更多"Examine the structure of the EMP_DE"的相关试题:
[单项选择] 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.
[单项选择] Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use?（）
A. SELECT e.last_name, d. department_name, d.location_id FROM employees e NATURAL JOIN departments D USING department_id ;
B. SELECT last_name, department_name, location_id FROM employees NATURAL JOIN departments WHERE e.department_id =d.department_id;
C. SELECT e.last_name, d.department_name, d.location_id FROM employees e NATURAL JOIN departments d;
D. SELECT e.last_name, d.department_name, d.location_id FROM employees e JOIN departments d USING (department_id );
[多项选择] Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables. EMPLOYEES NOT NULL, EMPLOYEE_ID NUMBER Primary Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, Primary DEPARTMENT_ID NUMBER Key VARCHAR2 DEPARTMENT_NAME (30) References NGR_ID MGR_ID NUMBER column of the EMPLOYEES table Foreign key to LOCATION_ID NUMBER LOCATION_ID column of the LOCATIONS table LOCATIONS NOT NULL, Primary LOCATION_ID NUMBER Key VARCHAR2 CITY |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;
[单项选择] Examine the structure of the EMPLOYEES table: Column name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key LAST_NAME VARCNAR2(30) FIRST_NAME VARCNAR2(30) JOB_ID NUMBER SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER You need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task? （）
A. CREATE INDEX NAME _IDX (first_name, last_name);
B. CREATE INDEX NAME _IDX (first_name, AND last_name)
C. CREATE INDEX NAME_IDX ON (First_name, last_name);
D. CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);
E. CREATE INDEX NAME_IDX ON employees (First_name, last_name);
F. CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
[单项选择] Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE Which INSERT statement is valid?（）
A. INSERT INTO employees (employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01/01/01);
B. INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01 january 01');
C. INSERT INTO employees(employee_id, first_name, last_name, Hire_date) VALUES (1000, 'John', 'smith', To_ date ('01/01/01));
D. INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01-Jan-01');
[单项选择] Examine the structure of the STUDENTS table: STUDENT_ID NUMBER NOT NULL, Primary Key STUDENT_NAME VARCHAR2(30) COURSE_ID VARCHAR2(10) NOT NULL MARKS NUMBER START_DATE DATE FINISH_DATE DATE You 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?（）
A. SELECT student_ id, marks, ROWNUM "Rank" FROM students WHERE ROWNUM <= 10 AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99 AND course_id = 'INT_SQL' ORDER BY mark DESC;
B. SELECT student_id, marks, ROWID "Rank" FROM students WHERE ROWID <= 10 AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99' AND course_id = 'INT_SQL' ORDER BY mark;
C. SELECT student_id, marks, ROWNUM "Rank" FROM (SELECT student_id, marks FROM students WHERE ROWNUM <= 10 AND finish_date BETWEEN '01-JAN-99' AND '31-DEC- 99' AND course_id = 'INT_SQL' ORDER BY mark DESC;
D. SELECT student_id, marks, ROWNUM "Rank" FROM (SELECT student_id, marks FROM students WHERE (finish_date BETWEEN '01-JAN-99 AND '31-DEC-99' AND course_id = 'INT_SQL' ORDER BY marks DESC) WHERE ROWNUM <= 10;
E. SELECT student id, marks, ROWNUM "Rank" FROM (SELECT student_id, marks FROM students ORDER BY marks) WHERE ROWNUM <= 10 AND finish date BETWEEN '01-JAN-99' AND '31-DEC-99' AND course _ id 'INT_SQL';
