更多"The EMP table contains these column"的相关试题:
[单项选择] The EMP table contains these columns: EMPLOYEE_ID NUMBER(4) EMPNAME VARCHAR2 (25) SALARY NUMBER(9,2) HIRE_DATE DATE You query the database with this SQL statement: SELECT empname,hire_date HIREDATE, salary FROM EMP ORDER BY hire_date; How will the results be sorted?（）
A. randomly
B. ascending by date
C. descending by date
D. ascending alphabetically
E. descending alphabetically
[单项选择] The EMP table contains these columns: LAST_NAME VARCHAR2 (25) SALARY NUMBER (6,2) DEPARTMENT_ID NUMBER (6) You need to display the employees who have not been assigned to any department. You write the SELECT statement: SELECT LAST_NAME, SALARY, DEPARTMENT_ID FROM EMP WHERE DEPARTMENT_ID = NULL; What is true about this SQL statement ?（）
A. The SQL statement displays the desired results.
B. The column in the WHERE clause should be changed to display the desired results.
C. The operator in the WHERE clause should be changed to display the desired results.
D. The WHERE clause should be changed to use an outer join to display the desired results.
[单项选择] The EMPLOYEES table contains these columns: EMPLOYEE_ID NUMBER(4) ENAME VARCHAR2 (25) JOB_ID VARCHAR2(10) Which SQL statement will return the ENAME, length of the ENAME, and the numeric position of the letter "a" in the ENAME column, for those employees whose ENAME ends with a the letter "n"?（）
A. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, 'a') FROM EMPLOYEES WHERE SUBSTR (ENAME, -1,1) = 'n';
B. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, ,-1,1) FROM EMPLOYEES WHERE SUBSTR (ENAME, -1,1) = 'n';
C. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR (ENAME, 1,1) = 'n';
D. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR (ENAME, -1,1) = 'n';
[单项选择] The EMPLOYEES table contains these columns: LAST_NAME VARCHAR2 (25) SALARY NUMBER (6,2) COMMISSION_PCT NUMBER (6) You need to write a query that will produce these results: 1.Display the salary multiplied by the commission_pct. 2.Exclude employees with a zero commission_pct. 3.Display a zero for employees with a null commission value. Evaluate the SQL statement: SELECT LAST_NAME, SALARY*COMMISSION_PCT FROM EMPLOYEES WHERE COMMISSION_PCT IS NOT NULL; What does the statement provide?（）
A. all of the desired results
B. two of the desired results
C. one of the desired results
D. an error statement
[单项选择] The EMP table has these columns: ENAME VARCHAR2(35) SALARY NUMBER(8,2) HIRE_DATE DATE Management wants a list of names of employees who have been with the company for more than five years. Which SQL statement displays the required results?（）
A. SELECT ENAME FROM EMP WHERE SYSDATE-HIRE_DATE >5;
B. SELECT ENAME FROM EMP WHERE HIRE_DATE-SYSDATE >5;
C. SELECT ENAME FROM EMP WHERE (SYSDATE_HIRE_DATE)/365 >5;
D. SELECT ENAME FROM EMP WHERE (SYSDATE_HIRE_DATE)*/365 >5;
[多项选择] Examine the data from the EMP table: EMP_ID DEPT_ID COMMISSION 1 10 500 2 20 1000 3 10 4 10 600 5 30 800 6 30 200 7 10 8 20 300 The COMMISSION column shows the monthly commission earned by the employee. Which three tasks would require subqueries or joins in order to perform in a single step? （）
A. Deleting the records of employees who do not earn commission.
B. Increasing the commission of employee 3 by the average commission earned in department 20.
C. Finding the number of employees who do NOT earn commission and are working for department 20.
D. Inserting into the table a new employee 10 who works for department 20 and earns a commission that is equal to the commission earned by employee 3.
E. Creating a table called COMMISSION that has the same structure and data as the columns EMP_ID and COMMISSIONS of the EMP table.
F. Decreasing the commission by 150 for the employees who are working in department 30 and earning a commission of more then 800.
[单项选择] The STUDENT_GRADES table has these columns STUDENT_ID NUMBER(12) SEMESTER_END DATE GPA NUMBER(4,3) Which statement finds students who have a grade point average (GPA) greater than 3.0 for the calendar year 2001?（）
A. SELECT student_id, gpa FROM student_grades WHERE semester_end BETWEEN '01-JAN-2001' AND '31-DEC-2001' OR gpa > 3.;
B. SELECT student_id, gpa FROM student_grades WHERE semester_end BETWEEN '01-JAN-2001' AND '31-DEC-2001' AND gpa gt 3.0;
C. SELECT student_id, gpa FROM student_grades WHERE semester_end BETWEEN '01-JAN-2001' AND '31-DEC-2001' AND gpa > 3.0;
D. SELECT student_id, gpa FROM student_grades WHERE semester_end > '01-JAN-2001' OR semester_end < '31-DEC-2001' AND gpa >=s 3.0;
[单项选择] The BOOKINGS table contains online booking information. When a booking is confirmed, the details are transferred to an archival table BOOKINGS_HIST and deleted from the BOOKINGS table. There is no fixed time interval between each online booking and its confirmation. Because sufficient space is not always available from the delete operations the high-water mark (HWM) is moved up and many rows are inserted below the HWM of the table. The BOOKINGS table has Automatic Segment Space Management (ASSM) and row movement enabled. The table is accessible in 24x7 mode. What is the most efficient method to reclaim the space released by the delete operations in the BOOKINGS table？（）
A. Perform EXPORT, DROP, and IMPORT operations on the BOOKINGS table sequentially
B. Shrink the BOOKINGS table by using the ALTER TABLE SHRINK SPACE command
C. Move the BOOKINGS table to a different location by using the ALTER TABLE MOVE command
D. Deallocate the space in the BOOKINGS table by using the ALTER TABLE DEALLOCATE UNUSED command
[单项选择] The EMPLOYEE table contains the following information: EMPNO NAME WORKDEPT 101 SAM A11 102 JOHN C12 103 JANE -104 PAT Remote 105 ANNE -106 BOB A11 The MANAGER table contains the following information: MGRID NAME DEPTNO EMPCOUNT 1 WU B01 - 2 JONES A11 - 3 CHEN - - 4 SMITH - -5 THOMAS C12 - After this statement is executed: UPDATE manager m SET empcount = (SELECT COUNT(workdept) FROM employee e WHERE workdept=m.deptno) What is the result of the following query?（） SELECT mgrid, empcount FROM MANAGER WHERE empcount IS NOT NULL ORDER BY mgrid
A. MGRID EMPCOUNT ----- -------- 1 0 2 2 5 1
B. MGRID EMPCOUNT ----- -------- 1 0 2 2 3 0 4 0 5 1
C. MGRID EMPCOUNT ----- -------- 1 3 2 3 3 3 4 3 5 3
D. MGRID EMPCOUNT ----- -------- 1 0 2 2 3 2 4 2 5 1
