Updating mysql table dating minor california

Yeah ,of course it'll recompile itself when it called next time. There is no logical column to do partition., I guess the insert into a new table will take considerable time with 27 mil records.. November 12, 2002 - am UTC wait 10 days so that you are deleting 30 million records from a 60 million record table and then this will be much more efficient. 3 million records on an indexed table will take considerable time.There is a chance that INSERT /* append */ select Tom, Recently I had conducted a interview in which one the dba mentioned that they had a table that might conatin 10 million records or might be 1 million.He meant to say they delete the records and some time later the table will be populated again and viceversa.Tom according to you do you consider partitions for such tables and if yes which type of partition.. November 13, 2002 - pm UTC hard to tell -- is the data deleted by something that is relatively constant (eg: the value in that column doesn't change - so the row doesn't need to move from partition to partition).In response to the Jack Silvey (from Richardson, TX ) review, where he wrote "It is the only way to fly.We also have an absolutely incredible stored procedure that rebuilds all of our indexes concurrently after the load, using the Oracle job scheduler as the mechanism of allowing separate threads in pl/sql": Could you provide more information about that procedure and how to rebuild multiple same-table indexes concurrently using Oracle job scheduler? November 19, 2002 - pm UTC instead of begin execute immediate 'alter index idx1 rebuild'; execute immediate 'alter index idx2 rebuild';end;you can code:declare l_job number;begin dbms_job.submit( l_job, 'execute immediate ''alter index idx1 rebuild'';' ); commit; dbms_job.submit( l_job, 'execute immediate ''alter index idx2 rebuild'';' ); commit;end; Now, just set job_queue_processes Thanks, Tom.I recorded the time right after the commit statement at the end of PL/SQL block - that's the start time.Then I kept querying user_jobs view every 2 - 3 sec, until the last of the 5 jobs were gone. The last question on this topic: is user_jobs view is the right place to look in order to determine that's rebuilding is done and how long it took?

If you are at the conference, drop into the Groundbreaker area and say Hello. I want to update and commit every time for so many records ( say 10,000 records). Fortunately, you are probably using partitioning so you can do this easily in parallel -- bit by bit.", database="mydatabase")mycursor = mydb.cursor()sql = "UPDATE customers SET address = 'Canyon 123' WHERE address = 'Valley 345'"mycursor.execute(sql)mydb.commit() print(mycursor.rowcount, "record(s) affected") It is considered a good practice to escape the values of any query, also in update statements.This is to prevent SQL injections, which is a common web hacking technique to destroy or misuse your database.The only one difference between your code and mine is that I issue just one commit at the end. Here is the numbers I've got: rebuilding indexes sequentually consistently took 76 sec., while using dbms_job.submit() calls took around 40 - 42 sec.I said "around", because the technique I used may not be perfect, though it served the purpose.

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