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Defining several variables at once (copy-paste-execute in shell)

How to print shell variables and values to be able to copy/paste them?When should one use $( ) in defining variablesSed command usage without defining variables in shell scriptbash- defining variables with VAR=${[number]:-default}Execution of a remote Bash script in GitHub fails with various methods - Maybe due to Windows10Shell Script : Variables correct formatUsing a bash array in an awk and also quoted variable: conflicting syntax issueRunning script line by line automatically yet being asked before each line from second line onwards

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

0

I have a list of variables I copy-paste one by one from a file to my shell and hit Enter to execute. For example:

var1="myvar1"

var2="myvar2"

Because in practice there are more variables, instead copy-pasting them and execute them one by one I desire to do all in one operation to save time.

I tried copy-pasting:

(

var1="myvar1"

var2="myvar2"

)

echo $var1 outputs an empty line instead date in relevant format.

How could I execute a few variables in one operation?

shell-script variable syntax

share|improve this question

edited yesterday

JohnDoea

asked yesterday

JohnDoeaJohnDoea

7211 gold badge1111 silver badges4848 bronze badges

add a comment | 

0

I have a list of variables I copy-paste one by one from a file to my shell and hit Enter to execute. For example:

var1="myvar1"

var2="myvar2"

Because in practice there are more variables, instead copy-pasting them and execute them one by one I desire to do all in one operation to save time.

I tried copy-pasting:

(

var1="myvar1"

var2="myvar2"

)

echo $var1 outputs an empty line instead date in relevant format.

How could I execute a few variables in one operation?

shell-script variable syntax

share|improve this question

edited yesterday

JohnDoea

asked yesterday

JohnDoeaJohnDoea

7211 gold badge1111 silver badges4848 bronze badges

add a comment | 

I have a list of variables I copy-paste one by one from a file to my shell and hit Enter to execute. For example:

var1="myvar1"

var2="myvar2"

Because in practice there are more variables, instead copy-pasting them and execute them one by one I desire to do all in one operation to save time.

I tried copy-pasting:

(

var1="myvar1"

var2="myvar2"

)

echo $var1 outputs an empty line instead date in relevant format.

How could I execute a few variables in one operation?

shell-script variable syntax

share|improve this question

edited yesterday

JohnDoea

asked yesterday

JohnDoeaJohnDoea

7211 gold badge1111 silver badges4848 bronze badges

var1="myvar1"

var2="myvar2"

(

var1="myvar1"

var2="myvar2"

)

share|improve this question

edited yesterday

JohnDoea

asked yesterday

JohnDoeaJohnDoea

7211 gold badge1111 silver badges4848 bronze badges

share|improve this question

edited yesterday

JohnDoea

asked yesterday

JohnDoeaJohnDoea

7211 gold badge1111 silver badges4848 bronze badges

asked yesterday

JohnDoeaJohnDoea

7211 gold badge1111 silver badges4848 bronze badges

asked yesterday

JohnDoeaJohnDoea

7211 gold badge1111 silver badges4848 bronze badges

2 Answers
2

active

oldest

votes

3

You don't have copy them one by one, you can paste all the lines together and newlines will work as Enter.

The reason that

(

var1="myvar1"

var2="myvar2"

)

doesn't work is that because it's executed in a subshell. It would work if you printed contents of the variable before the final
):

(

var1="myvar1"

var2="myvar2"

echo $var2

)

It's explained in Commpound Commands section in man bash:

Compound Commands

A compound command is one of the following. In most cases a list in a
command's description may be sepa- rated from the rest of the command
by one or more newlines, and may be followed by a newline in place of
a semicolon.

(list) list is executed in a subshell environment (see COMMAND
EXECUTION ENVIRONMENT below). Variable assignments and builtin
commands that affect the shell's environment do not remain in effect
after the command completes. The return status is the exit status
of list.

Later on it also says:

{ list; }
list is simply executed in the current shell environment.
list must be terminated with a newline or semicolon. This is known
as a group command. The return status is the exit status of list.
Note that unlike the metacharacters ( and ), { and } are reserved
words and must occur where a reserved word is permitted to be
recognized. Since they do not cause a word break, they must be
separated from list by whitespace or another shell metacharacter.

So that would work as commands would not be executed in a subshell but
in a current shell environment:

{

var1="myvar1"

var2="myvar2";

}

I see you're asking a lot of simple questions about Bash. Consider
reading some Bash tutorials, man bash and learn about common Bash
pitfalls.

share|improve this answer

edited yesterday

answered yesterday

Arkadiusz DrabczykArkadiusz Drabczyk

9,08833 gold badges2020 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Some of the facts you mention I learned a few years back but some I forgot because I am not a professional sysadmin or doing sysadministration on a daily basis... Sometimes I prefer to ask to have direct documentation for me and others. I left my home country and left my books behind and not in a comfortable situation to read huge documentation. I had a Bash book from 2005 I used to read in train; I should buy the new edition in the right time for me...
  
