Unable to assign env variable using a command's output
up vote
1
down vote
favorite
I can set environment variables like this.
➤ foo=bar env | grep foo
foo=bar
But, what if, I can only get foo=bar string after I execute a command (which is my use-case). The simplest way to emulate this is using a simple echo command.
➤ `echo foo=bar` env | grep foo
zsh: command not found: foo=bar
In this, zsh/bash starts interpreting it as a command. How do I fix this?
bash zsh
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
add a comment |
up vote
1
down vote
favorite
I can set environment variables like this.
➤ foo=bar env | grep foo
foo=bar
But, what if, I can only get foo=bar string after I execute a command (which is my use-case). The simplest way to emulate this is using a simple echo command.
➤ `echo foo=bar` env | grep foo
zsh: command not found: foo=bar
In this, zsh/bash starts interpreting it as a command. How do I fix this?
bash zsh
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
1
It's not clear what you are asking.
– Emily E.
Nov 22 at 13:24
1
FWIW, I find this question very clear. I think it's a good example of How to create a Minimal, Complete, and Verifiable example (MCVE).
– pjh
Nov 22 at 19:18
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I can set environment variables like this.
➤ foo=bar env | grep foo
foo=bar
But, what if, I can only get foo=bar string after I execute a command (which is my use-case). The simplest way to emulate this is using a simple echo command.
➤ `echo foo=bar` env | grep foo
zsh: command not found: foo=bar
In this, zsh/bash starts interpreting it as a command. How do I fix this?
bash zsh
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
I can set environment variables like this.
➤ foo=bar env | grep foo
foo=bar
But, what if, I can only get foo=bar string after I execute a command (which is my use-case). The simplest way to emulate this is using a simple echo command.
➤ `echo foo=bar` env | grep foo
zsh: command not found: foo=bar
In this, zsh/bash starts interpreting it as a command. How do I fix this?
bash zsh
bash zsh
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
asked Nov 22 at 12:50
Abhijeet Rastogi
9,9191964108
9,9191964108
1
It's not clear what you are asking.
– Emily E.
Nov 22 at 13:24
1
FWIW, I find this question very clear. I think it's a good example of How to create a Minimal, Complete, and Verifiable example (MCVE).
– pjh
Nov 22 at 19:18
add a comment |
1
It's not clear what you are asking.
– Emily E.
Nov 22 at 13:24
1
FWIW, I find this question very clear. I think it's a good example of How to create a Minimal, Complete, and Verifiable example (MCVE).
– pjh
Nov 22 at 19:18
1
1
It's not clear what you are asking.
– Emily E.
Nov 22 at 13:24
It's not clear what you are asking.
– Emily E.
Nov 22 at 13:24
1
1
FWIW, I find this question very clear. I think it's a good example of How to create a Minimal, Complete, and Verifiable example (MCVE).
– pjh
Nov 22 at 19:18
FWIW, I find this question very clear. I think it's a good example of How to create a Minimal, Complete, and Verifiable example (MCVE).
– pjh
Nov 22 at 19:18
add a comment |
3 Answers
3
active
oldest
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up vote
3
down vote
The problem is that the shell looks for the form var1=val1 var2=val2 ... command before doing expansions, including command substitution (`...` or $(...)).
One way to work around the problem is to use eval to cause the shell to do its parsing after the command substitution is done:
eval "$(echo foo=bar)" env | grep foo
(See What is the benefit of using $() instead of backticks in shell scripts? for an explanation of why I've changed `...` to $(...).)
Unfortunately, eval is potentially dangerous. It should not be used unless there is no other alternative. See Why should eval be avoided in Bash, and what should I use instead?.
Another alternative is to use export in a subshell. One way is:
{ export "$(echo foo=bar)" ; env ; } | grep foo
Since the export is done in a pipeline (and not the last part of a pipeline, which would make a difference for some shells or some modes) it doesn't affect the variable settings in the current shell. If the command is not in a pipeline, then it would be necessary to explicitly run it in a subshell to get the same effect. E.g.
( export "$(echo foo=bar)" ; env ) > tmpfile
grep foo tmpfile
answered Nov 22 at 19:43
pjh
1,584611
add a comment |
up vote
0
down vote
It is working for me as expected. Have you tried like below?
env `echo foo=bar` | grep foo
answered Nov 22 at 13:02
Mohit Kumar
584420
add a comment |
up vote
0
down vote
I'm not sure if I understood your question, but try this:
$ echo "hello" && foo=bar env | grep foo
hello
foo=bar
answered Nov 22 at 13:03
downtheroad
535
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The problem is that the shell looks for the form var1=val1 var2=val2 ... command before doing expansions, including command substitution (`...` or $(...)).
One way to work around the problem is to use eval to cause the shell to do its parsing after the command substitution is done:
eval "$(echo foo=bar)" env | grep foo
(See What is the benefit of using $() instead of backticks in shell scripts? for an explanation of why I've changed `...` to $(...).)
Unfortunately, eval is potentially dangerous. It should not be used unless there is no other alternative. See Why should eval be avoided in Bash, and what should I use instead?.
Another alternative is to use export in a subshell. One way is:
{ export "$(echo foo=bar)" ; env ; } | grep foo
Since the export is done in a pipeline (and not the last part of a pipeline, which would make a difference for some shells or some modes) it doesn't affect the variable settings in the current shell. If the command is not in a pipeline, then it would be necessary to explicitly run it in a subshell to get the same effect. E.g.
( export "$(echo foo=bar)" ; env ) > tmpfile
grep foo tmpfile
answered Nov 22 at 19:43
pjh
1,584611
add a comment |
up vote
3
down vote
The problem is that the shell looks for the form var1=val1 var2=val2 ... command before doing expansions, including command substitution (`...` or $(...)).
