Multiline command : comment out one line












7















I like to use the following format in scripts for commands with a lot of parameters (for readability):



docker run 
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


But, sometimes I'd like to comment one of these parameters out like:



# -p 8080:8080  


This doesn't work, as the EOL is interpreted as return and the command fails. Tried this too:



 # -p 8080:8080


which also didn't work.



Question: Is there a way to comment out the parameter, so it's still on its own line, but I'd be able to execute the script?










share|improve this question























  • If any of the answers solved your problem, please accept it by clicking the checkmark next to it. Thank you!

    – Jeff Schaller
    Feb 17 at 13:46
















7















I like to use the following format in scripts for commands with a lot of parameters (for readability):



docker run 
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


But, sometimes I'd like to comment one of these parameters out like:



# -p 8080:8080  


This doesn't work, as the EOL is interpreted as return and the command fails. Tried this too:



 # -p 8080:8080


which also didn't work.



Question: Is there a way to comment out the parameter, so it's still on its own line, but I'd be able to execute the script?










share|improve this question























  • If any of the answers solved your problem, please accept it by clicking the checkmark next to it. Thank you!

    – Jeff Schaller
    Feb 17 at 13:46














7












7








7








I like to use the following format in scripts for commands with a lot of parameters (for readability):



docker run 
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


But, sometimes I'd like to comment one of these parameters out like:



# -p 8080:8080  


This doesn't work, as the EOL is interpreted as return and the command fails. Tried this too:



 # -p 8080:8080


which also didn't work.



Question: Is there a way to comment out the parameter, so it's still on its own line, but I'd be able to execute the script?










share|improve this question














I like to use the following format in scripts for commands with a lot of parameters (for readability):



docker run 
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


But, sometimes I'd like to comment one of these parameters out like:



# -p 8080:8080  


This doesn't work, as the EOL is interpreted as return and the command fails. Tried this too:



 # -p 8080:8080


which also didn't work.



Question: Is there a way to comment out the parameter, so it's still on its own line, but I'd be able to execute the script?







shell-script shell scripting






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Feb 6 at 15:46









ChirloChirlo

25816




25816













  • If any of the answers solved your problem, please accept it by clicking the checkmark next to it. Thank you!

    – Jeff Schaller
    Feb 17 at 13:46



















  • If any of the answers solved your problem, please accept it by clicking the checkmark next to it. Thank you!

    – Jeff Schaller
    Feb 17 at 13:46

















If any of the answers solved your problem, please accept it by clicking the checkmark next to it. Thank you!

– Jeff Schaller
Feb 17 at 13:46





If any of the answers solved your problem, please accept it by clicking the checkmark next to it. Thank you!

– Jeff Schaller
Feb 17 at 13:46










2 Answers
2






active

oldest

votes


















11














You can't comment out a piece of a line.



Note that since the newlines are escaped, the command is actually a single line (to the shell parser), and there's no way to comment out a part of a single line (except for at the very end).



Instead, maybe just make a copy of the original command in an editor and comment it out completely while keeping the modified command uncommented:



docker run 
--rm
-u root
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean

# Was originally:
# docker run
# --rm
# -u root
# -p 8080:8080
# -v jenkins-data:/var/jenkins_home
# -v /var/run/docker.sock:/var/run/docker.sock
# -v "$HOME":/home
# jenkinsci/blueocean




Alternatively, if you want to occasionally delete or change the -p option and its argument (assuming bash or a shell with the same array syntax):



port=( -p 8080:8080 )

docker run
--rm
-u root
"${port[@]}"
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


Then just change or comment out the assignment to port.



Taking this further:



docker_run_args=(
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"


Inside the array assignment, there are no issues with commenting out a line:



docker_run_args=(
--rm
-u root
# -p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"





share|improve this answer





















  • 5





    Putting all the arguments in an array is also a great way to do it, since you can have a multi-line array definition with no ``s and you can comment individual lines... Wanna include that into your answer too?

    – filbranden
    Feb 6 at 16:02








  • 1





    @filbranden Will do as it's a good extension of what I already had written. Thanks.

    – Kusalananda
    Feb 6 at 16:04



















8














You could substitute an empty command substitution:



docker run 
--rm
-u root
$(: -p 8080:8080 )
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean





share|improve this answer
























  • Not very readable (and Vim doesn't recognise : as the empty command or a comment) but works.

    – tricasse
    Feb 6 at 18:07











  • I like that with this solution I can still copy it with the mouse, paste it and run it. Now just need a vi macro to be able to do/undo this change easy, thanks!

    – Chirlo
    Feb 6 at 18:36











  • @tricasse an alternative to : would be true

    – Jeff Schaller
    Feb 6 at 20:27











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes









11














You can't comment out a piece of a line.



Note that since the newlines are escaped, the command is actually a single line (to the shell parser), and there's no way to comment out a part of a single line (except for at the very end).



