UNIX for Dummies Questions & Answers

I`m not sure, but I found one file there`s so paragraph:
"Deleting Older Records from the Database
When the dbage function is run periodically, it should be run twice: once to delete
records for nonexistent conections with the -dd (dead data) option, and once to
force the deletion of outdated records.
For example, to delete dead data records and then delete report and raw data
records according to device definitions, enter the commands:
OLCmd dbage -n OracleStatsDB -dd
OLCmd dbage -n OraclStatsDB -d
The dbage function can be automated using cron under Solaris, or at under
Windows. See Rebuilding the Indexes on page 11 for a cron example.
For a complete description of the dbage command, see the OpenLane SLM
Administrator's Guide."
Maybe it could help?...

But I am not deleteing anything from database !
I need to trim the logfile..

As far as I can see you've got 2 problems
1) you must calculate the date on unix (not so easy)
By the way, how many days do you want to go back ?
You wrote "the backlog from some tables. Now that logfile is growing too big and I need to housekeep it! effectively I want to keep last 30 days data in that file and move rest to archived file.
The file contents are as below.
the algorithm I want is as follow :

take input file my.log
go to first line, check if this is more than 90 days
"
2) you must read your file line by line and act appropriatly.

I think I can find/make something that does the trick but, not in sed. Is regular korn shell ok for you ?

All,
Out of woods finally !
Have a look at the code.
I got the date-90 in the required format from the sql itself :-)

and following is the korn code :
-------------------------------------------------------------------------------
DATE=`grep -v SQL /$HOME/dba/perfmon/siebel/date.lst | tail -2`
#this will get me line number of first occurence
grep -n $DATE mon_txn_rtr_backlog.log > /$HOME/dba/perfmon/siebel/tmp.log 2>&1
#Grep doesn't like my date parameter and might throw and error
#even if it's happy with that and deletes first 2 occurences, we run the script
#24 times a day so we are getting the correct date anyway!
sed "1,2 d" /$HOME/dba/perfmon/siebel/tmp.log > /$HOME/dba/perfmon/siebel/tmp1.l
og
LINE=`head -1 tmp1.log | cut -d":" -f2`
LINE=` expr $LINE - 1 `
if test $LINE != 0
then
# because we want to delete from 1 line early from first occurence
sed "1,$LINE d" /$HOME/dba/perfmon/siebel/mon_txn_rtr_backlog.log > /$HOME/dba/p
erfmon/siebel/mon_tmp.log
mv /$HOME/dba/perfmon/siebel/mon_tmp.log /$HOME/dba/perfmon/siebel/mon_txn_rtr_b
acklog.log
rm /$HOME/dba/perfmon/siebel/tmp.log
rm /$HOME/dba/perfmon/siebel/tmp1.log
fi
rm /$HOME/dba/perfmon/siebel/tmp.log
rm /$HOME/dba/perfmon/siebel/tmp1.log
-------------------------------------------------------------------------------

it works.

I still wonder how can I find out the date ( today-90 ) using shell !

Thanks anyway,
cheers

All,
This may sound elementary but I am tired getting this at work!!
This piece of code is giving me a problem, in the sense when I run it manually it runs well, but when I run this though cron it gives me an error for expr syntax.

sed "1,2 d" /$HOME/dba/perfmon/siebel/tmp.log > /HOME/dba/perfmon/siebel/tmp1.log
LINE=`head -1 tmp1.log | cut -d":" -f2`
LINE=` expr $LINE - 1 `
if test $LINE != 0
then
.
.
.....

this `expr` is not running fine , can somebody please throw light on this please ? I know it all looks okay, but still ??
LINE=` expr $LINE - 1 `

Inside your script, in between these two lines:

LINE=`head -1 tmp1.log | cut -d":" -f2`
LINE=` expr $LINE - 1 `

place a line that says "echo $LINE".. i'm betting that $LINE doesn't have a numerical value when the expr command is executed..

echo $LINE gives me proper number when I execute manually, through cron, tyhe logfile says "expr: Syntax error"


Please help !