CODEWITHSUNDEEP
bashunixxargs
user2939398
user2939398

Reputation: 33

How to truncate trailing space in xargs

I would like to use xargs to list the contents of some files based on the output of command A. Xargs replace-str seem to be adding a space to the end and causing the command to fail. Any suggestions? I know this can be worked around using for loop. But curious to know how to do this using xargs.

lsscsi |awk -F\/ '/ATA/ {print $NF}' | xargs -L 1 -I % cat /sys/block/%/queue/scheduler
cat: /sys/block/sda /queue/scheduler: No such file or directory

Upvotes: 1

Views: 2132

Answers (2)

mklement0
mklement0

Reputation: 439228

The problem is not with xargs -I, which does not append a space to each argument, which can be verified as follows:

$ echo 'sda' | xargs -I % echo '[%]'
[sda]

Incidentally, specifying -L 1 in addition to -I is pointless: -I implies line-by-line processing.

Therefore, it must be the output from the command that provides input to xargs that contains the trailing space.
You can adapt your awk command to fix that:

lsscsi | 
  awk -F/ '/ATA/ {sub(/ $/,"", $NF); print $NF}' | 
    xargs -I % cat '/sys/block/%/queue/scheduler'
  • sub(/ $/,"", $NF) replaces a trailing space in field $NF with the empty string, thereby effectively removing it.
  • Note how I've (single-)quoted cat's argument so as to make it work even with filenames with spaces.

Upvotes: 2

Jay Kumar R
Jay Kumar R

Reputation: 537

lsscsi |awk -F\/ '/ATA/ {print $NF}'| awk '{print $NF}' | xargs -L 1 -I % cat /sys/block/%/queue/scheduler

The first awk stmt splits by "/" so anything else is considered as field. In this is case "sda " becomes whole field including a space at the end. But by default, awk removes space . So after the pipe, the second awk prints $NF (which is last word of the line) and leaves out " " space as delimiter. awk { print $1 } will do the same because we have only one word, "sda" which is both first and last.

Upvotes: 0

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