a. Query for the given instruction: select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name. Explanation: Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join. Use the natural outer join between instructor and teaches tables. Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

i would like to get help to resolve the following case

Challenge: Assume that the assigned network addresses are correct. Can you deduce (guess) what the network subnet masks are? Explain while providing subnet mask bits for each subnet mask. [Hint: Look at the addresses in binary and consider the host ids]

What is the main difference between Static routing and Dynamic routing (OSPF)? in terms of either wildcard mask or subnet mask, or especially for increasing the number of networks, explaining the reason while providing a specific example like what command they use in CLI

Answer 2

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Answer 8

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Answer 9

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Answer 10

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Answer 11

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Answer 12

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Answer 13

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Answer 14

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Answer 15

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Answer 16

Expert Solution & Answer

Answer 17

Expert Solution & Answer

Answer 18

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 19

Ch. 4 - Prob. 1PE Ch. 4 - Prob. 2PE Ch. 4 - Prob. 3PE Ch. 4 - Prob. 4PE Ch. 4 - Prob. 5PE Ch. 4 - Prob. 6PE Ch. 4 - Prob. 7PE Ch. 4 - Prob. 8PE Ch. 4 - SQL allows a foreign-key dependency to refer to...Ch. 4 - Given the relations a(name, address, title) and...

Concept explainers

Explanation of Solution

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Want to see the full answer?

Chapter 4 Solutions

Concept explainers

Explanation of Solution

Explanation of Solution

b. Query without using outer join: Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I. Explanation: Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

Explanation of Solution

d. Query for the given instruction: select dept_name, count (ID) from department natural left outer join instructor group by dept_name Explanation: Initially perform a natural left outer join between department and instructor table...

Want to see the full answer?

Chapter 4 Solutions

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...