Review this SQL Statement: SELECT ename, emp_number, salary FROM employee WHERE dept_number = (SELECT dept_number FROM department WHERE location IN('CHICAGO','ATLANTA')); Why may this statement return an error?
Question
Review this SQL Statement: SELECT ename, emp_number, salary FROM employee WHERE dept_number = (SELECT dept_number FROM department WHERE location IN('CHICAGO','ATLANTA')); Why may this statement return an error?
Solution
This SQL statement might return an error due to the subquery in the WHERE clause. The subquery is supposed to return a single value because it's being used in a comparison operation (=). However, if there are multiple departments located in either 'CHICAGO' or 'ATLANTA', the subquery will return multiple department numbers, causing an error because you can't compare a single value (dept_number from the main query) with multiple values (dept_number(s) from the subquery).
To fix this, you could modify the subquery to ensure it only returns a single department number. If you're interested in retrieving employees from all departments in 'CHICAGO' or 'ATLANTA', you could use the IN operator instead of the equals (=) operator. The modified SQL statement would look like this:
SELECT ename, emp_number, salary FROM employee WHERE dept_number IN (SELECT dept_number FROM department WHERE location IN('CHICAGO','ATLANTA'));
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Time left 0:18:35Question 1Not yet answeredMarked out of 1.00Flag questionTipsQuestion textWhich SQL statement produces an error? Select one:a.SELECT department_id, job_id, AVG(salary)FROM emp_dept_vuGROUP BY department_id, job_id;b.SELECT job_id, SUM(salary)FROM emp_dept_vuWHERE department_id IN (10,20)GROUP BY job_idHAVING SUM(salary) > 20000;c.None of the statements produce an error; all are valid. d.SELECT department_id, SUM(salary)FROM emp_dept_vuGROUP BY department_id;e.SELECT *FROM emp_dept_vu;Clear my choiceQuestion 2Not yet answeredMarked out of 1.00Flag questionTipsQuestion textWhich statements would execute successfully?Select one or more:a.SELECT student_name,SUM(subject1) FROM marks WHERE student_name LIKE 'R%'; b.SELECT SUM (subject1+subject2+subject3)FROM marksWHERE student_name IS NULLc.SELECT student_name,subject1 FROM marks WHERE subject1 > AVG(subject1); d.SELECT SUM (DISTINCT NVL(subject1,0)),MAX(subject1)FROM marksWHERE subject1 > subject2;Question 3Not yet answeredMarked out of 1.00Flag questionTipsQuestion textSELECT cust_city, COUNT(cust_last_name) FROM customersWHERE cust_credit_limit > 1000 GROUP BY cust_city HAVING AVG(cust_credit_limit) BETWEEN 5000 AND 6000;Which statement is true regarding the outcome of the above query? Select one:a.It executes successfully.b.It returns an error because WHERE and HAVING clauses cannot be used in the same SELECT statement.c.It returns an error because the BETWEEN operator cannot be used in the HAVING clause.d.It returns an error because WHERE and HAVING clauses cannot be used to apply conditions on the same column. e.Date functionsClear my choiceQuestion 4Not yet answeredMarked out of 1.00Flag questionTipsQuestion textConsider the below tables:Promotions TableColumn NameDatatypeConstraintPromo_idNumberPKPromo_nameVarchar Promo_begin_dateDate Promo_end_dateDate Sales TableColumn NameDatatypeConstraintPromo_idNumberFKCust_idNumberFKTime_idDate Customer TableColumn NameDatatypeConstraintcust_idNumberPKcust_nameVarchar The Below query will generate a report showing the promo name along with the customer name for all products that were sold during their promo campaign and before 30th October 2007.SELECT promo_name,cust_name FROM promotions p JOIN sales s ON(time_id BETWEEN promo_begin_date AND promo_end_date) JOIN customer c ON (s.cust_id = c.cust_id) AND time_id < '30-oct-2007'; Which statement is true regarding the above query? Select one:a.It produces an error because the join order of the tables is incorrect.b.It produces an error because equijoin and nonequijoin conditions cannot be used in the samec.It executes successfully but does not give the required result.d.It executes successfully and gives the required result.Clear my choiceQuestion 5Not yet answeredMarked out of 1.00Flag questionTipsQuestion textThe COMMISSION column shows the monthly commission earned by the employee. Emp_IdDept_IdCommission1105002201000310 410600530800630200710 820300 Which tasks would require sub queries or joins in order to be performed in a single step?Select one or more:a.Finding the number of employees who earn a commission that is higher than the average commission of the company b.Finding the total commission earned by the employees in department 10c.Listing the employees who earn the same amount of commission as employee 3d.Listing the departments whose average commission is more that 600 e.Listing the employees whose annual commission is more than 6000 f.Listing the employees who do not earn commission and who are working for department 20 in descending order of the employee ID
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