A. The sort is in ascending by order by default.
B. The sort is in descending order by default.
C. The ORDER BY clause must precede the WHERE clause.
D. The ORDER BY clause is executed on the client side.
E. The ORDER BY clause comes last in the SELECT statement.
F. The ORDER BY clause is executed first in the query execution.
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A. Convert 10 to ‘TEN’
B. Convert ‘10’ to 10
C. Convert ‘10’ to ‘10’
D. Convert ‘TEN’ to 10
E. Convert a date to a character expression
F. Convert a character expression to a date
A. the use of rowid
B. a GROUP BY clause
C. an ORDER BY clause
D. only an inline view
E. an inline view and an outer query
The CUSTOMERS table has these columns:
The CUSTOMER_ID column is the primary key for the table.Which two statements find the number of customers? ()
A. SELECT TOTAL(*) FROM customers;
B. SELECT COUNT(*) FROM customers;
C. SELECT TOTAL(customer_id) FROM customers;
D. SELECT COUNT(customer_id) FROM customers;
E. SELECT COUNT(customers) FROM customers;
F. SELECT TOTAL(customer_name) FROM customers;
Examine the description of the EMPLOYEES table:
EMP_ID NUMBER(4) NOT NULL
LAST_NAME VARCHAR2(30) NOT NULL
FIRST_NAME VARCHAR2(30)
DEPT_ID NUMBER(2)
Which statement produces the number of different departments that have employees with last name Smith?()
A. SELECT COUNT (*) FROM employees WHERE last _name='smith';
B. SELECT COUNT (dept_id) FROM employees WHERE last _name='smith';
C. SELECT DISTINCT (COUNT (dept_id) FROM employees WHERE last _name='smith';
D. SELECT COUNT (DISTINCT dept_id) FROM employees WHERE last _name='smith';
E. SELECT UNIQE (dept_id) FROM employees WHERE last _name='smith';
A. INSERT
B. UPDATE
C. SELECT
D. DESCRIBE
E. DELETE
F. RENAME
A. DROP emp_dept_uv;
B. DELETE emp_dept_uv;
C. REMOVE emp_dept_uv;
D. DROP VIEW emp_dept_uv;
E. DELETE VIEW emp_dept_uv;
F. REMOVE VIEW emp_dept_uv;
Examine the data from the EMP table:
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.
A. CASCADE
B. UNIQUE
C. NONUNIQUE
D. CHECK
E. PRIMARY KEY
F. CONSTANT
G. NOT NULL
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';
Examine the structure of the EMP_DEPT_VU view:
Column Name Type Remarks
EMPLOYEE_ID NUMBER From the EMPLOYEES table
EMP_NAME VARCHAR2(30) From the EMPLOYEES table
JOB_ID VARCHAR2(20) From the EMPLOYEES table
SALARY NUMBER From the EMPLOYEES table
DEPARTMENT_ID NUMBER From the DEPARTMENTS table
DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table
Which SQL statement produces an error?()
A. SELECT * FROM emp_dept_vu;
B. SELECT department_id, SUM(salary) FROM emp_dept_vu GROUP BY department _ id;
C. SELECT department_id, job_id, AVG(salary) FROM emp_dept_vu GROUP BY department _ id, job_id;
D. SELECT job_id, SUM(salary) FROM emp_dept_vu WHERE department_id IN (10,20) GROUP BY job_id HAVING SUM (salary) > 20000
E. None of the statements produce an error; all are valid.
最新试题
The PRODUCTS table has these columns:PRODUCT_ID NUMBER(4)PRODUCT_NAME VARCHAR2(45)PRICE NUMBER(8,2)Evaluate this SQL statement:SELECT *FROM PRODUCTSORDER BY price, product _ name;What is true about the SQL statement? ()
Which three statements about subqueries are true? ()
Which one is a system privilege? ()
Which SQL statement displays the date March 19, 2001 in a format that appears as "Nineteenth of March 2001 12:00:00 AM"? ()
In which two cases would you use an outer join? ()
Which two are true about aggregate functions?()
Which statement accomplish this? ()
Examine the structure of the STUDENTS table:STUDENT_ID NUMBER NOT NULL, Primary KeySTUDENT_NAME VARCHAR2(30)COURSE_ID VARCHAR2(10) NOT NULLMARKS NUMBERSTART_DATE DATEFINISH_DATE DATEYou need to create a report of the 10 students who achieved the highest ranking in the course INT SQL and who completed the course in the year 1999.Which SQL statement accomplishes this task? ()
What is true about sequences? ()
You need to design a student registration database that contains several tables storing academic information.The STUDENTS table stores information about a student. The STUDENT_GRADES table storesinformation about the student's grades. Both of the tables have a column named STUDENT_ID. The STUDENT_ID column in the STUDENTS table is a primary key.You need to create a foreign key on the STUDENT_ID column of the STUDENT_GRADES table thatpoints to the STUDENT_ID column of the STUDENTS table. Which statement creates the foreign key?()