Oracle null test question

Create Table Table1 (ID varchar2 (10) Not null, grzhye number (10, 2), gmsfhm varchar2 (18), rylb varchar2 (10), cardno varchar2 (20 )); comment on column table1.id is 'personal number'; Comment on column table1.grzhye is 'personal account balance '; Comment on column table1.gmsfhm is 'citizenship number '; comment on column table1.rylb is 'personnel class'; Comment on column table1.cardno is 'Card number'; alter table Table1 add constraint pk_table1 primary key (ID); Create index idx_table1_gmsfhm on Table1 (gmsfhm) tablespace Yb; Create index idx_table1_cardno on Table1 (cardno); The data in the table is as follows: Id grzhye gmsfhm rylb cardno 1,100 123456770707771 01 1401000001 2, null 123456770707772 null 3,200 123456770707773 03 1401000003

Each of the first two questions has 10 points, and each other has 5 points. A total of 18 questions have a full score of 100.

1. Select count (*) from Table1 where 1 = 2;

Result ()

A. null B. 0 C. 1 d. An error is reported.

2. Select sum (grzhye) from Table1 where 1 = 2;

Result ()

A. null B. 0 C. 1 d. An error is reported.

3. Select sum (grzhye) from Table1;

Result ()

A. null B. 0 C. 300 D. An error is reported.

4. Select count (*) from (select sum (grzhye) from Table1 where 1 = 2 );

Result ()

A. 0 B. 1 C. null D. An error is reported.

5. Select AVG (grzhye) from Table1;

Result ()

A. 100 B. 0 C. null D. 150

6. Execute the following statement ()

Alter table Table1

Add constraint udx_table1_cardno unique (cardno );

A. Successful B. Error

7. Select * From Table1 where cardno is null; whether the idx_table1_cardno index () is used if the optimization method is based on rules ()

A. Yes B. No

8. Select * From Table1 where cardno = '20160301'; how to optimize the rules and determine whether the idx_table1_cardno index () is used ()

A. Yes B. No

9. Select min (grzhye) from Table1;

The result is ()

A. null B. 100 C. Error

10. Select ID | cardno from Table1 where id = '2 ';

The result will be :()

A. null B. 2 C. Error

11. Select 100 + null from dual; the result is ()

A. null B. 100 C. Error

12. Select 100 * null from dual; the result is ()

A. null B. 100 C. 0 d. Error

13. Select 100/null from dual; the result is ()

A. null B. 100 C. 0 d. Error

14. Select null/0 from dual; the result is ()

A. null B. 0 C. Error

15. Select rylb, sum (grzhye)/count (rylb) from Table1 group by rylb;

() Records will be found

A. 0 B. 2 C. 3 d. Error

16. Select 100/sum (grzhye) from Table1 where id = '2 ';

Result :()

A. null B. 0 C. 100 D. Error

17. Update Table1 set cardno = NULL where id = '2 ';

Update Table1 set cardno = ''where id = '2 ';

The above two sentences ,()

A. The results are the same B. Only the first sentence is successful C. Only the second sentence is successful

18. Select * From Table1 where cardno = '';

Several records will be found ()

A. 0 B. 1 C. Error