单选题Exhibit: Examine the data in the EMPLOYEES table. Examine the subquery: SELECT last_name FROM employees WHERE salary IN (SELECT MAX(salary) FROM employees GROUP BY department_id); Which statement is true?()A The SELECT statement is syntactically accurate.B The SELECT statement does not work because there is no HAVING clause.C The SELECT statement does not work because the column specified in the GROUP BY clause is not in the SELECT list.D The SELECT statement does not work because the GROUP BY clause should be in the main query and not in the subquery.

单选题
Exhibit: Examine the data in the EMPLOYEES table. Examine the subquery: SELECT last_name FROM employees WHERE salary IN (SELECT MAX(salary) FROM employees GROUP BY department_id); Which statement is true?()
A

The SELECT statement is syntactically accurate.

B

The SELECT statement does not work because there is no HAVING clause.

C

The SELECT statement does not work because the column specified in the GROUP BY clause is not in the SELECT list.

D

The SELECT statement does not work because the GROUP BY clause should be in the main query and not in the subquery.


参考解析

解析: 暂无解析

相关考题:

View the Exhibit and examine the Data Pump architecture. Identify the numbered components.() A. 1 - Oracle Loader, 2 - Oracle Data Pump, 3 - Direct Path APIB. 1 - Oracle Data Pump, 2 - Direct Path API, 3 - Oracle LoaderC. 1 - Direct Path API, 2 - Oracle Loader, 3 - Oracle Data PumpD. 1 - Oracle Loader, 2 - Direct Path API, 3 - Oracle Data Pump

Exhibit:Examine the data in the EMPLOYEES table.Examine the subquery:SELECT last_nameFROM employeesWHERE salary IN (SELECT MAX(salary)FROM employeesGROUP BY department_id);Which statement is true?() A. The SELECT statement is syntactically accurate.B. The SELECT statement does not work because there is no HAVING clause.C. The SELECT statement does not work because the column specified in the GROUP BY clause is not in the SELECT list.D. The SELECT statement does not work because the GROUP BY clause should be in the main query and not in the subquery.

Examine the structure of the EMPLOYEES table:What is the correct syntax for an inline view?() A.B.C.D.

Examine the data in the EMPLOYEES table.On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. The JOB_ID column is a NOT NULL column.Evaluate this DELETE statement:Why does the DELETE statement fail when you execute it?()A. There is no row with dept_id 90 in the EMPLOYEES table.B. You cannot delete the JOB_ID column because it is a NOT NULL column.C. You cannot specify column names in the DELETE clause of the DELETE statement.D. You cannot delete the EMPLOYEE_ID column because it is the primary key of the table.

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.

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.

Examine the structure if the EMPLOYEES and NEW EMPLOYEES tables:Which MERGE statement is valid?() A.B.C.D.

Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:Which UPDATE statement is valid?() A.B.C.D.

Examine the data in the EMPLOYEES and DEPARTMENTS tables:Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables:On the EMPLOYEES table, EMPLOYEE_ID is the primary key.MGR_ID is the ID of managers and refers to the EMPLOYEE_ID.DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table.On the DEPARTMENTS table, DEPARTMENT_ID is the primary key.Examine this DELETE statement:What happens when you execute the DELETE statement?()A. Only the row with department ID 40 is deleted in the DEPARTMENTS table.B. The statement fails because there are child records in the EMPLOYEES table with department ID 40.C. 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.D. 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.E. The row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.F. The statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.

