Question

I want to process cron file with the time and cron entry into different columns of DB.

cat root | awk '{print $1, $2, $3, $4, $5, $6}'

47 * * * * string=`find somefile`; echo $string > success.txt 2> err.txt

It is easy to awk the first 5 placeholders of the cron using awk. But how do I select the actual cron entry?
In the above example I want to select every thing from “string” to “$string”
I do also want to select the standard and error out file paths. i.e. success.txt and err.txt

update:

awk '{print $NF}'

The above works for the last variable, but the following does not to find the second-last.

awk '{print $NF-2}'

Answer 1

A simple approach to get rid of the initial fields is to use cut:

cut -d' ' --complement -f1-5

I believe this requires the GNU version of cut, as the --complement flag is an extension.

I’d use awk to split the rest into fields:

awk '{ 
    outfile=$(NF-2)
    errfile=$NF
    for(n=NF; n>(NF-4); n--) { $n = ""}
    printf("command: %s\noutput: %s\nerror: %s\n", $0, outfile, errfile) 
}'

Note that this approach involves a few assumptions:

• whitespace can be standardized to single spaces
• both stdout and stderr are redirected
• no spaces are present in the names for the output and error files