how to set parameters when pipe bash script to bash












3














How to execute bash script with parameters:



./foo.sh a b c


When it's compressed (e.g. using xz).



 xzcat foo.sh | bash <<how_to_supply_here_parameters?>>


Specific usecase:



I produced very big rmlint.sh file and store it compressed:



time rmlint -o sh:stdout -c sh:hardlink|tee >( xz > rmlint.sh.xz )


Therefore I would normally execute 



./rmlint.sh -d -x -p


However, file is too big to be uncompressed. Therefore I would love to do same by pipe-ing it to bash:



xzcat rmlint.sh.xz | bash ...





bash shell-script pipe







share|improve this question




















  • 2




    How big is this script, is it a world record? Did you use functions, or paste/copy code all over the place?
    – ctrl-alt-delor
    2 days ago
















3














How to execute bash script with parameters:



./foo.sh a b c


When it's compressed (e.g. using xz).



 xzcat foo.sh | bash <<how_to_supply_here_parameters?>>


Specific usecase:



I produced very big rmlint.sh file and store it compressed:



time rmlint -o sh:stdout -c sh:hardlink|tee >( xz > rmlint.sh.xz )


Therefore I would normally execute 



./rmlint.sh -d -x -p


However, file is too big to be uncompressed. Therefore I would love to do same by pipe-ing it to bash:



xzcat rmlint.sh.xz | bash ...





bash shell-script pipe







share|improve this question




















  • 2




    How big is this script, is it a world record? Did you use functions, or paste/copy code all over the place?
    – ctrl-alt-delor
    2 days ago














3












3








3







How to execute bash script with parameters:



./foo.sh a b c


When it's compressed (e.g. using xz).



 xzcat foo.sh | bash <<how_to_supply_here_parameters?>>


Specific usecase:



I produced very big rmlint.sh file and store it compressed:



time rmlint -o sh:stdout -c sh:hardlink|tee >( xz > rmlint.sh.xz )


Therefore I would normally execute 



./rmlint.sh -d -x -p


However, file is too big to be uncompressed. Therefore I would love to do same by pipe-ing it to bash:



xzcat rmlint.sh.xz | bash ...





bash shell-script pipe







share|improve this question















How to execute bash script with parameters:



./foo.sh a b c


When it's compressed (e.g. using xz).



 xzcat foo.sh | bash <<how_to_supply_here_parameters?>>


Specific usecase:



I produced very big rmlint.sh file and store it compressed:



time rmlint -o sh:stdout -c sh:hardlink|tee >( xz > rmlint.sh.xz )


Therefore I would normally execute 



./rmlint.sh -d -x -p


However, file is too big to be uncompressed. Therefore I would love to do same by pipe-ing it to bash:



xzcat rmlint.sh.xz | bash ...





bash shell-script pipe




bash shell-script pipe






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









SouravGhosh

455311




455311










asked 2 days ago









Grzegorz Wierzowiecki

5,2101361104




5,2101361104








  • 2




    How big is this script, is it a world record? Did you use functions, or paste/copy code all over the place?
    – ctrl-alt-delor
    2 days ago














  • 2




    How big is this script, is it a world record? Did you use functions, or paste/copy code all over the place?
    – ctrl-alt-delor
    2 days ago








2




2




How big is this script, is it a world record? Did you use functions, or paste/copy code all over the place?
– ctrl-alt-delor
2 days ago




How big is this script, is it a world record? Did you use functions, or paste/copy code all over the place?
– ctrl-alt-delor
2 days ago










1 Answer
1






active

oldest

votes


















7














You should use the -s option and -- to separate arguments you want to pass:



echo 'echo "$@"' | sh -s 3 4 5



echo 'printf "{%s}" "$0"; printf " {%s}" "$@"; echo' |

    sh -s -- -d -x -p --foo=bar

{sh} {-d} {-x} {-p} {--foo=bar}


This should work with any POSIX shell, not just bash. From susv4:




-s
Read commands from the standard input.



If there are no operands and the -c option is not specified, the -s
option shall be assumed.







share|improve this answer























  • There is a problem with this, as my parameters have "-" prefix, therefore I call bash -s -d -x -p and get in return : bash: -d: invalid option ( pastebin.com/JXRiUuLR )
    – Grzegorz Wierzowiecki
    yesterday






  • 2




    You use the -- end of options marker, as usual. See the 2nd example.
    – mosvy
    yesterday










  • Thank you a lot for help and apologise for stupid overlook! You helped me to find what I was overlooking (to short sleep Today), which was -- - Thank you! Now it works perfectly!
    – Grzegorz Wierzowiecki
    yesterday










  • Added -- to first line of your solution as it was something I overlooked, so I hope it will help some others, who may overlook it as well
    – Grzegorz Wierzowiecki
    23 hours ago











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









7














You should use the -s option and -- to separate arguments you want to pass:



echo 'echo "$@"' | sh -s 3 4 5



echo 'printf "{%s}" "$0"; printf " {%s}" "$@"; echo' |

    sh -s -- -d -x -p --foo=bar

{sh} {-d} {-x} {-p} {--foo=bar}


This should work with any POSIX shell, not just bash. From susv4:




-s
Read commands from the standard input.



