您要对EMPLOYEES表的FIRST_NAME和LAST_NAME列创建一个组合索引。以下哪条语句将完成此任务()
第1题:
“雇员”表包含以下列: EMPLOYEE_ID NOT NULL, Primary Key SSNUM NOT NULL, Unique LAST_NAME VARCHAR2(25) FIRST_NAME VARCHAR2(25) DEPARTMENT_ID NUMBER 部门表中 DEPARTMENT_ID 列的外键 SALARY NUMBER(8,2) 如果执行以下语句: CREATE INDEX emp_name_idx ON employees(last_name, first_name); 以下哪个说法是的()
第2题:
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?()
第3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?()
第4题:
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?()
第5题:
Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE Which UPDATE statement is valid?()
第6题:
You get an error because of a primary key violation.
The data and structure of the EMPLOYEES table are deleted.
The data in the EMPLOYEES table is deleted but not the structure.
You get an error because the statement is not syntactically correct.
第7题:
SELECT COUNT(*) FROM employees WHERE last_name='Smith';
SELECT COUNT(dept_id) FROM employees WHERE last_name='Smith';
SELECT DISTINCT(COUNT(dept_id)) FROM employees WHERE last_name='Smith';
SELECT COUNT(DISTINCT dept_id) FROM employees WHERE last_name='Smith';
SELECT UNIQUE(dept_id) FROM employees WHERE last_name='Smith';
第8题:
INSERT INTO employees VALUES ( NULL, 'John', 'Smith');
INSERT INTO employees( first_name, last_name) VALUES( 'John', 'Smith');
INSERT INTO employees VALUES ( '1000', 'John', NULL);
INSERT INTO employees (first_name, last_name, employee_id) VALUES ( 1000, 'John', 'Smith');
INSERT INTO employees (employee_id) VALUES (1000);
INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'John', ' ');
第9题:
UPDATE employees SET first_name = 'John' SET last_name = 'Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John', SET last_name = 'Smoth' WHERE employee_id = 180;
UPDATE employee SET first_name = 'John' AND last_name = 'Smith' WHERE employee_id = 180;
UPDATE employee SET first_name = 'John', last_name = 'Smith' WHERE employee_id = 180;
第10题:
ALTER INDEX ln_idx ON employees(first_name)
ALTER INDEX ln_idx TO employees(first_name)
ALTER INDEX ln_idx TO fn_idx ON employees(first_name)
以上都不能;您无法变更索引
第11题:
INSERT INTO employees (employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01/01/01);
INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01 january 01');
INSERT INTO employees(employee_id, first_name, last_name, Hire_date) VALUES (1000, 'John', 'smith', To_ date ('01/01/01));
INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01-Jan-01');
第12题:
CREATE INDEX NAME _IDX (first_name, last_name);
CREATE INDEX NAME _IDX (first_name, AND last_name)
CREATE INDEX NAME_IDX ON (First_name, last_name);
CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);
CREATE INDEX NAME_IDX ON employees (First_name, last_name);
CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
第13题:
评估此CREATE VIEW 语句: CREATE VIEW pt_view AS (SELECT first_name, last_name, status, courseid, subject, term FROM faculty f, course c WHERE f.facultyid = c.facultyid); 此语句将创建什么类型的视图()
第14题:
Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) Which three statements inserts a row into the table? ()
第15题:
Which three statements inserts a row into the table?()
第16题:
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? ()
第17题:
Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2 (25) LAST_NAME VARCHAR2 (25) HIRE_DATE DATE Which UPDATE statement is valid? ()
第18题:
此语句会创建一个基于函数的索引
因为语法错误,此语句将失败
该语句将创建一个组合唯一索引
该语句将创建一个组合非唯一索引
第19题:
INSERT INTO employees VALUES ( NULL, ‘John’,‘Smith’);
INSERT INTO employees( first_name, last_name) VALUES(‘John’,‘Smith’);
INSERT INTO employees VALUES (‘1000’,‘John’,NULL);
INSERT INTO employees(first_name,last_name, employee_id) VALUES ( 1000, ‘John’,‘Smith’);
INSERT INTO employees (employee_id) VALUES (1000);
INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, ‘John’,‘’);
第20题:
UPDATE employees SET first_name = 'John' SET last_name = 'Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John', SET last_name = 'Smoth' WHERE employee_id = 180;
UPDATE employee SET first_name = 'John' AND last_name = 'Smith' WHERE employee_id = 180;
UPDATE employee SET first_name = 'John', last_name = 'Smith' WHERE employee_id = 180;
第21题:
嵌套
简单
内嵌
复杂
第22题:
UPDATE employees SET first_name = 'John' SET last_name ='Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John', SET last_name ='Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John' AND last_name ='Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John', last_name ='Smith' WHERE employee_id = 180;
第23题:
INSERT INTO employees VALUES (NULL, 'JOHN','Smith');
INSERT INTO employees( first_name, last_name) VALUES ('JOHN','Smith');
INSERT INTO employees VALUES ('1000','JOHN','NULL');
INSERT INTO employees(first_name,last_name, employee_id) VALUES ('1000, 'john','Smith');
INSERT INTO employees (employee_id) VALUES (1000);
INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'john',);