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How to get 2nd last column of the line- UNIX?

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# 1  
Old 09-21-2016
How to get 2nd last column of the line- UNIX?
I want to retrieve Status from below example. Columns numbers will be dynamic but Status will always be 2nd last-

Code:
JobName StartTime EndTime Status ExitCode

Code:
autorep -j $jobName | grep '^FR' | awk -F' ' '{print $2}'

The above code gives me the 2nd column from start of the line.
Last edited by Don Cragun; 09-21-2016 at 06:34 AM.. Reason: Change B tags to CODE tags.
# 2  
Old 09-21-2016
Hello Tanu,

Please use code tags as per forum rules for your commands/codes/Inputs which you are using into your posts. Following may help you in same.
Code:
autorep -j $jobName | grep '^FR' | awk  '{print $(NF-1)}'

Where $NFdefines last field of each record/line. So $(NF-1)defines the second last field of each line. Similarly you could get 3rd field from last by $(NF-2)and so on.

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
Tanu
# 3  
Old 09-21-2016
Quote:
Originally Posted by Tanu
I want to retrieve Status from below example. Columns numbers will be dynamic but Status will always be 2nd last-

Code:
JobName StartTime EndTime Status ExitCode

Code:
autorep -j $jobName | grep '^FR' | awk -F' ' '{print $2}'

The above code gives me the 2nd column from start of the line.
Then
Code:
autorep -j $jobName | grep '^FR' | awk -F' ' '{print $NF}'

will give you the last field and:
Code:
autorep -j $jobName | grep '^FR' | awk -F' ' '{print $(NF - 1)}'

will give you the next to the last field.
This User Gave Thanks to Don Cragun For This Post:
Tanu
# 4  
Old 09-21-2016
If they are space separated and you can be certain that Status will always be the penultimate one, you could use variable substitution:-
Code:
status_line=$(autorep -j $jobname | grep -E "^FR" )                # Catch the whole messages
status="${status_line#${status_line% * *}"                         # Chop off everything except the last two fields
status="${status% *}"                                              # Chop off the last field


There is also the awk way to do this, but it could be expensive if you call this in a loop. The above is all within the shell process.

How will you be running this?




I hope that this helps,
Robin



~~~~~~~~~~~~~~
Bah, Don has beaten me to the punch! Smilie
Last edited by rbatte1; 09-21-2016 at 06:42 AM.. Reason: Don replied faster than I did.
This User Gave Thanks to rbatte1 For This Post:
Tanu
# 5  
Old 09-21-2016
Thanks R.Singh. I'll make sure I'm using code/tags for my next posts.
The solution provided by you worked and I have learned something (NF). Cheers Smilie

---------- Post updated at 04:48 AM ---------- Previous update was at 04:46 AM ----------

@rbatte1 -Thanks for explaining. Really helpful
# 6  
Old 09-21-2016
Quote:
Originally Posted by rbatte1
If they are space separated and you can be certain that Status will always be the penultimate one, you could use variable substitution:-
Code:
status_line=$(autorep -j $jobname | grep -E "^FR" )        # Catch the whole messages
status="${status_line#${status_line% * *}"                 # Chop off everything except the last two fields
status="${status% *}"                                      # Chop off the last field


There is also the awk way to do this, but it could be expensive if you call this in a loop. The above is all within the shell process.

How will you be running this?




I hope that this helps,
Robin



~~~~~~~~~~~~~~
Bah, Don has beaten me to the punch! Smilie
Hi Robin,
It looks like I beat you by about 3 minutes; but Ravinder also beat me by a few seconds.

What you suggested will be faster than what Ravinder and I suggested as long as there is only one line of output selected by the grep. But, I think you had a slight typo... The line:
Code:
status="${status_line#${status_line% * *}"                 # Chop off everything except the last two fields

has mismatched braces. I think you meant:
Code:
status="${status_line#${status_line% * *}}"                # Chop off everything except the last two fields

If more than one line is selected by the grep the awk solutions will yield the next to the last field on each selected line while your suggestion will just yield the next to the last field on the last line. I have no idea whether or not the command:
Code:
autorep -j "$jobname" | grep -E "^FR"

could produce more than one line of output.
# 7  
Old 09-21-2016
Yes, indeed it was left off in error.

I suppose to deal with multi-record output, one could:-
Code:
while read status_line
   status="${status_line#${status_line% * *}}"                # Chop off everything except the last two fields
   status="${status% *}"                                      # Chop off the last field
   .... do whatever here ....
done< <(autorep -j $jobname | grep -E "^FR" )

Would that be any better?
I could make the substitution a single line, but it would get awfully complicated and a not great to support.


Robin
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