Only the row with department ID 40 is deleted in the DEPARTMENTS table.
The statement fails because there are child records in the EMPLOYEES table with department ID 40.
The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 110 and 106 are deleted from the EMPLOYEES table.
The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 106 and 110 and the employees working under employee 110 are deleted from the EMPLOYEES table.
The row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.
The statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.
第1题:
Examine the data in the EMPLOYEES and EMP_HIST tables:The EMP_HIST table is updated at the end of every year. The employee ID, name, job ID, and salary of each existing employee are modified with the latest data. New employee details are added to the table.Which statement accomplishes this task?()
A.
B.
C.
D.
第2题:
Examine the data of the EMPLOYEES table.EMPLOYEES (EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID)Which statement lists the ID, name, and salary of the employee, and the ID and name of the employee‘s manager, for all the employees who have a manager and earn more than 4000?()
A.
B.
C.
D.
E.
第3题:
A.
B.
C.
D. The query fails because the subquery returns more than one row.
E. The query fails because the outer query and the inner query are using different tables.
第4题:
A.
B.
C.
D.
第5题:
A.
B.
C.
D.
第6题:
Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables.EMPLOYEESNOT NULL, PrimaryEMPLOYEE_ID NUMBERKeyVARCHAR2EMP_NAME(30)VARCHAR2JOB_ID(20)SALARY NUMBERReferencesMGR_ID NUMBEREMPLOYEE_IDcolumnDEPARTMENT_ID NUMBER Foreign key toDEPARTMENT_IDcolumn ofthe DEPARTMENTStableDEPARTMENTSNOT NULL,DEPARTMENT_ID NUMBERPrimary KeyVARCHAR2DEPARTMENT_NAME|30|ReferencesMGR_ID columnMGR_ID NUMBERof theEMPLOYEES tableTAXMIN_SALARY NUMBERMAX_SALARY NUMBERTAX_PERCENT NUMBERFor which situation would you use a nonequijoin query?()
A. To find the tax percentage for each of the employees.
B. To list the name, job id, and manager name for all the employees.
C. To find the name, salary, and department name of employees who are not working with Smith.
D. To find the number of employees working for the Administrative department and earning less then 4000.
E. To display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.
第7题:
A.
B.
C.
D.
第8题:
A. To find the tax percentage for each of the employees.
B. To list the name, job id, and manager name for all the employees.
C. To find the name, salary, and department name of employees who are not working with Smith.
D. To find the number of employees working for the Administrative department and earning less then 4000.
E. To display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.
第9题:
A.
B.
C.
D.
第10题:
Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.EMPLOYEESNOT NULL,EMPLOYEE_ID NUMBERPrimary KeyVARCHAR2EMP_NAME(30)VARCHAR2JOB_ID(20)SALARY NUMBERReferencesMGR_ID NUMBEREMPLOYEE_IDcolumnDEPARTMENT_ID NUMBER Foreign key toDEPARTMENT_IDcolumn of theDEPARTMENTStableDEPARTMENTSNOT NULL, PrimaryDEPARTMENT_ID NUMBERKeyVARCHAR2DEPARTMENT_NAME(30)References NGR_IDMGR_ID NUMBERcolumn ofthe EMPLOYEES tableForeign key toLOCATION_ID NUMBERLOCATION_IDcolumn of theLOCATIONS tableLOCATIONSNOT NULL, PrimaryLOCATION_ID NUMBERKeyVARCHAR2CITY|30)Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?()
A. SELECT emp_name, department_name, city FROM employees e JOIN departments d USING (department_id) JOIN locations 1 USING (location_id) WHERE salary > 10000;
B. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 JOIN ON (e.department_id = d.department id) AND (d.location_id =1.location_id) AND salary > 10000;
C. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE salary > 10000;
D. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE e.department_id = d.department_id AND d.location_id = 1.location_id AND salary > 10000;
E. SELECT emp_name, department_name, city FROM employees e NATURAL JOIN departments, locations WHERE salary > 10000;