Click the Exhibit button and examine the data in the EMPLOYEES table.Which three subqueries work? () A.SELECT * FROM employees where salary (SELECT MIN(salary) FROM employees GROUP BY department_id);B.SELECT * FROM employees WHERE salary = (SELECT AVG(salary) FROM employees GROUP BY department_id);C.SELECT distinct department_id FROM employees WHERE salary ANY (SELECT AVG(salary) FROM employees GROUP BY department_id);D.SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);E.SELECT last_name FROM employees WHERE salary ANY (SELECT MAX(salary) FROM employees GROUP BY department_id);F.SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY AVG(SALARY));

Click the Exhibit button to examine the structures of the EMPLOYEES, DEPARTMENTS, and TAX tables.For which situation would you use a nonequijoin query?() A.to find the tax percentage for each of the employeesB.to list the name, job_id, and manager name for all the employeesC.to find the name, salary, and the department name of employees who are not working with SmithD.to find the number of employees working for the Administrative department and earning less than 4000E.to display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned

Click the Exhibit button and examine the data from the EMP table.The COMMISSION column shows the monthly commission earned by the employee.Which two tasks would require subqueries or joins in order to be performed in a single step? ()A.listing the employees who earn the same amount of commission as employee 3B.finding the total commission earned by the employees in department 10C.finding the number of employees who earn a commission that is higher than the average commission of the companyD.listing the departments whose average commission is more than 600E.listing the employees who do not earn commission and who are working for department 20 in descending order of the employee IDF.listing the employees whose annual commission is more than 6000

Click the Exhibit button and examine the data in the EMPLOYEES table.Examine the subquery:SELECT last_nameFROM employeesWHERE salary IN (SELECT MAX(salary)FROM employeesGROUP BY department_id);Which statement is true?()A.The SELECT statement is syntactically accurate.B.The SELECT statement does not work because there is no HAVING clause.C.The SELECT statement does not work because the column specified in the GROUP BY clause is not in the SELECT list.D.The SELECT statement does not work because the GROUP BY clause should be in the main query and not in the subquery.

You are the administrator of a SQL Server 2000 computer. The server contains database named Sales. Users report that they cannot add new data to the database. You examine the properties of the database. The database is configured as shown in the Sales Properties exhibit.You examine drive E. The hard disk is configured as shown in the Local Disk Properties exhibit.You want the users to be able to add data, and you want to minimize administrative overhead. What should you do?A.Increase the maximum file size of Sales_Data to 1,500MB.B.Use the DBCC SHRINKDATABASE statement.C.Set automatic file growth to 10 percent.D.Create another data file named Sales_Data2 in a new SECONDARY filegroup.

Click the Exhibit button to examine the data of the EMPLOYEES table. Evaluate this SQL statement:SELECT e.employee_id Emp_id, e.emp_name Employee, e.salary, m.employee_id Mgr_id, m.emp_name ManagerFROM employees e JOIN employees m ON (e.mgr_id = m.employee_id)AND e.salary 4000;What is its output?()A.AB.BC.CD.DE.E

Click the Exhibit button and examine the data in the EMPLOYEES table.On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. The JOB_ID column is a NOT NULL column.Evaluate this DELETE statement:DELETE employee_id, salary, job_idFROM employeesWHERE dept_id = 90;Why does the DELETE statement fail when you execute it?()A.There is no row with dept_id 90 in the EMPLOYEES table.B.You cannot delete the JOB_ID column because it is a NOT NULL column.C.You cannot specify column names in the DELETE clause of the DELETE statement.D.You cannot delete the EMPLOYEE_ID column because it is the primary key of the table.

Click the Exhibit button and examine the data from the ORDERS and CUSTOMERS tables.Evaluate the SQL statement:SELECT *FROM ordersWHERE cust_id = (SELECT cust_idFROM customersWHERE cust_name = ‘Smith‘);What is the result when the query is executed?()A.AB.BC.CD.DE.E

He rolled up his trouer leg to exhibit his wounded knee.A:spread B:openC:showD:examine

To examine the Exhibit, press the Exhibit button. Which of the following addresses should be reachable without adding any routes?()A、129.35.57.22B、129.35.54.22C、129.35.58.22D、129.35.254.22

To examine the Exhibit, press the Exhibit button. Which of the following addresses will be reachable without adding any routes?()A、129.35.54.21B、129.35.57.21C、129.35.58.21D、129.35.55.21

