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The Oracle 1z0-071 (Oracle Database SQL) Certification Exam is designed to test the knowledge and skills of professionals in the field of database management. Oracle is a leading provider of database management systems, and this certification exam is one of the most widely recognized and respected credentials in the industry. It is an entry-level exam that covers the basic concepts and principles of SQL programming, as well as the features and functionality of the Oracle Database.
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Oracle Database SQL Sample Questions (Q232-Q237):
NEW QUESTION # 232
You must create a table EMPLOYEES in which the values in the columns EMPLOYEES_ID and LOGIN_ID must be unique and not null.
Which two SQL statements would create the required table? (Choose two.) CREATE TABLE employees
- A. (employee_id NUMBER,
login_id NUMBER,
employee_name VARCHAR2(100),
hire_date DATE,
CONSTRAINT emp_id_uk UNIQUE (employee_id, login_id));
CREATE TABLE employees - B. (employee_id NUMBER CONSTRAINT emp_id_pk PRIMARY KEY,
login_id NUMBER UNIQUE,
employee_name VARCHAR2(25),
hire_date DATE);
CREATE TABLE employees - C. (employee_id NUMBER,
login_id NUMBER,
employee_name VARCHAR2(25),
hire_date DATE,
CONSTRAINT emp_id_pk PRIMARY KEY (employee_id, login_id));
CREATE TABLE employees - D. (employee_id NUMBER CONSTRAINT emp_id_nn NOT NULL,
login_id NUMBER CONSTRAINT login_id_nn NOT NULL,
employee_name VARCHAR2(100),
hire_date DATE,
CONSTRAINT emp_num_id_uk UNIQUE (employee_id, login_id)); - E. (employee_id NUMBER,
login_id NUMBER,
employee_name VARCHAR2(100),
hire_date DATE,
CONSTRAINT emp_id_uk UNIQUE (employee_id, login_id);
CONSTRAINT emp_id_nn NOT NULL (employee_id, login_id));
CREATE TABLE employees
Answer: C,D
Explanation:
Explanation
NEW QUESTION # 233
Examine the data in the CUST_NAME column of the CUSTOMERS table.
You want to extract only those customer names that have three names and display the * symbol in place of the first name as follows:
Which two queries give the required output?
- A. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')),LENGTH(cust_name),'*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,2)<>0;
- B. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')),LENGTH(cust_name),'*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',-1,2)<>0;
- C. SELECT LPAD(SUBSTR(cust_name, INSTR (cust_name ' ')),LENGTH(cust_name) - INSTR(cust_name, ' '), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,2)<>0;
- D. SELECT LPAD(SUBSTR(cust_name, INSTR (cust_name ' ')),LENGTH(cust_name) - INSTR(cust_name, ' '), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,-2)<>0;
Answer: A,B
NEW QUESTION # 234
View the Exhibit and examine the structure of the ORDERS table.
Which UPDATE statement is valid?
- A. UPDATE ordersSET order_date = TO_DATE('12-mar-2007','dd-mon-yyyy'),SET order_total = TO_NUMBER (NULL)WHERE order_id = 2455;
- B. UPDATE ordersSET order_date = '12-mar-2007',order_total = NULLWHERE order_id = 2455;
- C. UPDATE ordersSET order_date = '12-mar-2007',order_total IS NULLWHERE order_id = 2455;
- D. UPDATE ordersSET order_date = '12-mar-2007',AND order_total = TO_NUMBER(NULL)WHERE order_id = 2455;
Answer: B
NEW QUESTION # 235
Which two statements are true about Oracle synonyms?
- A. All private synonym names must be unique in the database.
- B. Any user can create a PUBLIC synonym.
- C. A synonym has an object number.
- D. A synonym can be created on an object in a package.
- E. A synonym can have a synonym.
Answer: C,E
NEW QUESTION # 236
Which two statements are true about the DUAL table?
- A. It can display multiple rows but only a single column.
- B. It can be accessed by any user who has the SELECT privilege in any schema
- C. It consists of a single row and single column of VARCHAR2 data type.
- D. It can be used to display only constants or pseudo columns.
- E. It can display multiple rows and columns.
- F. It can be accessed only by the SYS user.
Answer: B,E
NEW QUESTION # 237
......
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