How to determine the approximate amount of SQL work left in Oracle?

Problem:

You want to know how much longer a long running SQL might ake to finish.

Solution

We can use “V$SESSION_LONGOPS" view to know the approximate time of a query left to execute. "V$SESSION_LONGOPS” view displays the status of various database operations that have been running for longer than six seconds. Please note that this view give you only a rough estimate of when a SQL might complete.

Example

select
  a.username
 ,a.opname
 ,b.sql_text
 ,to_char(a.start_time,'DD-MON-YY HH24:MI') start_time
 ,a.elapsed_seconds how_long
 ,a.time_remaining secs_left
 ,a.sofar
 ,a.totalwork
 ,round(a.sofar/a.totalwork*100,2) percent
from v$session_longops a
    ,v$sql             b
where a.sql_address    = b.address
and   a.sql_hash_value = b.hash_value
and   a.sofar <> a.totalwork
and   a.totalwork != 0;

In order, to view the results properly via SQLPLUS, set he below parameters first.

Example

SET LINESIZE 141 TRIMSPOOL ON PAGES 66
COL username   FORMAT A8        HEAD "User|Name"
COL opname     FORMAT A16       HEAD "Operation|Type"
COL sql_text   FORMAT A33       HEAD "SQL|Text" TRUNC
COL start_time FORMAT A15       HEAD "Start|Time"
COL how_long   FORMAT 99,990    HEAD "Time|Run"
COL secs_left  FORMAT 99,990    HEAD "Appr.|Secs Left"
COL sofar      FORMAT 9,999,990 HEAD "Work|Done"
COL totalwork  FORMAT 9,999,990 HEAD "Total|Work"
COL percent    FORMAT 999.90    HEAD "%|Done"
select
  a.username
 ,a.opname
 ,b.sql_text
 ,to_char(a.start_time,'DD-MON-YY HH24:MI') start_time
 ,a.elapsed_seconds how_long
 ,a.time_remaining secs_left
 ,a.sofar
 ,a.totalwork
 ,round(a.sofar/a.totalwork*100,2) percent
from v$session_longops a
    ,v$sql             b
where a.sql_address    = b.address
and   a.sql_hash_value = b.hash_value?
and   a.sofar <> a.totalwork
and   a.totalwork != 0;

Kiran P

e

Updated on: 05-Dec-2020

732 Views

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