616.The EMP table has some discrepancy in data entry with a particular employee ID. You execute the query as shown in the Exhibit to retrieve all versions of the row that exist between two SCNs.View the Exhibit.
Which two statements about the results of the query shown in the Exhibit are correct? (Choose two.)
Exhibit:

A. The LAST_SCN value in the first row is NULL, which means that the versions of the row still exist at SCN 6636300.
B. The LAST_SCN value in the second row in NULL, which means that the version of the row still exists at SCN 6636300.
C. The LAST_SCN value in the third row is 6636280, which means that the version of row exists above SCN 6636280.
D. The LAST_SCN value in the second row is NULL, which means that the version of the row no longer exists because it was deleted.
Answer: AD

参考：http://docs.oracle.com/cd/E11882_01/appdev.112/e41502/adfns_flashback.htm#ADFNS1006

SQL> create table emp (ename varchar2(10));

表已创建。

SQL> insert into emp values ('Jones');

已创建 1 行。

SQL> commit;

提交完成。

SQL> delete emp where ename='Jones';

已删除 1 行。

SQL> commit;

提交完成。

SQL> insert into emp values ('Smith');

已创建 1 行。

SQL> commit;

提交完成。

SQL> select versions_xid as vxid,
2 versions_startscn as f_scn,
3 versions_endscn as L_scn,
4 versions_operation as oper,ename
5 from emp
6 versions between scn minvalue and maxvalue
7 ;

VXID                  F_SCN      L_SCN O ENAME
---------------- ---------- ---------- - ----------
01001F00A30B0000    4643139            I Smith
03000C004A0C0000    4643121            D Jones
05000F005F0C0000    4643112    4643121 I Jones

ITPub博客

1Z0-053 V13.02 Q616

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