  – JohnDoea
  yesterday
  

  
  
  
  

add a comment | 

0

The parenthesis you used causes the enclosed commands to be executed by a subshell; hence once that shell exits, de variables are still not set in the current shell.

I'm at a loss why you think those parenthesis would set those variables "in one go", or why it would be necessary to them them "in one go"; however you could place those assignments on one line such as:

var1="myvar1" var2="myvar2"

Of course there's no reason not to copy and paste those lines all together in one paste, as Arkadiusz Drabczyk commented, if that's easier and all that you want to accomplish.

share|improve this answer

edited yesterday

ilkkachu

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answered yesterday

wurtelwurtel

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add a comment | 

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3

You don't have copy them one by one, you can paste all the lines together and newlines will work as Enter.

The reason that

(

var1="myvar1"

var2="myvar2"

)

doesn't work is that because it's executed in a subshell. It would work if you printed contents of the variable before the final
):

(

var1="myvar1"

var2="myvar2"

echo $var2

)

It's explained in Commpound Commands section in man bash:

Compound Commands

A compound command is one of the following. In most cases a list in a
command's description may be sepa- rated from the rest of the command
by one or more newlines, and may be followed by a newline in place of
a semicolon.

(list) list is executed in a subshell environment (see COMMAND
EXECUTION ENVIRONMENT below). Variable assignments and builtin
commands that affect the shell's environment do not remain in effect
after the command completes. The return status is the exit status
of list.

Later on it also says:

{ list; }
list is simply executed in the current shell environment.
list must be terminated with a newline or semicolon. This is known
as a group command. The return status is the exit status of list.
Note that unlike the metacharacters ( and ), { and } are reserved
words and must occur where a reserved word is permitted to be
recognized. Since they do not cause a word break, they must be
separated from list by whitespace or another shell metacharacter.

So that would work as commands would not be executed in a subshell but
in a current shell environment:

{

var1="myvar1"

var2="myvar2";

}

I see you're asking a lot of simple questions about Bash. Consider
reading some Bash tutorials, man bash and learn about common Bash
pitfalls.

share|improve this answer

edited yesterday

answered yesterday

Arkadiusz DrabczykArkadiusz Drabczyk

9,08833 gold badges2020 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Some of the facts you mention I learned a few years back but some I forgot because I am not a professional sysadmin or doing sysadministration on a daily basis... Sometimes I prefer to ask to have direct documentation for me and others. I left my home country and left my books behind and not in a comfortable situation to read huge documentation. I had a Bash book from 2005 I used to read in train; I should buy the new edition in the right time for me...
  
  – JohnDoea
  yesterday
  

  
  
  
  

add a comment | 

3

You don't have copy them one by one, you can paste all the lines together and newlines will work as Enter.

The reason that

(

var1="myvar1"

var2="myvar2"

)

doesn't work is that because it's executed in a subshell. It would work if you printed contents of the variable before the final
):

(

var1="myvar1"

var2="myvar2"

echo $var2

)

It's explained in Commpound Commands section in man bash:

Compound Commands

A compound command is one of the following. In most cases a list in a
command's description may be sepa- rated from the rest of the command
by one or more newlines, and may be followed by a newline in place of
a semicolon.

(list) list is executed in a subshell environment (see COMMAND
EXECUTION ENVIRONMENT below). Variable assignments and builtin
commands that affect the shell's environment do not remain in effect
after the command completes. The return status is the exit status
of list.

Later on it also says:

{ list; }
list is simply executed in the current shell environment.
list must be terminated with a newline or semicolon. This is known
as a group command. The return status is the exit status of list.
Note that unlike the metacharacters ( and ), { and } are reserved
words and must occur where a reserved word is permitted to be
recognized. Since they do not cause a word break, they must be
separated from list by whitespace or another shell metacharacter.

So that would work as commands would not be executed in a subshell but
in a current shell environment:

{

var1="myvar1"

var2="myvar2";

}

I see you're asking a lot of simple questions about Bash. Consider
reading some Bash tutorials, man bash and learn about common Bash
pitfalls.

share|improve this answer

edited yesterday

answered yesterday

Arkadiusz DrabczykArkadiusz Drabczyk

9,08833 gold badges2020 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Some of the facts you mention I learned a few years back but some I forgot because I am not a professional sysadmin or doing sysadministration on a daily basis... Sometimes I prefer to ask to have direct documentation for me and others. I left my home country and left my books behind and not in a comfortable situation to read huge documentation. I had a Bash book from 2005 I used to read in train; I should buy the new edition in the right time for me...
  
  – JohnDoea
  yesterday
  

  
  
  
  

add a comment | 

You don't have copy them one by one, you can paste all the lines together and newlines will work as Enter.