Click the Exhibit button and examine the data in the EMPLOYEES table. Which three subqueries work?()A、SELECT * FROM employees where salary (SELECT MIN(salary) FROM employees GROUP BY department_id);B、SELECT * FROM employees WHERE salary = (SELECT AVG(salary) FROM employees GROUP BY department_id);C、SELECT distinct department_id FROM employees WHERE salary ANY (SELECT AVG(salary) FROM employees GROUP BY department_id);D、SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);E、SELECT last_name FROM employees WHERE salary ANY (SELECT MAX(salary) FROM employees GROUP BY department_id);F、SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY AVG(SALARY));

Examine the data in the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMP_NAME DEPT_ID MGR_ID JOB_ID SALARY EMPLOYEE_ID 101 Smith 20 120 SA_REP 4000 102 Martin 10 105 CLERK 2500 103 Chris 20 120 IT_ADMIN 4200 104 John 30 108 HR_CLERK 2500 105 Diana 30 108 IT_ADMIN 5000 106 Smith 40 110 AD_ASST 3000 108 Jennifer 30 110 HR_DIR 6500 110 Bob 40 EX_DIR 8000 120 Ravi 20 110 SA*DIR 6500 DEPARTMENTS DEPARTMENT_ID DEPARTMENT_NAME 10 Admin 20 Education 30 IT 40 Human Resources Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables: CREATE TABLE departments (department_id NUMBER PRIMARY KEY, department _ name VARCHAR2(30)); CREATE TABLE employees (EMPLOYEE_ID NUMBER PRIMARY KEY, EMP_NAME VARCHAR2(20), DEPT_ID NUMBER REFERENCES departments(department_id), MGR_ID NUMBER REFERENCES employees(employee id), MGR_ID NUMBER REFERENCES employees(employee id), JOB_ID VARCHAR2(15). SALARY NUMBER); ON the EMPLOYEES, On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table. On the DEPARTMENTS table, DEPARTMENT_ID is the primary key. Examine this DELETE statement: DELETE FROM departments WHERE department id = 40; What happens when you execute the DELETE statement?()A、Only the row with department ID 40 is deleted in the DEPARTMENTS table.B、The statement fails because there are child records in the EMPLOYEES table with department ID 40.C、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.D、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.E、The row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.F、The statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.

Exhibit, Viewing the Exhibit and examine the Flashback Database architecture. Identify the missing component (shown with a "?`"= in the Flashback Database architecture.()A、 DBWnB、 RVWRC、 ARCHD、 RECOE、 TRWR

单选题View the Exhibit and examine the details of the EMPLOYEES table.  The query was written to format the PHONE_NUMBER for the employees.   Which option would be the correct format in the output?()A  xxx-xxx-xxxxB  (xxx) xxxxxxxC  (xxx) xxx-xxxxD  xxx-(xxx)-xxxx

单选题Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees, whether or not they have matching departments in the departments table. Which query would you use?()ASELECT last_name, department_name FROM employees NATURAL JOIN departments;BSELECT last_name, department_name FROM employees JOIN departments ;CSELECT last_name, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id);DSELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);ESELECT last_name, department_name FROM employees FULL JOIN departments ON (e.department_id = d.department_id);FSELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);

单选题To examine the Exhibit, press the Exhibit button. Which of the following addresses should be reachable without adding any routes?()A129.35.57.22B129.35.54.22C129.35.58.22D129.35.254.22

单选题EXHIBIT, Emp Table Exhibit A Exhibit B Examine the data from the EMP table. Evaluate this SQL statement: SELECT * FROM emp WHERE emp _ id = 3); WHERE commission = (SELECT commission FROM emp What is the result when the query is executed?()A Exhibit AB Exhibit BC The query returns no rowsD The query fails because the outer query is retrieving more than one columnE The query fails because both the inner and outer queries are retrieving data from the same table.