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Passing arguments from one script to another



2019 Community Moderator ElectionPass command line arguments to bash scriptBash globbing and argument passingAdd arguments to 'bash -c'Passing arguments from one command into the nextHow to extract unknown arguments within a shell script?functions argumentsCall one shell script with anotherBASH: how to pass a default argument if no arguments after the first were passedBash script to pass arguments to a scriptHow to pass arguments to a script that were generated by another script










1















Let's say I am in a Bash script and am sourcing a file. How do I pass variables to another script after I source the file like this.



sed -i 's/ = /=/' $file
source $file


Let's say file contains



variable1=10
variable2=apple


If I want to use these in another script, how do I pass these arguments to the other script, then run the script in my current Bash script.










share|improve this question



















  • 2





    please also read stackoverflow.com/q/5228345/4023950

    – αғsнιη
    4 hours ago















1















Let's say I am in a Bash script and am sourcing a file. How do I pass variables to another script after I source the file like this.



sed -i 's/ = /=/' $file
source $file


Let's say file contains



variable1=10
variable2=apple


If I want to use these in another script, how do I pass these arguments to the other script, then run the script in my current Bash script.










share|improve this question



















  • 2





    please also read stackoverflow.com/q/5228345/4023950

    – αғsнιη
    4 hours ago













1












1








1








Let's say I am in a Bash script and am sourcing a file. How do I pass variables to another script after I source the file like this.



sed -i 's/ = /=/' $file
source $file


Let's say file contains



variable1=10
variable2=apple


If I want to use these in another script, how do I pass these arguments to the other script, then run the script in my current Bash script.










share|improve this question
















Let's say I am in a Bash script and am sourcing a file. How do I pass variables to another script after I source the file like this.



sed -i 's/ = /=/' $file
source $file


Let's say file contains



variable1=10
variable2=apple


If I want to use these in another script, how do I pass these arguments to the other script, then run the script in my current Bash script.







bash scripting variable arguments






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









ctrl-alt-delor

12k42360




12k42360










asked 4 hours ago









appleapple

274




274







  • 2





    please also read stackoverflow.com/q/5228345/4023950

    – αғsнιη
    4 hours ago












  • 2





    please also read stackoverflow.com/q/5228345/4023950

    – αғsнιη
    4 hours ago







2




2





please also read stackoverflow.com/q/5228345/4023950

– αғsнιη
4 hours ago





please also read stackoverflow.com/q/5228345/4023950

– αғsнιη
4 hours ago










1 Answer
1






active

oldest

votes


















3














You would pass them pretty much the same as you would pass arguments in any other way:



sed -i 's/ = /=/' "$file"
source "$file"

/path/to/another/script.sh "$variable1" "$variable2"


Obviously using the appropriate command line switches (or not if applicable).



If using the code as above, the value of $variable1 will be available in the other script as $1 (the 1st command line argument), while $variable2 will be available as $2.



To keep the original names in your new script you would need to reassign them using the positional parameters, ie:



variable1=$1
variable2=$2


However this may not be the most efficient way to do this, you might be better off with the suggestion below:




It sounds like you may actually want to source your file within the second script and not the first. In which case you may want to do the following:



script1.sh:



sed -i 's/ = /=/' "$file"
/path/to/another/script2.sh "$file"


script2.sh:



file=$1
source "$file"
printf '%sn' "$variable1"
printf '%sn' "$variable2"


Related recommended reading: 3.4.1 Positional Parameters




Note: assigning $1 to the file variable is not necessary, you could also simply source "$1" but I have written it this way in an attempt to show how positional parameters are handled







share|improve this answer

























  • I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

    – apple
    4 hours ago











  • Is there something I need to do in the second script to receive $variable1 ?

    – apple
    4 hours ago











  • Many thanks! Working now.

    – apple
    3 hours ago










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














You would pass them pretty much the same as you would pass arguments in any other way:



sed -i 's/ = /=/' "$file"
source "$file"

/path/to/another/script.sh "$variable1" "$variable2"


Obviously using the appropriate command line switches (or not if applicable).



If using the code as above, the value of $variable1 will be available in the other script as $1 (the 1st command line argument), while $variable2 will be available as $2.



To keep the original names in your new script you would need to reassign them using the positional parameters, ie:



variable1=$1
variable2=$2


However this may not be the most efficient way to do this, you might be better off with the suggestion below:




It sounds like you may actually want to source your file within the second script and not the first. In which case you may want to do the following:



script1.sh:



sed -i 's/ = /=/' "$file"
/path/to/another/script2.sh "$file"


script2.sh:



file=$1
source "$file"
printf '%sn' "$variable1"
printf '%sn' "$variable2"


Related recommended reading: 3.4.1 Positional Parameters




Note: assigning $1 to the file variable is not necessary, you could also simply source "$1" but I have written it this way in an attempt to show how positional parameters are handled







share|improve this answer

























  • I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

    – apple
    4 hours ago











  • Is there something I need to do in the second script to receive $variable1 ?

