评估此CREATE TABLE语句的执行结果： CREATE TABLE part( part_id NUMBER, part_name VARCHAR2(25), manufacturer_id NUMBER(9), retail_price NUMBER(7,2) NOT NULL, CONSTRAINT part_id_pk PRIMARY KEY(part_id), CONSTRAINT cost_nn NOT NULL(cost), CONSTRAINT FOREIGN KEY (manufac

第1题：

数据库中的中的PROFITS表中PRODUCT_NAME与SALE_PERIOD列建立主键。下列哪个语句无法定义这个主键？（）

• A、create table profits（product_name varchar2（10），sale_perion varchar2（10），profit number， constraint pk_profits_01 primary key（product_name，sale_period））
• B、alter table profits add constraint pk_profits_01 primary key（product_name，sale_period）deferrabl
• C、以上两者都是

第2题：

When defining a referential constraint between the parent table T2 and the dependent table T1, which of the following is true?（）

• A、The list of column names in the FOREIGN KEY clause can be a subset of the list of column names in the primary key of T2 or a UNIQUE constraint that exists on T2.
• B、The list of column names in the FOREIGN KEY clause can be a subset of the list of column names in the primary key of T1 or a UNIQUE constraint that exists on T1.
• C、The list of column names in the FOREIGN KEY clause must be identical to the list of column names in the primary key of T2 or a UNIQUE constraint that exists on T2.
• D、The list of column names in the FOREIGN KEY clause must be identical to the list of column names in the primary key of T1 or a UNIQUE constraint that exists on T1.

第3题：

Which two statements are true about the primary key constraint in a table？ （）

• A、It is not possible to disable the primary key constraint.
• B、It is possible to have more than one primary key constraint in a single table.
• C、The primary key constraint can be referred by only one foreign key constraint.
• D、The primary key constraint can be imposed by combining more than one column.
• E、The non-deferrable primary key constraint creates an unique index on the primary key column if it is not already indexed.

第4题：

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 stores information 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 that points to the STUDENT_ID column of the STUDENTS table. Which statement creates the foreign key?（）

• A、CREATE TABLE student_grades (student_id NUMBER(12),semester_end DATE, gpa NUMBER(4,3), CONSTRAINT student_id_fk REFERENCES (student_id) FOREIGN KEY student (student_id));
• B、CREATE TABLE student_grades(student_id NUMBER(12),semester_end DATE, gpa NUMBER(4,3), student_id_fk FOREIGN KEY (student_id) REFERENCES students (student_id));
• C、CREATE TABLE student_grades(student_id NUMBER(12),semester_end DATE, gpa NUMBER(4,3), CONSTRAINT FOREIGN KEY (student_id) REFERENCES student (student_id));
• D、CREATE TABLE student_grades(student_id NUMBER(12),semester_end DATE, gpa NUMBER(4,3), CONSTRAINT student_id_fk FOREIGN KEY (student_id) REFERENCES students (student_id));

第5题：

Examine the command that is used to create a table：  SQL> CREATE TABLE orders（oid NUMBER（6） PRIMARY KEY， odate DATE， ccode NUMBER（6）， oamt NUMBER（10，2）） TABLESPACE users；  Which two statements are true about the effect of the above command） （）

• A、A CHECK constraint is created on the OID column.
• B、A NOT NULL constraint is created on the OID column.
• C、The ORDERS table is the only object created in the USERS tablespace.
• D、The ORDERS table and a unique index are created in the USERS tablespace.
• E、The ORDERS table is created in the USERS tablespace and a unique index is created on the OIDcolumn in the SYSTEM tablespace.

第6题：

Which SQL statement defines a FOREIGN KEY constraint on the DEPTNO column of the EMP table?（）

• A、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk FOREIGN KEY deptno REFERENCES dept deptno);
• B、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
• C、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk REFERENCES dept (deptno) FOREIGN KEY (deptno));
• D、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) FOREIGN KEY CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));

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第13题：

Given the following requirements: Create a table to contain employee data, with a unique numeric identifier automatically assigned when a row is added, has an EDLEVEL column that permits only the values 'C', 'H' and 'N', and permits inserts only when a corresponding value for the employee's department exists in the DEPARTMENT table. Which of the following CREATE statements will successfully create this table?（）