If there are no operands and the -c option is not specified, the -s
option shall be assumed.







share|improve this answer























  • There is a problem with this, as my parameters have "-" prefix, therefore I call bash -s -d -x -p and get in return : bash: -d: invalid option ( pastebin.com/JXRiUuLR )
    – Grzegorz Wierzowiecki
    yesterday






  • 2




    You use the -- end of options marker, as usual. See the 2nd example.
    – mosvy
    yesterday










  • Thank you a lot for help and apologise for stupid overlook! You helped me to find what I was overlooking (to short sleep Today), which was -- - Thank you! Now it works perfectly!
    – Grzegorz Wierzowiecki
    yesterday










  • Added -- to first line of your solution as it was something I overlooked, so I hope it will help some others, who may overlook it as well
    – Grzegorz Wierzowiecki
    23 hours ago
















7














You should use the -s option and -- to separate arguments you want to pass:



echo 'echo "$@"' | sh -s 3 4 5



echo 'printf "{%s}" "$0"; printf " {%s}" "$@"; echo' |

    sh -s -- -d -x -p --foo=bar

{sh} {-d} {-x} {-p} {--foo=bar}


This should work with any POSIX shell, not just bash. From susv4:




-s
Read commands from the standard input.



If there are no operands and the -c option is not specified, the -s
option shall be assumed.







share|improve this answer























  • There is a problem with this, as my parameters have "-" prefix, therefore I call bash -s -d -x -p and get in return : bash: -d: invalid option ( pastebin.com/JXRiUuLR )
    – Grzegorz Wierzowiecki
    yesterday






  • 2




    You use the -- end of options marker, as usual. See the 2nd example.
    – mosvy
    yesterday










  • Thank you a lot for help and apologise for stupid overlook! You helped me to find what I was overlooking (to short sleep Today), which was -- - Thank you! Now it works perfectly!
    – Grzegorz Wierzowiecki
    yesterday










  • Added -- to first line of your solution as it was something I overlooked, so I hope it will help some others, who may overlook it as well
    – Grzegorz Wierzowiecki
    23 hours ago














7












7








7






You should use the -s option and -- to separate arguments you want to pass:



echo 'echo "$@"' | sh -s 3 4 5



echo 'printf "{%s}" "$0"; printf " {%s}" "$@"; echo' |

    sh -s -- -d -x -p --foo=bar

{sh} {-d} {-x} {-p} {--foo=bar}


This should work with any POSIX shell, not just bash. From susv4:




-s
Read commands from the standard input.



If there are no operands and the -c option is not specified, the -s
option shall be assumed.







share|improve this answer














You should use the -s option and -- to separate arguments you want to pass:



echo 'echo "$@"' | sh -s 3 4 5



echo 'printf "{%s}" "$0"; printf " {%s}" "$@"; echo' |

    sh -s -- -d -x -p --foo=bar

{sh} {-d} {-x} {-p} {--foo=bar}


This should work with any POSIX shell, not just bash. From susv4:




-s
Read commands from the standard input.



If there are no operands and the -c option is not specified, the -s
option shall be assumed.








share|improve this answer














share|improve this answer



share|improve this answer








edited 23 hours ago









Grzegorz Wierzowiecki

5,2101361104




5,2101361104










answered 2 days ago









mosvy

5,8781325




5,8781325












  • There is a problem with this, as my parameters have "-" prefix, therefore I call bash -s -d -x -p and get in return : bash: -d: invalid option ( pastebin.com/JXRiUuLR )
    – Grzegorz Wierzowiecki
    yesterday






  • 2




    You use the -- end of options marker, as usual. See the 2nd example.
    – mosvy
    yesterday










  • Thank you a lot for help and apologise for stupid overlook! You helped me to find what I was overlooking (to short sleep Today), which was -- - Thank you! Now it works perfectly!
    – Grzegorz Wierzowiecki
    yesterday










  • Added -- to first line of your solution as it was something I overlooked, so I hope it will help some others, who may overlook it as well
    – Grzegorz Wierzowiecki
    23 hours ago


















  • There is a problem with this, as my parameters have "-" prefix, therefore I call bash -s -d -x -p and get in return : bash: -d: invalid option ( pastebin.com/JXRiUuLR )
    – Grzegorz Wierzowiecki
    yesterday






  • 2




    You use the -- end of options marker, as usual. See the 2nd example.
    – mosvy
    yesterday










  • Thank you a lot for help and apologise for stupid overlook! You helped me to find what I was overlooking (to short sleep Today), which was -- - Thank you! Now it works perfectly!
    – Grzegorz Wierzowiecki
    yesterday










  • Added -- to first line of your solution as it was something I overlooked, so I hope it will help some others, who may overlook it as well
    – Grzegorz Wierzowiecki
    23 hours ago
















There is a problem with this, as my parameters have "-" prefix, therefore I call bash -s -d -x -p and get in return : bash: -d: invalid option ( pastebin.com/JXRiUuLR )
– Grzegorz Wierzowiecki
yesterday




There is a problem with this, as my parameters have "-" prefix, therefore I call bash -s -d -x -p and get in return : bash: -d: invalid option ( pastebin.com/JXRiUuLR )
– Grzegorz Wierzowiecki
yesterday




2




2




You use the -- end of options marker, as usual. See the 2nd example.
– mosvy
yesterday




You use the -- end of options marker, as usual. See the 2nd example.
– mosvy
yesterday












Thank you a lot for help and apologise for stupid overlook! You helped me to find what I was overlooking (to short sleep Today), which was -- - Thank you! Now it works perfectly!
– Grzegorz Wierzowiecki
yesterday




Thank you a lot for help and apologise for stupid overlook! You helped me to find what I was overlooking (to short sleep Today), which was -- - Thank you! Now it works perfectly!
– Grzegorz Wierzowiecki
yesterday












Added -- to first line of your solution as it was something I overlooked, so I hope it will help some others, who may overlook it as well
– Grzegorz Wierzowiecki
23 hours ago




Added -- to first line of your solution as it was something I overlooked, so I hope it will help some others, who may overlook it as well
– Grzegorz Wierzowiecki
23 hours ago


















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