of tables should be in singular form, not plural
Related
I have 4 tables, I would like to select one column from each table, but only if the department has both 'Mick' and 'Dave working in it (must have both names, not one or the other). But it does not seem to be working properly:
SELECT SCHOOL_NAME, TOWN, COUNTY
FROM STUDENTS
NATURAL JOIN SCHOOLS NATURAL JOIN TOWNS NATURAL JOIN
COUNTIES
WHERE FIRST_NAME IN ('Mick','Dave)
/
I'm going wrong somewhere (probably lots of places :( ). Any help would be great
Don't use NATURAL JOIN. It is an abomination, because it does not take properly declared foreign key relationships into account. It only looks at the names of columns. This can introduce really hard to find errors.
Second, what you want is aggregation:
select sc.SCHOOL_NAME, t.TOWN, c.COUNTY
from STUDENTS st join
SCHOOLS sc
on st.? = sc.? join
TOWNS t
on t.? = ? join
COUNTIES c
on c.? = t.?
where FIRST_NAME in ('Mick', 'Dave')
group by sc.SCHOOL_NAME, t.TOWN, c.COUNTY
having count(distinct st.first_name) = 2;
The ? are placeholders for table and column names. If you are learning SQL, it is all the more important that you understand how columns line up for joins in different tables.
A where clause can only check the values in a single row. There is a separate row for each student, so there is no way -- with just a where -- to find both students. That is where the aggregation comes in.
You need at least three Join conditions, and properly end the string Dave with quote :
SELECT SCHOOL_NAME, TOWN, COUNTY
FROM SCHOOLS h
JOIN TOWNS t ON (t.id=h.town_id)
JOIN COUNTIES c ON (t.county_id=c.id)
WHERE EXISTS ( SELECT school_id
FROM STUDENTS s
WHERE s.first_name in ('Mick','Dave')
AND school_id = h.id
GROUP BY school_id
HAVING count(1)>1
);
SQL Fiddle Demo
You can use an analytic function in a sub-query to count the students who have the name Mick or Dave for each school_id (assuming that is your identifier for a school):
SELECT SCHOOL_NAME, TOWN, COUNTY
FROM ( SELECT *
FROM (
SELECT d.*,
COUNT(
DISTINCT
CASE WHEN FIRST_NAME IN ( 'Mick', 'Dave' ) THEN FIRST_NAME END
) OVER( PARTITION BY school_id )
AS num_matched
FROM STUDENTS d
)
WHERE num_matched = 2
)
NATURAL JOIN SCHOOLS
NATURAL JOIN TOWNS
NATURAL JOIN COUNTIES;
SQLFiddle
You would also be better to use an INNER JOIN and explicitly specify the join condition rather than relying on NATURAL JOIN.
Hi I am evaluating a SQL query with the following schema:
Student(sid, name, age, gender, dept, GPA)
Faculty(fid, name, age, office, dept)
Course(cid, name, description)
Teach(fid, cid, term)
Enrollment(sid, cid, term, grade, final grade)
The question to base query on is:
Find the students who took at least one course with Tom
The query looks like:
SELECT *
FROM Student S1
WHERE NOT EXISTS (SELECT E1.cid
FROM Enrolment E1, Student S2
WHERE E1.sid = S2.sid
AND S2.name = 'Tom'
MINUS
SELECT E2.Cid
FROM E2 Enrollment
WHERE E2.sid = S1.sid)
This query is not making sense to me because to me it seems like the NOT EXISTS condition would only select a student when the subquery is empty, and this would only occur in the case when student s1 has taken all the same courses as Tom. However the question is to find the students who took at least one course, not all courses as Tom. Maybe I am interpreting wrong but I could use some clarification.
Try this
DECLARE #TomID nvarchar(MAX)
set #TomID = (SELECT TOP 1 ID from Students where Name = 'Tom')
select distinct E1.SID from Enrollment E1
where E1.SID != #TomID AND E1.CID in
(select distinct E.CID from Enrollment E
where E.SID=#TomID)
Now there are two tables student(sid, name) and course(cid, name), which relation is many-to-many. So there is another table student_course(sid, cid) that stores the information about what courses have been chosen by whom.
How to write the sql that can get the courses that have been chosen by all the students?
Standard solution: use the NOT EXISTS ( ... NOT EXISTS (...)) construct:
find all courses in which all students take part
==>> there must not exist a student that does not take part in this course
SELECT * FROM course c
WHERE NOT EXISTS (
SELECT * from student s
WHERE NOT EXISTS (
SELECT * from student_course cs
WHERE cs.sid = s.sid
AND cs.cid = c.cid
)
)
;
This query is often faster (given appropiate indexes) than the count() == count() variant. The reason for this: you do not have to count all the (distinct) records; once you found one student that does not take this course you can omit this course from your list of suspects. Also: ANTI-joins often can make use of indexes [so can a count(), but that still has to count all the (distinct) keyvalues in the index]
Select c.cid, c.name
From course c where
(select count(1) from student) = (select count(1) from student_course sc where sc.cid = c.cid);
See SQL Fiddle
It finds all courses where the count of entries for that course in the student_course table matches the number of students
CID NAME
1 Test Course1
4 Test Course4
I have three tables:
1) Students: studentID (KEY), name, surname, address
2) Exams: examID (KEY), examName
3) Grades: studenID (KEY), examID(KEY), grade
How to write SQL query to show the best students (for example those with average grade above 9)?
