native barcode generator for crystal reports crack What Generates the Most and Least Undo in Objective-C

Encode Data Matrix 2d barcode in Objective-C What Generates the Most and Least Undo

This is a frequently asked but easily answered question. The presence of indexes (or the fact that a table is an index-organized table) may affect the amount of undo generated dramatically, as indexes are complex data structures and may generate copious amounts of undo information. That said, an INSERT will, in general, generate the least amount of undo, since all Oracle needs to record for this is a rowid to delete. An UPDATE is typically second in the race (in most cases). All that needs to be recorded are the changed bytes. It is most common that you UPDATE some small fraction of the entire row s data. Therefore, a small fraction of the row must be remembered in the undo. Many of the previous examples run counter to this rule of thumb, but that s because they update large, fixedsized rows and they update the entire row. It is much more common to UPDATE a row and change a small percentage of the total row. A DELETE will, in general, generate the most undo. For a DELETE, Oracle must record the entire row s before image into the undo segment. The previous temporary table example, with regard to redo generation, demonstrated that fact: the DELETE generated the most redo, and since the only logged element of the DML operation on a temporary table is the undo, we in fact observed that the DELETE generated the most undo. The INSERT generated very little undo that needed to be logged. The UPDATE generated an amount equal to the before image of the data that was changed, and the DELETE generated the entire set of data written to the undo segment. As previously mentioned, you must also take into consideration the work performed on an index. You ll find that an update of an unindexed column not only executes much faster, it also tends to generate significantly less undo than an update of an indexed column. For example, we ll create a table with two columns, both containing the same information, and index one of them: ops$tkyte%ORA11GR2> create table t 2 as 3 select object_name unindexed, 4 object_name indexed 5 from all_objects

6 / Table created. ops$tkyte%ORA11GR2> create index t_idx on t(indexed); Index created. ops$tkyte%ORA11GR2> exec dbms_stats.gather_table_stats(user,'T'); PL/SQL procedure successfully completed. Now we ll update the table, first updating the unindexed column and then the indexed column. We ll need a new V$ query to measure the amount of undo we ve generated in each case. The following query accomplishes this for us. It works by getting our session ID (SID) from V$MYSTAT, using that to find our record in the V$SESSION view, and retrieving the transaction address (TADDR). It uses the TADDR to pull up our V$TRANSACTION record (if any) and selects the USED_UBLK column the number of used undo blocks. Since we currently are not in a transaction, we expect it to return 0 rows right now: ops$tkyte%ORA11GR2> update t set unindexed = lower(unindexed); 72077 rows updated. ops$tkyte%ORA11GR2> select 2 from v$transaction 3 where addr = (select 4 from 5 where 6 7 8 9 ) 10 / USED_UBLK ---------1214 ops$tkyte%ORA11GR2> commit; Commit complete. That UPDATE used 1,214 blocks to store its undo. The commit would free that up, or release it, so if we rerun the query against V$TRANSACTION, it would once again show us no rows selected. When we update the same data only indexed this time we ll observe the following: ops$tkyte%ORA11GR2> update t set indexed = lower(indexed); 72077 rows updated. ops$tkyte%ORA11GR2> select 2 from v$transaction 3 where addr = (select 4 from 5 where 6 7 8 9 ) used_ublk taddr v$session sid = (select sid from v$mystat where rownum = 1 ) used_ublk taddr v$session sid = (select sid from v$mystat where rownum = 1 )