    – apple
    4 hours ago











  • Many thanks! Working now.

    – apple
    3 hours ago















3














You would pass them pretty much the same as you would pass arguments in any other way:



sed -i 's/ = /=/' "$file"
source "$file"

/path/to/another/script.sh "$variable1" "$variable2"


Obviously using the appropriate command line switches (or not if applicable).



If using the code as above, the value of $variable1 will be available in the other script as $1 (the 1st command line argument), while $variable2 will be available as $2.



To keep the original names in your new script you would need to reassign them using the positional parameters, ie:



variable1=$1
variable2=$2


However this may not be the most efficient way to do this, you might be better off with the suggestion below:




It sounds like you may actually want to source your file within the second script and not the first. In which case you may want to do the following:



script1.sh:



sed -i 's/ = /=/' "$file"
/path/to/another/script2.sh "$file"


script2.sh:



file=$1
source "$file"
printf '%sn' "$variable1"
printf '%sn' "$variable2"


Related recommended reading: 3.4.1 Positional Parameters




Note: assigning $1 to the file variable is not necessary, you could also simply source "$1" but I have written it this way in an attempt to show how positional parameters are handled







share|improve this answer

























  • I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

    – apple
    4 hours ago











  • Is there something I need to do in the second script to receive $variable1 ?

    – apple
    4 hours ago











  • Many thanks! Working now.

    – apple
    3 hours ago













3












3








3







You would pass them pretty much the same as you would pass arguments in any other way:



sed -i 's/ = /=/' "$file"
source "$file"

/path/to/another/script.sh "$variable1" "$variable2"


Obviously using the appropriate command line switches (or not if applicable).



If using the code as above, the value of $variable1 will be available in the other script as $1 (the 1st command line argument), while $variable2 will be available as $2.



To keep the original names in your new script you would need to reassign them using the positional parameters, ie:



variable1=$1
variable2=$2


However this may not be the most efficient way to do this, you might be better off with the suggestion below:




It sounds like you may actually want to source your file within the second script and not the first. In which case you may want to do the following:



script1.sh:



sed -i 's/ = /=/' "$file"
/path/to/another/script2.sh "$file"


script2.sh:



file=$1
source "$file"
printf '%sn' "$variable1"
printf '%sn' "$variable2"


Related recommended reading: 3.4.1 Positional Parameters




Note: assigning $1 to the file variable is not necessary, you could also simply source "$1" but I have written it this way in an attempt to show how positional parameters are handled







share|improve this answer















You would pass them pretty much the same as you would pass arguments in any other way:



sed -i 's/ = /=/' "$file"
source "$file"

/path/to/another/script.sh "$variable1" "$variable2"


Obviously using the appropriate command line switches (or not if applicable).



If using the code as above, the value of $variable1 will be available in the other script as $1 (the 1st command line argument), while $variable2 will be available as $2.



To keep the original names in your new script you would need to reassign them using the positional parameters, ie:



variable1=$1
variable2=$2


However this may not be the most efficient way to do this, you might be better off with the suggestion below:




It sounds like you may actually want to source your file within the second script and not the first. In which case you may want to do the following:



script1.sh:



sed -i 's/ = /=/' "$file"
/path/to/another/script2.sh "$file"


script2.sh:



file=$1
source "$file"
printf '%sn' "$variable1"
printf '%sn' "$variable2"


Related recommended reading: 3.4.1 Positional Parameters




Note: assigning $1 to the file variable is not necessary, you could also simply source "$1" but I have written it this way in an attempt to show how positional parameters are handled








share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago

























answered 4 hours ago









Jesse_bJesse_b

13.6k23371




13.6k23371












  • I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

    – apple
    4 hours ago











  • Is there something I need to do in the second script to receive $variable1 ?

    – apple
    4 hours ago











  • Many thanks! Working now.

    – apple
    3 hours ago

















  • I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

    – apple
    4 hours ago











  • Is there something I need to do in the second script to receive $variable1 ?

    – apple
    4 hours ago











  • Many thanks! Working now.

    – apple
    3 hours ago
















I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

– apple
4 hours ago





I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

– apple
4 hours ago













Is there something I need to do in the second script to receive $variable1 ?

– apple
4 hours ago





Is there something I need to do in the second script to receive $variable1 ?

– apple
4 hours ago













Many thanks! Working now.

– apple
3 hours ago





Many thanks! Working now.

– apple
3 hours ago

















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