One way to work around the problem is to use eval to cause the shell to do its parsing after the command substitution is done:
eval "$(echo foo=bar)" env | grep foo
(See What is the benefit of using $() instead of backticks in shell scripts? for an explanation of why I've changed `...` to $(...).)
Unfortunately, eval is potentially dangerous. It should not be used unless there is no other alternative. See Why should eval be avoided in Bash, and what should I use instead?.
Another alternative is to use export in a subshell. One way is:
{ export "$(echo foo=bar)" ; env ; } | grep foo
Since the export is done in a pipeline (and not the last part of a pipeline, which would make a difference for some shells or some modes) it doesn't affect the variable settings in the current shell. If the command is not in a pipeline, then it would be necessary to explicitly run it in a subshell to get the same effect. E.g.
( export "$(echo foo=bar)" ; env ) > tmpfile
grep foo tmpfile
answered Nov 22 at 19:43
pjh
1,584611
add a comment |
up vote
3
down vote
up vote
3
down vote
The problem is that the shell looks for the form var1=val1 var2=val2 ... command before doing expansions, including command substitution (`...` or $(...)).
One way to work around the problem is to use eval to cause the shell to do its parsing after the command substitution is done:
eval "$(echo foo=bar)" env | grep foo
(See What is the benefit of using $() instead of backticks in shell scripts? for an explanation of why I've changed `...` to $(...).)
Unfortunately, eval is potentially dangerous. It should not be used unless there is no other alternative. See Why should eval be avoided in Bash, and what should I use instead?.
Another alternative is to use export in a subshell. One way is:
{ export "$(echo foo=bar)" ; env ; } | grep foo
Since the export is done in a pipeline (and not the last part of a pipeline, which would make a difference for some shells or some modes) it doesn't affect the variable settings in the current shell. If the command is not in a pipeline, then it would be necessary to explicitly run it in a subshell to get the same effect. E.g.
( export "$(echo foo=bar)" ; env ) > tmpfile
grep foo tmpfile
answered Nov 22 at 19:43
pjh
1,584611
The problem is that the shell looks for the form var1=val1 var2=val2 ... command before doing expansions, including command substitution (`...` or $(...)).
One way to work around the problem is to use eval to cause the shell to do its parsing after the command substitution is done:
eval "$(echo foo=bar)" env | grep foo
(See What is the benefit of using $() instead of backticks in shell scripts? for an explanation of why I've changed `...` to $(...).)
Unfortunately, eval is potentially dangerous. It should not be used unless there is no other alternative. See Why should eval be avoided in Bash, and what should I use instead?.
Another alternative is to use export in a subshell. One way is:
{ export "$(echo foo=bar)" ; env ; } | grep foo
Since the export is done in a pipeline (and not the last part of a pipeline, which would make a difference for some shells or some modes) it doesn't affect the variable settings in the current shell. If the command is not in a pipeline, then it would be necessary to explicitly run it in a subshell to get the same effect. E.g.
( export "$(echo foo=bar)" ; env ) > tmpfile
grep foo tmpfile
answered Nov 22 at 19:43
pjh
1,584611
answered Nov 22 at 19:43
pjh
1,584611
answered Nov 22 at 19:43
pjh
1,584611
answered Nov 22 at 19:43
pjh
1,584611
1,584611
add a comment |
add a comment |
up vote
0
down vote
It is working for me as expected. Have you tried like below?
env `echo foo=bar` | grep foo
answered Nov 22 at 13:02
Mohit Kumar
584420
add a comment |
up vote
0
down vote
It is working for me as expected. Have you tried like below?
env `echo foo=bar` | grep foo
answered Nov 22 at 13:02
Mohit Kumar
584420
add a comment |
up vote
0
down vote
up vote
0
down vote
It is working for me as expected. Have you tried like below?
env `echo foo=bar` | grep foo
answered Nov 22 at 13:02
Mohit Kumar
584420
It is working for me as expected. Have you tried like below?
env `echo foo=bar` | grep foo
answered Nov 22 at 13:02
Mohit Kumar
584420
answered Nov 22 at 13:02
Mohit Kumar
584420
answered Nov 22 at 13:02
Mohit Kumar
584420
answered Nov 22 at 13:02
Mohit Kumar
584420
584420
add a comment |
add a comment |
up vote
0
down vote
I'm not sure if I understood your question, but try this:
$ echo "hello" && foo=bar env | grep foo
hello
foo=bar
answered Nov 22 at 13:03
downtheroad
535
add a comment |
up vote
0
down vote
I'm not sure if I understood your question, but try this:
$ echo "hello" && foo=bar env | grep foo
hello
foo=bar
answered Nov 22 at 13:03
downtheroad
535
add a comment |
up vote
0
down vote
up vote
0
down vote
I'm not sure if I understood your question, but try this:
$ echo "hello" && foo=bar env | grep foo
hello
foo=bar
answered Nov 22 at 13:03
downtheroad
535
I'm not sure if I understood your question, but try this:
$ echo "hello" && foo=bar env | grep foo
hello
foo=bar
answered Nov 22 at 13:03
downtheroad
535
answered Nov 22 at 13:03
downtheroad
535
answered Nov 22 at 13:03
downtheroad
535
answered Nov 22 at 13:03
downtheroad
535
535
add a comment |
add a comment |
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1
It's not clear what you are asking.
– Emily E.
Nov 22 at 13:24
1
FWIW, I find this question very clear. I think it's a good example of How to create a Minimal, Complete, and Verifiable example (MCVE).
– pjh
Nov 22 at 19:18