Hi,

Does anyone know how to get the date automatically placed in the name of a log file? I have log file generated daily and I want to be able to differentiate and archive easily.

Thanks for you help.

there are plenty of ways to insert a date:

date=$(date +%d%m%y)

and then use as you see fit... you need to give mroe info about how these files are created if this isn't what you're after.

thanks,

that's a good starting point.

There must be something else required as it is not working. Here is the script I am running

#!/bin/sh
#
echo "---------------------------------------" >> ./backup_logs_$ORACLE_SID_%Y-%m-%d.log
echo "Starting rman for ORACLE_SID: $ORACLE_SID" >> ./backup_logs_$ORACLE_SID_%Y-%m-%d.log
date >> ./backup_logs_$ORACLE_SID_%Y-%m-%d.log
#
rman <<EOF
connect target backup_admin2/backup2pa;
spool log to $HOME/rman_scripts/backup_logs_$ORACLE_SID_%Y-%m-%d.log append;
backup database plus archivelog;
delete noprompt obsolete;
quit;
EOF
#
echo "End of rman for $ORACLE_SID" >> ./backup_logs_$ORACLE_SID_%Y-%m-%d.log
date >> ./backup_logs_$ORACLE_SID_%Y-%m-%d.log
echo "---------------------------------------" >> ./backup_logs_$ORACLE_SID_%Y-%m-%d.log
exit

I'm trying to get a file name that would look like this:

backup_logs_GEN_06-06-02.log

but what I get is

backup_logs_%Y-%m-%d.log

any ideas?

It's better to put the date-part into a variable beforehand.
So:

OK I have adjusted the script as such:-


#!/bin/sh
#
log_date=$(date +%Y%m%d)
echo "---------------------------------------" >> ./backup_logs_$ORACLE_SID_{log_date}.log
echo "Starting rman for ORACLE_SID: $ORACLE_SID" >> ./backup_logs_$ORACLE_SID_{log_date}.log
date >> ./backup_logs_$ORACLE_SID_{log_date}.log
#
#rman <<EOF
#connect target backup_admin2/backup2pa;
#spool log to $HOME/rman_scripts/backup_logs_$ORACLE_SID_{log_date}.log append;
#backup database plus archivelog;
#delete noprompt obsolete;
#quit;
#EOF
#
echo "End of rman for $ORACLE_SID" >> ./backup_logs_$ORACLE_SID_{log_date}.log
date >> ./backup_logs_$ORACLE_SID_{log_date}.log
echo "---------------------------------------" >> ./backup_logs_$ORACLE_SID_{log_date}.log
exit

now I get a file called

backup_logs_{log_date}.log

notice how it also misses the $ORACLE_SID variable, why is that? it even misses the second _.

my mistake, I missed the $.

Thanks

echo "---------------------------------------" >> ./backup_logs_$ORACLE_SID_{log_date}.log
should be
echo "---------------------------------------" >> ./backup_logs_$ORACLE_SID_${log_date}.log

As always, take babysteps in the beginning so start by echoing the intended output of the variables:
echo "backup_logs_$ORACLE_SID_${log_date}.log"

Now you can see that maybe the ORACLE_SID variable is not set at all.
I don't know ORACLE_SID, the way you're using it suggests it should be set in your environment ...

I'm not sure but if the file does not exist beforehand it might be better to create it first by using:

Thanks,

$ORACLE_SID is an environment variable set in another script. I have found that by using curly brackets, as with the date variable, it now works fine.

i.e.

${ORACLE_SID}