how to set parameters when pipe bash script to bash
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
edited Dec 27 '18 at 11:06
SouravGhosh
493311
asked Dec 27 '18 at 9:13
Grzegorz Wierzowiecki
5,2101361104
add a comment |
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
edited Dec 27 '18 at 11:06
SouravGhosh
493311
asked Dec 27 '18 at 9:13
Grzegorz Wierzowiecki
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
Dec 27 '18 at 10:08
add a comment |
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
edited Dec 27 '18 at 11:06
SouravGhosh
493311
asked Dec 27 '18 at 9:13
Grzegorz Wierzowiecki
5,2101361104
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
edited Dec 27 '18 at 11:06
SouravGhosh
493311
asked Dec 27 '18 at 9:13
Grzegorz Wierzowiecki
5,2101361104
edited Dec 27 '18 at 11:06
SouravGhosh
493311
asked Dec 27 '18 at 9:13
Grzegorz Wierzowiecki
5,2101361104
edited Dec 27 '18 at 11:06
SouravGhosh
493311
edited Dec 27 '18 at 11:06
SouravGhosh
493311
edited Dec 27 '18 at 11:06
SouravGhosh
493311
493311
asked Dec 27 '18 at 9:13
Grzegorz Wierzowiecki
5,2101361104
asked Dec 27 '18 at 9:13
Grzegorz Wierzowiecki
5,2101361104
asked Dec 27 '18 at 9:13
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
Dec 27 '18 at 10:08
add a comment |
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
Dec 27 '18 at 10:08
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
Dec 27 '18 at 10:08
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
Dec 27 '18 at 10:08
add a comment |
1 Answer
1
active
oldest
votes
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
-coption is not specified, the-s
option shall be assumed.
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
answered Dec 27 '18 at 9:45
mosvy
6,0211425
There is a problem with this, as my parameters have "-" prefix, therefore I callbash -s -d -x -pand get in return :bash: -d: invalid option( pastebin.com/JXRiUuLR )
– Grzegorz Wierzowiecki
Dec 27 '18 at 10:47
2
You use the--end of options marker, as usual. See the 2nd example.
– mosvy
Dec 27 '18 at 10:57
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
Dec 27 '18 at 11:01
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
Dec 28 '18 at 10:42
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
-coption is not specified, the-s
option shall be assumed.
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
answered Dec 27 '18 at 9:45
mosvy
6,0211425
There is a problem with this, as my parameters have "-" prefix, therefore I callbash -s -d -x -pand get in return :bash: -d: invalid option( pastebin.com/JXRiUuLR )
– Grzegorz Wierzowiecki
Dec 27 '18 at 10:47
2
You use the--end of options marker, as usual. See the 2nd example.
– mosvy
Dec 27 '18 at 10:57
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
Dec 27 '18 at 11:01
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
Dec 28 '18 at 10:42
add a comment |
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
-coption is not specified, the-s
option shall be assumed.
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
answered Dec 27 '18 at 9:45
mosvy
6,0211425
There is a problem with this, as my parameters have "-" prefix, therefore I callbash -s -d -x -pand get in return :bash: -d: invalid option( pastebin.com/JXRiUuLR )
– Grzegorz Wierzowiecki
Dec 27 '18 at 10:47
2
You use the--end of options marker, as usual. See the 2nd example.
– mosvy
Dec 27 '18 at 10:57
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
Dec 27 '18 at 11:01
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
Dec 28 '18 at 10:42
add a comment |
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
-coption is not specified, the-s
option shall be assumed.
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
answered Dec 27 '18 at 9:45
mosvy
6,0211425
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
-coption is not specified, the-s
option shall be assumed.
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
answered Dec 27 '18 at 9:45
mosvy
6,0211425
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
edited Dec 28 '18 at 10:41
Grzegorz Wierzowiecki
5,2101361104
5,2101361104
answered Dec 27 '18 at 9:45
mosvy
6,0211425
answered Dec 27 '18 at 9:45
mosvy
6,0211425
answered Dec 27 '18 at 9:45
mosvy
6,0211425
6,0211425
There is a problem with this, as my parameters have "-" prefix, therefore I callbash -s -d -x -pand get in return :bash: -d: invalid option( pastebin.com/JXRiUuLR )
– Grzegorz Wierzowiecki
Dec 27 '18 at 10:47
2
You use the--end of options marker, as usual. See the 2nd example.
– mosvy
Dec 27 '18 at 10:57
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
Dec 27 '18 at 11:01
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
Dec 28 '18 at 10:42
add a comment |
There is a problem with this, as my parameters have "-" prefix, therefore I callbash -s -d -x -pand get in return :bash: -d: invalid option( pastebin.com/JXRiUuLR )
– Grzegorz Wierzowiecki
Dec 27 '18 at 10:47
2
You use the--end of options marker, as usual. See the 2nd example.
– mosvy
Dec 27 '18 at 10:57
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
Dec 27 '18 at 11:01
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
Dec 28 '18 at 10:42
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
Dec 27 '18 at 10:47
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
Dec 27 '18 at 10:47
2
2
You use the
-- end of options marker, as usual. See the 2nd example.– mosvy
Dec 27 '18 at 10:57
You use the
-- end of options marker, as usual. See the 2nd example.– mosvy
Dec 27 '18 at 10:57
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
Dec 27 '18 at 11:01
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
Dec 27 '18 at 11:01
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
Dec 28 '18 at 10:42
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
Dec 28 '18 at 10:42
add a comment |
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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
Dec 27 '18 at 10:08