UNIX for Dummies Questions & Answers

Code:

#!/bin/ksh
cd /idn/home/was/logs/ic_server
set da=date | nawk '{ print $2,$3-1}'
echo $da
ls -lrt SystemOut* | nawk '$0 ~ da {print $NF}'
#ls -lrt | nawk '/d/{print $NF}'

The following error is coming

Code:

nawk: empty regular expression
input record number 1
source line number 1

please help !!

Hi,

You have missed out at least one set of Quotes;

Code:

#!/bin/ksh
cd /idn/home/was/logs/ic_server
set da=`date | nawk '{ print $2,$3-1 }'`
echo $da
ls -lrt SystemOut* | nawk '$0 ~ da {print $NF}'
#ls -lrt | nawk '/d/{print $NF}'

But if that doesn't fix it - you'll have to wait for someone who can.

Regards

Dave

In which line i have missed the quote?

If you're trying to do what I think you're trying to do (print a list of files that were changed yesterday), you'll need something like:

Code:

#!/bin/ksh
cd /idn/home/was/logs/ic_server
set $(date)
ls -lrt SystemOut* | nawk -v mon=$3 -v day=$3 'BEGIN {
        day--
        printf "Month is %s, day was %d\n",mon,day
      }
$6 ~ mon && $7 ~ day {print $NF}'

Note however that this won't work on the first day of any month and it will also include files that are x years + 1 day old in the output; not just files that are one day old.

To make it work on the first of the month, you'll need to see if day is 0 after it is decremented and adjust mon and day to the last day of the previous month. (Don't forget to account for leap years!) The date -d option can help with this on some systems, but the meaning of the -d option varies from system to system.