今天做个试验，delete未提交rollback是否会改变行号？
创建测试表： create table dxmy (id number,name varchar2(100)); 插入数据 insert into dxmy values(1,'dxmy'); insert into dxmy values(2,'dxmy'); insert into dxmy values(3,'dxmy'); insert into dxmy values(4,'dxmy'); commit; 查看行号 SQL> select dbms_rowid.rowid_relative_fno(rowid) fno, 2 dbms_rowid.rowid_block_number(rowid) block_id, 3 dbms_rowid.rowid_row_number(rowid) row_id, 4 id 5 from dxmy; FNO BLOCK_ID ROW_ID ID ---------- ---------- ---------- ---------- 4 462 0 1 4 462 1 2 4 462 2 3 4 462 3 4 delete，但是不提交： SQL> delete dxmy where id = 1; 1 row deleted SQL> SQL> select dbms_rowid.rowid_relative_fno(rowid) fno, 2 dbms_rowid.rowid_block_number(rowid) block_id, 3 dbms_rowid.rowid_row_number(rowid) row_id, 4 id 5 from dxmy; FNO BLOCK_ID ROW_ID ID ---------- ---------- ---------- ---------- 4 462 1 2 4 462 2 3 4 462 3 4 rollback： SQL> rollback; Rollback complete SQL> select dbms_rowid.rowid_relative_fno(rowid) fno, 2 dbms_rowid.rowid_block_number(rowid) block_id, 3 dbms_rowid.rowid_row_number(rowid) row_id, 4 id 5 from dxmy; FNO BLOCK_ID ROW_ID ID ---------- ---------- ---------- ---------- 4 462 0 1 4 462 1 2 4 462 2 3 4 462 3 4 行号未改变。 那如果delete之后，我提交了，然后重新插入呢？ SQL> delete dxmy where id = 1; 1 row deleted SQL> commit; Commit complete SQL> insert into dxmy values(1,'dxmy'); 1 row inserted SQL> commit; Commit complete SQL> select dbms_rowid.rowid_relative_fno(rowid) fno, 2 dbms_rowid.rowid_block_number(rowid) block_id, 3 dbms_rowid.rowid_row_number(rowid) row_id, 4 id 5 from dxmy; FNO BLOCK_ID ROW_ID ID ---------- ---------- ---------- ---------- 4 462 1 2 4 462 2 3 4 462 3 4 4 463 0 1 <pre name="code" class="html">SQL> delete dxmy where id = 1; 1 row deleted SQL> commit; Commit complete
原文：http://blog.csdn.net/ghostliming/article/details/51736392