The reason that

(

var1="myvar1"

var2="myvar2"

)

doesn't work is that because it's executed in a subshell. It would work if you printed contents of the variable before the final
):

(

var1="myvar1"

var2="myvar2"

echo $var2

)

It's explained in Commpound Commands section in man bash:

Compound Commands

A compound command is one of the following. In most cases a list in a
command's description may be sepa- rated from the rest of the command
by one or more newlines, and may be followed by a newline in place of
a semicolon.

(list) list is executed in a subshell environment (see COMMAND
EXECUTION ENVIRONMENT below). Variable assignments and builtin
commands that affect the shell's environment do not remain in effect
after the command completes. The return status is the exit status
of list.

Later on it also says:

{ list; }
list is simply executed in the current shell environment.
list must be terminated with a newline or semicolon. This is known
as a group command. The return status is the exit status of list.
Note that unlike the metacharacters ( and ), { and } are reserved
words and must occur where a reserved word is permitted to be
recognized. Since they do not cause a word break, they must be
separated from list by whitespace or another shell metacharacter.

So that would work as commands would not be executed in a subshell but
in a current shell environment:

{

var1="myvar1"

var2="myvar2";

}

I see you're asking a lot of simple questions about Bash. Consider
reading some Bash tutorials, man bash and learn about common Bash
pitfalls.

share|improve this answer

edited yesterday

answered yesterday

Arkadiusz DrabczykArkadiusz Drabczyk

9,08833 gold badges2020 silver badges3636 bronze badges

(

var1="myvar1"

var2="myvar2"

)

(

var1="myvar1"

var2="myvar2"

echo $var2

)

{

var1="myvar1"

var2="myvar2";

}

share|improve this answer

edited yesterday

answered yesterday

Arkadiusz DrabczykArkadiusz Drabczyk

9,08833 gold badges2020 silver badges3636 bronze badges

answered yesterday

Arkadiusz DrabczykArkadiusz Drabczyk

9,08833 gold badges2020 silver badges3636 bronze badges

answered yesterday

Arkadiusz DrabczykArkadiusz Drabczyk

9,08833 gold badges2020 silver badges3636 bronze badges

0

The parenthesis you used causes the enclosed commands to be executed by a subshell; hence once that shell exits, de variables are still not set in the current shell.

I'm at a loss why you think those parenthesis would set those variables "in one go", or why it would be necessary to them them "in one go"; however you could place those assignments on one line such as:

var1="myvar1" var2="myvar2"

Of course there's no reason not to copy and paste those lines all together in one paste, as Arkadiusz Drabczyk commented, if that's easier and all that you want to accomplish.

share|improve this answer

edited yesterday

ilkkachu

67.3k1010 gold badges112112 silver badges193193 bronze badges

answered yesterday

wurtelwurtel

11.8k11 gold badge1616 silver badges2929 bronze badges

add a comment | 

0

The parenthesis you used causes the enclosed commands to be executed by a subshell; hence once that shell exits, de variables are still not set in the current shell.

I'm at a loss why you think those parenthesis would set those variables "in one go", or why it would be necessary to them them "in one go"; however you could place those assignments on one line such as:

var1="myvar1" var2="myvar2"

Of course there's no reason not to copy and paste those lines all together in one paste, as Arkadiusz Drabczyk commented, if that's easier and all that you want to accomplish.

share|improve this answer

edited yesterday

ilkkachu

67.3k1010 gold badges112112 silver badges193193 bronze badges

answered yesterday

wurtelwurtel

11.8k11 gold badge1616 silver badges2929 bronze badges

add a comment | 

The parenthesis you used causes the enclosed commands to be executed by a subshell; hence once that shell exits, de variables are still not set in the current shell.

I'm at a loss why you think those parenthesis would set those variables "in one go", or why it would be necessary to them them "in one go"; however you could place those assignments on one line such as:

var1="myvar1" var2="myvar2"

Of course there's no reason not to copy and paste those lines all together in one paste, as Arkadiusz Drabczyk commented, if that's easier and all that you want to accomplish.

share|improve this answer

edited yesterday

ilkkachu

67.3k1010 gold badges112112 silver badges193193 bronze badges

answered yesterday

wurtelwurtel

11.8k11 gold badge1616 silver badges2929 bronze badges

var1="myvar1" var2="myvar2"

share|improve this answer

edited yesterday

ilkkachu

67.3k1010 gold badges112112 silver badges193193 bronze badges

answered yesterday

wurtelwurtel

11.8k11 gold badge1616 silver badges2929 bronze badges

edited yesterday

ilkkachu

67.3k1010 gold badges112112 silver badges193193 bronze badges

edited yesterday

ilkkachu

67.3k1010 gold badges112112 silver badges193193 bronze badges

answered yesterday

wurtelwurtel

11.8k11 gold badge1616 silver badges2929 bronze badges

answered yesterday

wurtelwurtel

11.8k11 gold badge1616 silver badges2929 bronze badges