Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Chapter 3, Problem 32E
Explanation of Solution
Query:
The query for the given scenario:
SELECT distinct teaches.ID, Name FROM takes, teaches, instructor WHERE teaches.course_id = takes.course_id AND instructor.Id = teaches.Id and Grade is not NULL AND teaches.ID not in (SELECT distinct teaches.Id FROM takes, teaches WHERE teaches.course_id = takes.course_id and Grade="A");
Explanation:
- The subquery(inner select query) finds teacher Id's who have given Grade "A" in any course that they teach...
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Chapter 3 Solutions
Database System Concepts
Ch. 3 - Prob. 1PECh. 3 - Prob. 2PECh. 3 - Prob. 3PECh. 3 -
Suppose that we have a relation marks(ID, score)...Ch. 3 - Suppose that we have a relation marks(ID, score)...Ch. 3 -
The SQL like operator is case sensitive (in most...Ch. 3 - Prob. 7PECh. 3 - Prob. 8PECh. 3 - Prob. 9PECh. 3 - Prob. 10PE
Ch. 3 - Prob. 11ECh. 3 - Prob. 12ECh. 3 - Prob. 13ECh. 3 -
Consider the insurance database of Figure 3.17,...Ch. 3 - Prob. 15ECh. 3 - Prob. 16ECh. 3 - Prob. 17ECh. 3 - Prob. 18ECh. 3 -
List two reasons why null values might be...Ch. 3 - Prob. 20ECh. 3 - Prob. 21ECh. 3 - Prob. 22ECh. 3 - Prob. 23ECh. 3 - Prob. 24ECh. 3 - Prob. 25ECh. 3 - Prob. 26ECh. 3 - Using the university schema, write an SQL query to...Ch. 3 - Prob. 28ECh. 3 - Prob. 29ECh. 3 - Prob. 30ECh. 3 - Prob. 31ECh. 3 - Prob. 32ECh. 3 - Prob. 33ECh. 3 - Prob. 34ECh. 3 - Prob. 35E
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.Similar questions
- Create a query that displays each instructor schedule. Display the instructor name, course title, time slot information, for each course taught by the instructor. Also display the building and room number where the course is taught. Sort the results in a meaningful timetable way, so that each instructor can easily see their schedules.arrow_forwardShow a list of employee id, names, department, years and job of any employee in the staff table who makes a total amount more than their manager or has more years of service than their manager. Make sure to include both salary and commission when calculating the total amount someone makes. Exclude staff in department 10 from the query. Order the results by department then namearrow_forwardWrite a query that uses INNER JOIN TO connect the EMPLOYEE, EMPPROJACT, PROJACT and PROJECT tables. Include the Project number , Department number, Project start and end date and AC STAFF WHERE They belong to department D11 , Salary is more than or equal to 65 percent of $15,000 AND Salary is less than or equal to 130 percent of $40,000arrow_forward
- Write a query to produce the total number of hours and charges for each of the projects represented in the ASSIGNMENT table, sorted by project number.arrow_forwardWhy the following statement is erroneous? SELECT dept_name, ID, avg (salary) FROM instructor GROUP BY dept_name; a) Dept_id should not be used in group by clause b) Group by clause is not valid in this query c) Avg(salary) should not be selected d) Nonearrow_forwardAssuming that the CUSTOMER table contains a CUST_AGE attribute, write the query to update the values in that attribute. (Hint: Use the results of the previous query.)arrow_forward
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