Instead, maybe just make a copy of the original command in an editor and comment it out completely while keeping the modified command uncommented:



docker run 
--rm
-u root
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean

# Was originally:
# docker run
# --rm
# -u root
# -p 8080:8080
# -v jenkins-data:/var/jenkins_home
# -v /var/run/docker.sock:/var/run/docker.sock
# -v "$HOME":/home
# jenkinsci/blueocean




Alternatively, if you want to occasionally delete or change the -p option and its argument (assuming bash or a shell with the same array syntax):



port=( -p 8080:8080 )

docker run
--rm
-u root
"${port[@]}"
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


Then just change or comment out the assignment to port.



Taking this further:



docker_run_args=(
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"


Inside the array assignment, there are no issues with commenting out a line:



docker_run_args=(
--rm
-u root
# -p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"





share|improve this answer





















  • 5





    Putting all the arguments in an array is also a great way to do it, since you can have a multi-line array definition with no ``s and you can comment individual lines... Wanna include that into your answer too?

    – filbranden
    Feb 6 at 16:02








  • 1





    @filbranden Will do as it's a good extension of what I already had written. Thanks.

    – Kusalananda
    Feb 6 at 16:04
















11














You can't comment out a piece of a line.



Note that since the newlines are escaped, the command is actually a single line (to the shell parser), and there's no way to comment out a part of a single line (except for at the very end).



Instead, maybe just make a copy of the original command in an editor and comment it out completely while keeping the modified command uncommented:



docker run 
--rm
-u root
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean

# Was originally:
# docker run
# --rm
# -u root
# -p 8080:8080
# -v jenkins-data:/var/jenkins_home
# -v /var/run/docker.sock:/var/run/docker.sock
# -v "$HOME":/home
# jenkinsci/blueocean




Alternatively, if you want to occasionally delete or change the -p option and its argument (assuming bash or a shell with the same array syntax):



port=( -p 8080:8080 )

docker run
--rm
-u root
"${port[@]}"
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


Then just change or comment out the assignment to port.



Taking this further:



docker_run_args=(
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"


Inside the array assignment, there are no issues with commenting out a line:



docker_run_args=(
--rm
-u root
# -p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"





share|improve this answer





















  • 5





    Putting all the arguments in an array is also a great way to do it, since you can have a multi-line array definition with no ``s and you can comment individual lines... Wanna include that into your answer too?

    – filbranden
    Feb 6 at 16:02








  • 1





    @filbranden Will do as it's a good extension of what I already had written. Thanks.

    – Kusalananda
    Feb 6 at 16:04














11












11








11







You can't comment out a piece of a line.



Note that since the newlines are escaped, the command is actually a single line (to the shell parser), and there's no way to comment out a part of a single line (except for at the very end).



Instead, maybe just make a copy of the original command in an editor and comment it out completely while keeping the modified command uncommented:



docker run 
--rm
-u root
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean

# Was originally:
# docker run
# --rm
# -u root
# -p 8080:8080
# -v jenkins-data:/var/jenkins_home
# -v /var/run/docker.sock:/var/run/docker.sock
# -v "$HOME":/home
# jenkinsci/blueocean




Alternatively, if you want to occasionally delete or change the -p option and its argument (assuming bash or a shell with the same array syntax):



port=( -p 8080:8080 )

docker run
--rm
-u root
"${port[@]}"
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


Then just change or comment out the assignment to port.



Taking this further:



docker_run_args=(
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"


Inside the array assignment, there are no issues with commenting out a line:



docker_run_args=(
--rm
-u root
# -p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"





share|improve this answer















You can't comment out a piece of a line.



Note that since the newlines are escaped, the command is actually a single line (to the shell parser), and there's no way to comment out a part of a single line (except for at the very end).



Instead, maybe just make a copy of the original command in an editor and comment it out completely while keeping the modified command uncommented:



docker run 
--rm
-u root
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean

# Was originally:
# docker run
# --rm
# -u root
# -p 8080:8080
# -v jenkins-data:/var/jenkins_home
# -v /var/run/docker.sock:/var/run/docker.sock
# -v "$HOME":/home
# jenkinsci/blueocean




Alternatively, if you want to occasionally delete or change the -p option and its argument (assuming bash or a shell with the same array syntax):



port=( -p 8080:8080 )

docker run
--rm
-u root
"${port[@]}"
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean


Then just change or comment out the assignment to port.



Taking this further:



docker_run_args=(
--rm
-u root
-p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"


Inside the array assignment, there are no issues with commenting out a line:



docker_run_args=(
--rm
-u root
# -p 8080:8080
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean
)

docker run "${docker_run_args[@]}"






share|improve this answer














share|improve this answer



share|improve this answer








edited Feb 6 at 16:06

























answered Feb 6 at 15:57









KusalanandaKusalananda

134k17254418




134k17254418








  • 5





    Putting all the arguments in an array is also a great way to do it, since you can have a multi-line array definition with no ``s and you can comment individual lines... Wanna include that into your answer too?

