View the Exhibit and examine the data in EMPLOYEES andDEPARTMENTS tables. In the EMPLOYEES table EMPLOYEE_ID is the PRIM
Posted: Wed Jun 07, 2023 6:11 am
View the Exhibit and examine the data in EMPLOYEES andDEPARTMENTS tables. In the EMPLOYEES table EMPLOYEE_ID is the PRIMARY KEY and DEPARTMENT_ID is the FOREIGN KEY. In the DEPARTMENTS tableDEPARTMENT_ID is the PRIMARY KEY.Evaluate the following UPDATE statement:UPDATE employees a -SET department_jd =(SELECT department_idFROM departments -WHERE location_id = 2100),(salary, commission_pct) =(SELECT 1.1*AVG(salary), 1.5*AVG(commission_pct)FROM employees b -WHERE a. department_jd = b. department_id)WHERE first_name|| '||last_name = 'Amit Banda';What would be the outcome of the above statement?
A. It would execute successfully and update the relevant data.
B. It would not execute successfully because there is no LOCATION_ID 2100 in the DEPARTMENTS table.
C. It would not execute successfully because the condition specified with the concatenation operator is not valid.
D. It would not execute successfully because multiple columns (SALARY,COMMISSION_PCT)cannot be used in an UPDATE statement.
A. It would execute successfully and update the relevant data.
B. It would not execute successfully because there is no LOCATION_ID 2100 in the DEPARTMENTS table.
C. It would not execute successfully because the condition specified with the concatenation operator is not valid.
D. It would not execute successfully because multiple columns (SALARY,COMMISSION_PCT)cannot be used in an UPDATE statement.