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Passing arguments from one script to another

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.

bash scripting variable arguments

share|improve this question

edited Mar 16 at 18:41

ctrl-alt-delor

12.2k42561

asked Mar 16 at 16:01

appleapple

274

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  please also read stackoverflow.com/q/5228345/4023950
  
  – αғsнιη
  Mar 16 at 16:19
  

  
  
  
  

add a comment | 

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.

bash scripting variable arguments

share|improve this question

edited Mar 16 at 18:41

ctrl-alt-delor

12.2k42561

asked Mar 16 at 16:01

appleapple

274

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  please also read stackoverflow.com/q/5228345/4023950
  
  – αғsнιη
  Mar 16 at 16:19
  

  
  
  
  

add a comment | 

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

edited Mar 16 at 18:41

ctrl-alt-delor

12.2k42561

asked Mar 16 at 16:01

appleapple

274

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

source $file

variable1=10

variable2=apple

share|improve this question

edited Mar 16 at 18:41

ctrl-alt-delor

12.2k42561

asked Mar 16 at 16:01

appleapple

274

share|improve this question

edited Mar 16 at 18:41

ctrl-alt-delor

12.2k42561

asked Mar 16 at 16:01

appleapple

274

edited Mar 16 at 18:41

ctrl-alt-delor

12.2k42561

edited Mar 16 at 18:41

ctrl-alt-delor

12.2k42561

asked Mar 16 at 16:01

appleapple

274

asked Mar 16 at 16:01

appleapple

274

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

edited Mar 16 at 16:49

answered Mar 16 at 16:10

Jesse_bJesse_b

14.1k23572

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Mar 16 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is there something I need to do in the second script to receive $variable1 ?
  
  – apple
  Mar 16 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Many thanks! Working now.
  
  – apple
  Mar 16 at 16:40
  

  
  
  
  

add a comment | 

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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

edited Mar 16 at 16:49

answered Mar 16 at 16:10

Jesse_bJesse_b

14.1k23572

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Mar 16 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is there something I need to do in the second script to receive $variable1 ?
  
  – apple
  Mar 16 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Many thanks! Working now.
  
  – apple
  Mar 16 at 16:40
  

  
  
  
  

add a comment | 

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

edited Mar 16 at 16:49

answered Mar 16 at 16:10

Jesse_bJesse_b

14.1k23572

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Mar 16 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is there something I need to do in the second script to receive $variable1 ?
  
  – apple
  Mar 16 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Many thanks! Working now.
  
  – apple
  Mar 16 at 16:40
  

  
  
  
  

add a comment | 

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

edited Mar 16 at 16:49

answered Mar 16 at 16:10

Jesse_bJesse_b

14.1k23572

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

source "$file"



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

variable1=$1

variable2=$2

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

/path/to/another/script2.sh "$file"

file=$1

source "$file"

printf '%sn' "$variable1"

printf '%sn' "$variable2"

share|improve this answer

edited Mar 16 at 16:49

answered Mar 16 at 16:10

Jesse_bJesse_b

14.1k23572

answered Mar 16 at 16:10

Jesse_bJesse_b

14.1k23572

answered Mar 16 at 16:10

Jesse_bJesse_b

14.1k23572