Pipe commands at each xargs pass












0















Lets say I have this :



echo '/dev/sd'{a..d} | xargs -n 1 lsblk $1


But now I want to pipe each pass of the xargs as for instance :



echo '/dev/sd'{a..b} | xargs -n 1 lsblk $1 | tail -n +2


Now, that does not work obviously, because the pipe is applied to the entire xargs commands. So I might try this :



echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2"


But the problem is that now $1 has no value inside of the shell.



How can I over come this? Or basically, How can I pipe the execution being done by xargs at each pass?






bash pipe xargs







share|improve this question


















  • 1





    You need to pass the parameter to the shell, like so I believe: echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2" {}

    – bishop
    Nov 14 '18 at 17:15






  • 5





    Unless echo /dev/sd/{a..b} is just a place holder for some other command, I wouldn't bother with xargs at all here. Just use a loop: for f in /dev/sd{a..b}; do lsblk "$f" | tail -n +2; done. (Even if it is a placeholder, there's probably still a way to avoid xargs.)

    – chepner
    Nov 14 '18 at 17:24








  • 2





    You may use: xargs -I {} bash -c 'lsblk "$1" | tail -n +2' - {} < <(printf '%sn' '/dev/sd'{a..d}) but keep in mind that it will spawn bash subpricess for each entry.

    – anubhava
    Nov 14 '18 at 17:28
















0















Lets say I have this :



echo '/dev/sd'{a..d} | xargs -n 1 lsblk $1


But now I want to pipe each pass of the xargs as for instance :



echo '/dev/sd'{a..b} | xargs -n 1 lsblk $1 | tail -n +2


Now, that does not work obviously, because the pipe is applied to the entire xargs commands. So I might try this :



echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2"


But the problem is that now $1 has no value inside of the shell.



How can I over come this? Or basically, How can I pipe the execution being done by xargs at each pass?






bash pipe xargs







share|improve this question


















  • 1





    You need to pass the parameter to the shell, like so I believe: echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2" {}

    – bishop
    Nov 14 '18 at 17:15






  • 5





    Unless echo /dev/sd/{a..b} is just a place holder for some other command, I wouldn't bother with xargs at all here. Just use a loop: for f in /dev/sd{a..b}; do lsblk "$f" | tail -n +2; done. (Even if it is a placeholder, there's probably still a way to avoid xargs.)

    – chepner
    Nov 14 '18 at 17:24








  • 2





    You may use: xargs -I {} bash -c 'lsblk "$1" | tail -n +2' - {} < <(printf '%sn' '/dev/sd'{a..d}) but keep in mind that it will spawn bash subpricess for each entry.

    – anubhava
    Nov 14 '18 at 17:28














0












0








0








Lets say I have this :



echo '/dev/sd'{a..d} | xargs -n 1 lsblk $1


But now I want to pipe each pass of the xargs as for instance :



echo '/dev/sd'{a..b} | xargs -n 1 lsblk $1 | tail -n +2


Now, that does not work obviously, because the pipe is applied to the entire xargs commands. So I might try this :



echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2"


But the problem is that now $1 has no value inside of the shell.



How can I over come this? Or basically, How can I pipe the execution being done by xargs at each pass?






bash pipe xargs







share|improve this question














Lets say I have this :



echo '/dev/sd'{a..d} | xargs -n 1 lsblk $1


But now I want to pipe each pass of the xargs as for instance :



echo '/dev/sd'{a..b} | xargs -n 1 lsblk $1 | tail -n +2


Now, that does not work obviously, because the pipe is applied to the entire xargs commands. So I might try this :



echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2"


But the problem is that now $1 has no value inside of the shell.



How can I over come this? Or basically, How can I pipe the execution being done by xargs at each pass?






bash pipe xargs




bash pipe xargs






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 14 '18 at 17:12









Matias BarriosMatias Barrios

1,547317




1,547317








  • 1





    You need to pass the parameter to the shell, like so I believe: echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2" {}

    – bishop
    Nov 14 '18 at 17:15






  • 5





    Unless echo /dev/sd/{a..b} is just a place holder for some other command, I wouldn't bother with xargs at all here. Just use a loop: for f in /dev/sd{a..b}; do lsblk "$f" | tail -n +2; done. (Even if it is a placeholder, there's probably still a way to avoid xargs.)

    – chepner
    Nov 14 '18 at 17:24








  • 2





    You may use: xargs -I {} bash -c 'lsblk "$1" | tail -n +2' - {} < <(printf '%sn' '/dev/sd'{a..d}) but keep in mind that it will spawn bash subpricess for each entry.

    – anubhava
    Nov 14 '18 at 17:28














  • 1





    You need to pass the parameter to the shell, like so I believe: echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2" {}

    – bishop
    Nov 14 '18 at 17:15






  • 5





    Unless echo /dev/sd/{a..b} is just a place holder for some other command, I wouldn't bother with xargs at all here. Just use a loop: for f in /dev/sd{a..b}; do lsblk "$f" | tail -n +2; done. (Even if it is a placeholder, there's probably still a way to avoid xargs.)

    – chepner
    Nov 14 '18 at 17:24








  • 2





    You may use: xargs -I {} bash -c 'lsblk "$1" | tail -n +2' - {} < <(printf '%sn' '/dev/sd'{a..d}) but keep in mind that it will spawn bash subpricess for each entry.

    – anubhava
    Nov 14 '18 at 17:28








1




1





You need to pass the parameter to the shell, like so I believe: echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2" {}

– bishop
Nov 14 '18 at 17:15





You need to pass the parameter to the shell, like so I believe: echo '/dev/sd'{a..b} | xargs -n 1 bash -c "lsblk $1 | tail -n +2" {}

– bishop
Nov 14 '18 at 17:15




5




5





Unless echo /dev/sd/{a..b} is just a place holder for some other command, I wouldn't bother with xargs at all here. Just use a loop: for f in /dev/sd{a..b}; do lsblk "$f" | tail -n +2; done. (Even if it is a placeholder, there's probably still a way to avoid xargs.)

– chepner
Nov 14 '18 at 17:24







Unless echo /dev/sd/{a..b} is just a place holder for some other command, I wouldn't bother with xargs at all here. Just use a loop: for f in /dev/sd{a..b}; do lsblk "$f" | tail -n +2; done. (Even if it is a placeholder, there's probably still a way to avoid xargs.)

– chepner
Nov 14 '18 at 17:24






2




2





You may use: xargs -I {} bash -c 'lsblk "$1" | tail -n +2' - {} < <(printf '%sn' '/dev/sd'{a..d}) but keep in mind that it will spawn bash subpricess for each entry.

– anubhava
Nov 14 '18 at 17:28





You may use: xargs -I {} bash -c 'lsblk "$1" | tail -n +2' - {} < <(printf '%sn' '/dev/sd'{a..d}) but keep in mind that it will spawn bash subpricess for each entry.

– anubhava
Nov 14 '18 at 17:28












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