Need to print nth till last column of ls output using sed


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# 1  
Need to print nth till last column of ls output using sed

I wish to print first, third and sixth till the last column from the output of ls command

Code:
ls -ltr /app/deploy.yml
-rw-rw-r-- 1 user1 dba 27342 Aug 28 10:17 /app/deploy.yml

Desired Output:
Quote:
-rw-rw-r-- user1 Aug 28 10:17 /app/deploy.yml
Below command gives me the desired output.
Code:
ls -ltr /app/deploy.yml  | awk '{$2=$4=$5=""; print $0}'

However, I wish to use sed instead of awk as the code where I'm incorporating the awk does not like these braces "{}"

Any solution without these braces would be great.
# 2  
Why not save the awk script to a little file, and run it like
Code:
awk -f file

? No braces ...
# 3  
Quote:
Originally Posted by RudiC
Why not save the awk script to a little file, and run it like
Code:
awk -f file

? No braces ...
Unfortunately this will not help as this command would run remotely across servers and hence copying the file to each remote host will not be feasible.

I would more so like to avoid doule quotes "". Guess braces {} are fine.

Need to rid my awk and other solutions which off double quotes as highlighted in RED here awk '{$2=$4=$5=""; print $0}'. The double quotes are the problem in my code.

Last edited by mohtashims; 08-28-2019 at 06:15 AM..
# 4  
So use:
Code:
awk '{$2=$4=$5=x; print}'

This User Gave Thanks to Don Cragun For This Post:
# 5  
Works !! Thank you @Don Gragun
# 6  
no braces examples:
Code:
ls -ltr /app/deploy.yml | awk '$2=$4=$5=_;1'
ls -ltr /app/deploy.yml | sed 's/\([^ ]*\) *[^ ]* \([^ ]*\) *[^ ]* *[^ ]* */\1 \2 /'


Last edited by rdrtx1; 08-28-2019 at 12:37 PM..
# 7  
A very short one that even works with the Solaris /usr/bin/awk:
Code:
ls -lotr /app/deploy.yml | awk '($2=$4=x)==0'

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