Question
Evaluate the PL/SQL block below and determine the value of each of the following variables according to the rules of scoping. DECLARE weight NUMBER(3) :=
Evaluate the PL/SQL block below and determine the value of each of the following variables according to the rules of scoping. DECLARE weight NUMBER(3) := 600; message VARCHAR2(255) := 'Product 10012'; BEGIN DECLARE weight NUMBER(3) := 1; message VARCHAR2(255) := 'Product 11001'; new_locn VARCHAR2(50) := 'Europe'; BEGIN weight := weight + 1; new_locn := 'Western ' || new_locn; -- Position 1 -- 2 END; weight := weight + 1; message := message || ' is in stock'; -- Position 2 -- END; A. The value of weight at position 1 is: B. The value of new_locn at position 1 is: C. The value of weight at position 2 is: D. The value of message at position 2 is: E. The value of new_locn at position 2 is: 2. Enter and run the following PL/SQL block, which contains a nested block. Look at the output and answer the questions. DECLARE v_employee_id employees.employee_id%TYPE; v_job employees.job_id%TYPE; BEGIN SELECT employee_id, job_id INTO v_employee_id, v_job FROM employees WHERE employee_id = 100; DECLARE v_employee_id employees.employee_id%TYPE; v_job employees.job_id%TYPE; BEGIN SELECT employee_id, job_id INTO v_employee_id, v_job FROM employees WHERE employee_id = 103; DBMS_OUTPUT.PUT_LINE(v_employee_id || ' is a(n) ' || v_job); END; 3 DBMS_OUTPUT.PUT_LINE(v_employee_id || ' is a(n) ' || v_job); END; A. Why does the inner block display the job_id of employee 103, not employee 100? B. Why does the outer block display the job_id of employee 100, not employee 103? C. Modify the code to display the details of employee 100 in the inner block. Use block labels.
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