    – filbranden
    Feb 6 at 16:02








  • 1





    @filbranden Will do as it's a good extension of what I already had written. Thanks.

    – Kusalananda
    Feb 6 at 16:04














  • 5





    Putting all the arguments in an array is also a great way to do it, since you can have a multi-line array definition with no ``s and you can comment individual lines... Wanna include that into your answer too?

    – filbranden
    Feb 6 at 16:02








  • 1





    @filbranden Will do as it's a good extension of what I already had written. Thanks.

    – Kusalananda
    Feb 6 at 16:04








5




5





Putting all the arguments in an array is also a great way to do it, since you can have a multi-line array definition with no ``s and you can comment individual lines... Wanna include that into your answer too?

– filbranden
Feb 6 at 16:02







Putting all the arguments in an array is also a great way to do it, since you can have a multi-line array definition with no ``s and you can comment individual lines... Wanna include that into your answer too?

– filbranden
Feb 6 at 16:02






1




1





@filbranden Will do as it's a good extension of what I already had written. Thanks.

– Kusalananda
Feb 6 at 16:04





@filbranden Will do as it's a good extension of what I already had written. Thanks.

– Kusalananda
Feb 6 at 16:04













8














You could substitute an empty command substitution:



docker run 
--rm
-u root
$(: -p 8080:8080 )
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean





share|improve this answer
























  • Not very readable (and Vim doesn't recognise : as the empty command or a comment) but works.

    – tricasse
    Feb 6 at 18:07











  • I like that with this solution I can still copy it with the mouse, paste it and run it. Now just need a vi macro to be able to do/undo this change easy, thanks!

    – Chirlo
    Feb 6 at 18:36











  • @tricasse an alternative to : would be true

    – Jeff Schaller
    Feb 6 at 20:27
















8














You could substitute an empty command substitution:



docker run 
--rm
-u root
$(: -p 8080:8080 )
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean





share|improve this answer
























  • Not very readable (and Vim doesn't recognise : as the empty command or a comment) but works.

    – tricasse
    Feb 6 at 18:07











  • I like that with this solution I can still copy it with the mouse, paste it and run it. Now just need a vi macro to be able to do/undo this change easy, thanks!

    – Chirlo
    Feb 6 at 18:36











  • @tricasse an alternative to : would be true

    – Jeff Schaller
    Feb 6 at 20:27














8












8








8







You could substitute an empty command substitution:



docker run 
--rm
-u root
$(: -p 8080:8080 )
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean





share|improve this answer













You could substitute an empty command substitution:



docker run 
--rm
-u root
$(: -p 8080:8080 )
-v jenkins-data:/var/jenkins_home
-v /var/run/docker.sock:/var/run/docker.sock
-v "$HOME":/home
jenkinsci/blueocean






share|improve this answer












share|improve this answer



share|improve this answer










answered Feb 6 at 16:01









Jeff SchallerJeff Schaller

42.9k1159137




42.9k1159137













  • Not very readable (and Vim doesn't recognise : as the empty command or a comment) but works.

    – tricasse
    Feb 6 at 18:07











  • I like that with this solution I can still copy it with the mouse, paste it and run it. Now just need a vi macro to be able to do/undo this change easy, thanks!

    – Chirlo
    Feb 6 at 18:36











  • @tricasse an alternative to : would be true

    – Jeff Schaller
    Feb 6 at 20:27



















  • Not very readable (and Vim doesn't recognise : as the empty command or a comment) but works.

    – tricasse
    Feb 6 at 18:07











  • I like that with this solution I can still copy it with the mouse, paste it and run it. Now just need a vi macro to be able to do/undo this change easy, thanks!

    – Chirlo
    Feb 6 at 18:36











  • @tricasse an alternative to : would be true

    – Jeff Schaller
    Feb 6 at 20:27

















Not very readable (and Vim doesn't recognise : as the empty command or a comment) but works.

– tricasse
Feb 6 at 18:07





Not very readable (and Vim doesn't recognise : as the empty command or a comment) but works.

– tricasse
Feb 6 at 18:07













I like that with this solution I can still copy it with the mouse, paste it and run it. Now just need a vi macro to be able to do/undo this change easy, thanks!

– Chirlo
Feb 6 at 18:36





I like that with this solution I can still copy it with the mouse, paste it and run it. Now just need a vi macro to be able to do/undo this change easy, thanks!

– Chirlo
Feb 6 at 18:36













@tricasse an alternative to : would be true

– Jeff Schaller
Feb 6 at 20:27





@tricasse an alternative to : would be true

– Jeff Schaller
Feb 6 at 20:27


















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