• A、CREATE TABLE emp ( empno SMALLINT NEXTVAL GENERATED ALWAYS AS IDENTITY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30) NOT NULL, workdept CHAR(3) NOT NULL, edlevel CHAR(1), PRIMARY KEY emp_pk (empno), FOREIGN KEY emp_workdept_fk ON (workdept) REFERENCES department (deptno), CHECK edlevel_ck VALUES (edlevel IN ('C','H','N')), );
• B、CREATE TABLE emp ( empno SMALLINT NOT NULL GENERATED BY DEFAULT AS IDENTITY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30) NOT NULL, workdept CHAR(3), edlevel CHAR(1), CONSTRAINT emp_pk PRIMARY KEY (empno), CONSTRAINT emp_workdept_fk FOREIGN KEY (workdept) REFERENCES department (deptno), CONSTRAINT edlevel_ck CHECK edlevel VALUES ('C','H','N') );
• C、CREATE TABLE emp ( empno SMALLINT NEXTVAL GENERATED BY DEFAULT AS IDENTITY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30) NOT NULL, workdept CHAR(3) NOT NULL, edlevel CHAR(1) CHECK IN ('C','H','N')), CONSTRAINT emp_pk PRIMARY KEY (empno), CONSTRAINT emp_workdept_fk FOREIGN KEY department (deptno) REFERENCES (workdept) );
• D、CREATE TABLE emp ( empno SMALLINT NOT NULL GENERATED BY DEFAULT AS IDENTITY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30) NOT NULL, workdept CHAR(3), edlevel CHAR(1), CONSTRAINT emp_pk PRIMARY KEY (empno), CONSTRAINT emp_workdept_fk FOREIGN KEY (workdept) REFERENCES department (deptno), CONSTRAINT edlevel_ck CHECK (edlevel IN ('C','H','N')) );

第14题：

Evaluate the CREATE TABLE statement：   CREATE TABLE products   （product_id NUMBER （6）  CONSTRAINT prod_id_pk PRIMARY KEY，  product_name VARCHAR2 （15））；   Which statement is true regarding the PROD_ID_PK constraint？（）

• A、 It would be created only if a unique index is manually created first.
• B、 It would be created and would use an automatically created unique index.
• C、 It would be created and would use an automatically created nonunique index.
• D、 It would be created and remains in a disabled state because no index is specified in the command.

第15题：

Which two statements are true about constraints? （）

• A、The UNIQUE constraint does not permit a null value for the column.
• B、A UNIQUE index gets created for columns with PRIMARY KEY and UNIQUE constraints.
• C、The PRIMARY KEY and FOREIGN KEY constraints create a UNIQUE index.
• D、The NOT NULL constraint ensures that null values are not permitted for the column.

第16题：

Which SQL statement defines the FOREIGN KEY constraint on the DEPTNO column of the EMP table?（）

• A、CREATE TABLE EMP (empno NUMBER(4), ename VARCNAR2(35), deptno NUMBER(7,2) NOT NULL CONSTRAINT emp_deptno_fk FOREIGN KEY deptno REFERENCES dept deptno);
• B、CREATE TABLE EMP (empno NUMBER(4), ename VARCNAR2(35), deptno NUMBER(7,2) CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
• C、CREATE TABLE EMP (empno NUMBER(4) ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk REFERENCES dept (deptno) FOREIGN KEY (deptno));
• D、CREATE TABLE EMP (empno NUMBER(4), ename VARCNAR2(35), deptno NUMBER(7,2) FOREIGN KEY CONSTRAINT emp deptno fk REFERENCES dept (deptno));

第17题：

Examine the following command: CREATE TABLE (prod_id number(4), Prod_name varchar2 (20), Category_id number(30), Quantity_on_hand number (3) INVISIBLE); Which three statements are true about using an invisible column in the PRODUCTS table?（）

• A、The %ROWTYPE attribute declarations in PL/SQL to access a row will not display the invisible column in the output.
• B、The DESCRIBE commands in SQL *Plus will not display the invisible column in the output.
• C、Referential integrity constraint cannot be set on the invisible column.
• D、The invisible column cannot be made visible and can only be marked as unused.
• E、A primary key constraint can be added on the invisible column.

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