How to write SQL query to rank Cities (column address) according to the average grade of their students?
I'm a system engineer, working with Unix and Linux systems and I am new in SQL, I only know about SQL basics, and I was trying to do this for past three days, with no success, so please help me. I presume it's not a complex thing for one who's experienced in SQL. Thanks a lot.
your first query to show the best students :
SELECT student.surname, students.surename, students.address
FROM
students INNER JOIN Grades ON Grades.StudentID=Students.StudentID
INNER JOIN Exams ON Grades.examID=exams.examID WHERE Grades.grade=
(SELECT MAX(grade) FROM Grades WHERE examID=exams.examID)
your second query to rank cities:
SELECT students.address
FROM
students INNER JOIN Grades ON Grades.StudentID=Students.StudentID
INNER JOIN Exams ON Grades.examID=exams.examID order by grades.grade DESC
Refer the fiddle here:
LINK 1 : http://sqlfiddle.com/#!4/ab4de6/19
LINK 2 : http://sqlfiddle.com/#!4/ab4de6/32
Below Queries should help you in Oracle:
--List of Students having Average grade >=9
SELECT S.studentID, S.NAME, S.SURNAME, S.ADDRESS, A.AVG_GRADE FROM
STUDENTS S JOIN
(
SELECT studentID, AVG(GRADE) AVG_GRADE FROM GRADES
GROUP BY studentID
) A
ON S.studentID = A.studentID
AND A.AVG_GRADE >=9
ORDER BY A.AVG_GRADE, S.studentID;
--------------------------------------------------------------------
--- Rank cities
SELECT A.ADDRESS, A.AVG_GRADE, ROWNUM RANKING FROM
(
SELECT S.ADDRESS, AVG(G.GRADE) AVG_GRADE FROM
STUDENTS S JOIN GRADES G
ON S.STUDENTID = G.STUDENTID
GROUP BY S.ADDRESS
ORDER BY 2 DESC
) A;
You need to know about the following concepts.
INNER QUERY / SUB QUERY
JOINS
AGGREGATE FUNCTIONS (Average Calculations)
GROUP BY
ORDER BY
ROWNUM
I've got a student table and an enrollment table; a student could have multiple enrollment records that can be active or inactive.
I want to get a select that has a single student record and an indicator as to whether that student has active enrollments.
I thought about doing this in an inline UDF that uses the student ID in a join to the enrollment table, but I wonder if there's a better way to do it in a single select statement.
The UDF call might look something like:
Select Student_Name,Student_Email,isEnrolled(Student_ID) from Student
What might the alternative - with one SQL statement - look like?
select Student_Name,
Student_Email,
(select count(*)
from Enrollment e
where e.student_id = s.student_id
) Number_Of_Enrollments
from Student e
will get the number of enrollments, which should help.
Why not join to a secondary select? Unlike other solutions this isn't firing a subquery for every row returned, but gathers the enrollment data for everyone all at once. The syntax may not be quite correct, but you should get the idea.
SELECT
s.student_name,
s.student_email,
IsNull( e.enrollment_count, 0 )
FROM
Students s
LEFT OUTER JOIN (
SELECT
student_id,
count(*) as enrollment_count
FROM
enrollments
WHERE
active = 1
GROUP BY
student_id
) e
ON s.student_id = e.student_id
The select from enrollments could also be redone as a function which returns a table for you to join on.
CREATE FUNCTION getAllEnrollmentsGroupedByStudent()
RETURNS #enrollments TABLE
(
student_id int,
enrollment_count int
) AS BEGIN
INSERT INTO
#enrollments
(
student_id,
enrollment_count
) SELECT
student_id,
count(*) as enrollment_count
FROM
enrollments
WHERE
active = 1
GROUP BY
student_id
RETURN
END
SELECT
s.student_name,
s.student_email,
e.enrollment_count
FROM
Students s
JOIN
dbo.getAllEnrollmentsGroupedByStudent() e
ON s.student_id = e.student_id
Edit:
Renze de Waal corrected my bad SQL!
Try someting like this:
SELECT Student_Name, Student_Email, CAST((SELECT TOP 1 1 FROM Enrollments e WHERE e.student_id=s.student_id) as bit) as enrolled FROM Student s
I think you can also use the exists statement in the select but not positive
try to avoid using udfs or subqueries, they are performance killers. banjolity seems to havea good solution otherwise because it uses a derivd table instead of a UDF or subselect.
select students.name,
decode(count(1), 0, "no enrollments", "has enrollments")
from students, enrollments
where
students.id = enrollments.sutdent_id and
enrollments.is_active = 1 group by students.name
Of course, replace the decode with a function your database uses (or, a case statement).