Oracle null test question

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 category ';

Comment on column TABLE1.CARDNO is 'Card No ';

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

1,100,123 456770707771, 01,140 201701

2, null, 123456770707772, null, null

3,200,123 456770707773, 03,140 201703

1. select count (*) from table1 where 1 = 2; the result is ()

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

2. select sum (grzhye) from table1 where 1 = 2; the result is ()

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

3. select sum (grzhye) from table1; the result is ()

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

4. select count (*) from (select sum (grzhye) from table1 where 1 = 2); the result is ()

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

5. select avg (grzhye) from table1; the result is ()

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

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 is :()

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'; the result is :()

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

19. select * from table1 where cardno is null; several records will be found ()

A. 0 B. 1 C. Error

20. select count (cardno) from table1; several records () will be found ()

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

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