Questions: 1- Display the IDs of all instructors who have never taught a couse and display their names. (USE JOIN) 2- For all instructors in the university who have taught some course, find their names and the course ID of all courses they taught 3- Find all courses whose identifier starts with the string "CS-1"

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Questions:
1- Display the IDs of all instructors who have never taught a couse and
display their names. (USE JOIN)
2- For all instructors in the university who have taught some course, find
their names and the course ID of all courses they taught
3- Find all courses whose identifier starts with the string "CS-1"
4- Find the names of all students who have taken any Comp. Sci. course
ever (there should be no duplicate names). (USE subquery)
5- Find the number of instructors in each department who teach a course in
the Spring 2018 semester.
6- Find the names of all instructors, along with their department names and
department building name (USE JOIN)
7- Find the average salary in each department.
8- Find the total number of instructors who teach a course in the Spring
2018 semester.
9- For the student with ID 12345 (or any other value), show all course_id
and title of all courses registered for by the student. (USE JOIN)
10- Show the total number of credits for such courses (taken by that student).
Don't display the tot_creds value from the student table, you should use
SQL aggregation on courses taken by the student.( USE JOIN)
11- List all name courses for each student (use LISTAGG function) you must
use student's identifier to make grouping and show his name in result.
Like this
20170000 Ahmed cs-101,cs-319
20180000 Ahmed phy-200,
12- Find all sections that had the maximum enrollment (along with the
enrollment), using a subquery.
13- Find all sections that had the maximum enrollment (along with the
enrollment) also include sections with no students taking them; the
enrollment for such sections should be treated as 0. Using aggregation
on a left outer join (use the SQL natural left outer join syntax)
Transcribed Image Text:Questions: 1- Display the IDs of all instructors who have never taught a couse and display their names. (USE JOIN) 2- For all instructors in the university who have taught some course, find their names and the course ID of all courses they taught 3- Find all courses whose identifier starts with the string "CS-1" 4- Find the names of all students who have taken any Comp. Sci. course ever (there should be no duplicate names). (USE subquery) 5- Find the number of instructors in each department who teach a course in the Spring 2018 semester. 6- Find the names of all instructors, along with their department names and department building name (USE JOIN) 7- Find the average salary in each department. 8- Find the total number of instructors who teach a course in the Spring 2018 semester. 9- For the student with ID 12345 (or any other value), show all course_id and title of all courses registered for by the student. (USE JOIN) 10- Show the total number of credits for such courses (taken by that student). Don't display the tot_creds value from the student table, you should use SQL aggregation on courses taken by the student.( USE JOIN) 11- List all name courses for each student (use LISTAGG function) you must use student's identifier to make grouping and show his name in result. Like this 20170000 Ahmed cs-101,cs-319 20180000 Ahmed phy-200, 12- Find all sections that had the maximum enrollment (along with the enrollment), using a subquery. 13- Find all sections that had the maximum enrollment (along with the enrollment) also include sections with no students taking them; the enrollment for such sections should be treated as 0. Using aggregation on a left outer join (use the SQL natural left outer join syntax)
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