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CHAPTER 7 CONCURRENCY AND MULTI-VERSIONING
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When we run TKPROF and view the results, we ll see something like this (note that I removed the ELAPSED, CPU, and DISK columns from this report): select * from t call count ------- -----Parse 1 Execute 1 Fetch 2 ------- -----total 4 query current ------ ---------0 0 0 0 7 0 ------ ---------7 0 rows ---------0 0 1 ---------1
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update t t1 set x = x+1 call count ------- -----Parse 1 Execute 1 Fetch 0 ------- -----total 2 query current ------ ---------0 0 7 3 0 0 ------ ---------7 3 rows ---------0 1 0 ---------1
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update t t2 set x = x+1 call count ------- -----Parse 1 Execute 1 Fetch 0 ------- -----total 2 query current ------ ---------0 0 7 1 0 0 ------ ---------7 1 rows ---------0 1 0 ---------1
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So, during just a normal query, we incur seven query (consistent) mode gets. During the first UPDATE, we incur the same seven I/Os (the search component of the update involves finding all of the rows that are in the table when the update began, in this case) and three current mode gets as well. The current mode gets are performed in order to retrieve the table block as it exists right now, the one with the row on it, to get an undo segment block to begin our transaction, and an undo block. The second update has exactly one current mode get; since we did not have to do the undo work again, we had only the one current get on the block with the row we want to update. The very presence of the current mode gets tells us that a modification of some sort took place. Before Oracle will modify a block with new information, it must get the most current copy of it. So, how does read consistency affect a modification Well, imagine you were executing the following UPDATE statement against some database table: Update t set x = x+1 where y = 5; We understand that the WHERE Y=5 component, the read-consistent phase of the query, will be processed using a consistent read (query mode gets in the TKPROF report). The set of WHERE Y=5 records that was committed in the table at the beginning of the statement s execution are the records it will see (assuming READ COMMITTED isolation; if the isolation is SERIALIZABLE, it would be the set of WHERE Y=5 records that existed when the transaction began). This means if that UPDATE statement were to take five minutes to process from start to finish, and someone added and committed a new record to the table
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CHAPTER 7 CONCURRENCY AND MULTI-VERSIONING
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with a value of 5 in the Y column, then that UPDATE would not see it because the consistent read would not see it. This is expected and normal. But, the question is, what happens if two sessions execute the following statements in order: Update t set y = 10 where y = 5; Update t Set x = x+1 Where y = 5; Table 7-8 demonstrates the timeline. Table 7-8. Sequence of Updates
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Session 1
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Update t set y=10 where y=5;
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Session 2
Comment
This updates the one row that matches the criteria.
Update t Set x=x+1 Where y=5;
Using consistent reads, this will find the record session 1 modified, but it won t be able to update it since session 1 has it locked. Session 2 will block and wait for this row. This releases session 2; session 2 becomes unblocked. It can finally do the current read on the block containing this row, where Y was equal to 5 when session 1 began its update. The current read will show that Y is now equal to 10, not 5 anymore.
Commit;
So the record that was Y=5 when you began the UPDATE is no longer Y=5. The consistent read component of the UPDATE says, You want to update this record because Y was 5 when we began, but the current version of the block makes you think, Oh, no, I can t update this row because Y isn t 5 anymore. It would be wrong . If we just skipped this record at this point and ignored it, then we would have a nondeterministic update. It would be throwing data consistency and integrity out the window. The outcome of the update (how many and which rows were modified) would depend on the order in which rows got hit in the table and what other activity just happened to be going on. You could take the same exact set of rows and in two different databases, each one running the transactions in exactly the same mix, you could observe different results, just because the rows were in different places on the disk. In this case, Oracle will choose to restart the update. When the row that was Y=5 when you started is found to contain the value Y=10, Oracle will silently roll back your update (just the update, not any other part of the transaction) and restart it, assuming you are using READ COMMITTED isolation. If you are using SERIALIZABLE isolation, then at this point you would receive an ORA-08177: can't serialize access for this transaction error. In READ COMMITTED mode, after the transaction rolls back your update, the database will restart the update (i.e., change the point in time at which the update is as of ), and instead of updating the data again, it will go into SELECT FOR UPDATE mode and attempt to lock all of the rows WHERE Y=5 for your session. Once it does this, it will run the UPDATE against that locked set of data, thus ensuring this time that it can complete without restarting. But to continue on with the but what happens if. . . train of thought, what happens if, after restarting the update and going into SELECT FOR UPDATE mode (which has the same read-consistent and read current block gets going on as an update does), a row that was Y=5 when you started the SELECT FOR UPDATE is found to be Y=11 when you go to get the current version of it That SELECT FOR UDPDATE will restart and the cycle begins again.
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