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Script runs command fine, but fails when trying to save output to variable

0

I have a shell script with the following contents. I call it with ./script.sh. Echo successfully runs, but not when I try to save its output to a variable. The same is true for all commands I've tested so far; ls, pwd, node, etc…

#!/bin/zsh

echo foo
# foo

output=$("echo foo")
# command not found: echo foo

How come?

EDIT: Fix: the last echo is inside a string!

shell-script zsh path echo

share|improve this question

edited 2 days ago

Audun Olsen

asked 2 days ago

Audun OlsenAudun Olsen

133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The only way I'm able to reproduce this is if I quote echo foo like: output=$('echo foo'). Or if I escape the space like output=$(echo foo).
  
  – Jesse_b
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can't reproduce here; do you have something odd in your zsh profile?
  
  – Jeff Schaller
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm sorry, The last echo is supposed to reside in a string.
  
  – Audun Olsen
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  running "echo foo" would result in a similar error; what's your goal?
  
  – Jeff Schaller
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Simply saving the output to a variable. But I see that it being inside a string may be my mistake…
  
  – Audun Olsen
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

0

I have a shell script with the following contents. I call it with ./script.sh. Echo successfully runs, but not when I try to save its output to a variable. The same is true for all commands I've tested so far; ls, pwd, node, etc…

#!/bin/zsh

echo foo
# foo

output=$("echo foo")
# command not found: echo foo

How come?

EDIT: Fix: the last echo is inside a string!

shell-script zsh path echo

share|improve this question

edited 2 days ago

Audun Olsen

asked 2 days ago

Audun OlsenAudun Olsen

133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The only way I'm able to reproduce this is if I quote echo foo like: output=$('echo foo'). Or if I escape the space like output=$(echo foo).
  
  – Jesse_b
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can't reproduce here; do you have something odd in your zsh profile?
  
  – Jeff Schaller
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm sorry, The last echo is supposed to reside in a string.
  
  – Audun Olsen
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  running "echo foo" would result in a similar error; what's your goal?
  
  – Jeff Schaller
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Simply saving the output to a variable. But I see that it being inside a string may be my mistake…
  
  – Audun Olsen
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

I have a shell script with the following contents. I call it with ./script.sh. Echo successfully runs, but not when I try to save its output to a variable. The same is true for all commands I've tested so far; ls, pwd, node, etc…

#!/bin/zsh

echo foo
# foo

output=$("echo foo")
# command not found: echo foo

How come?

EDIT: Fix: the last echo is inside a string!

shell-script zsh path echo

share|improve this question

edited 2 days ago

Audun Olsen

asked 2 days ago

Audun OlsenAudun Olsen

133

#!/bin/zsh

echo foo
# foo

output=$("echo foo")
# command not found: echo foo

share|improve this question

edited 2 days ago

Audun Olsen

asked 2 days ago

Audun OlsenAudun Olsen

133

share|improve this question

edited 2 days ago

Audun Olsen

asked 2 days ago

Audun OlsenAudun Olsen

133

asked 2 days ago

Audun OlsenAudun Olsen

133

asked 2 days ago

Audun OlsenAudun Olsen

133

2 Answers
2

active

oldest

votes

1

It seems like you found the issue. You should quote only the string passed to echo and not the entire command itself:

output=$("echo foo")

Should instead be:

output=$(echo "foo")

If you want the output variable to contain the literal string echo foo then you should just set it as:

output='echo foo'

but could also do:

output=$(echo 'echo foo')

share|improve this answer

answered 2 days ago

Jesse_bJesse_b

14k23572

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, this works. An additional question; if i replace echo with node, which was my initial goal, calling a node-script. I get command not found: node despite node being inside my $PATH variable. Which I checked by echoing path from inside the script. Thoughts?
  
  – Audun Olsen
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Without seeing the command you're attempting to execute and your path I can't really speculate.
  
  – Jesse_b
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  node script.js succeeds, output=$(node script.js) fails. Falls perfectly in line with the question title, actually.
  
  – Audun Olsen
  2 days ago
  

  
  
  
  

add a comment | 

0

As one might expect commands inside parentheses creates a subshell. But $() is not what you might think. It does not return the output of a command as a string it substitutes the output of a command into the shell. If you want proof try this: $(echo "echo 123")

Instead you want to use double quotes to capture it into a string instead of the array (the output split by spaces) being run on the shell.

output="$(echo foo)"

Note this isn't an issue specific to zsh it applies to bash as well.

share|improve this answer

answered 2 days ago

jdwolfjdwolf

2,833217

add a comment | 

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1

It seems like you found the issue. You should quote only the string passed to echo and not the entire command itself:

output=$("echo foo")

Should instead be:

output=$(echo "foo")

If you want the output variable to contain the literal string echo foo then you should just set it as:

output='echo foo'

but could also do:

output=$(echo 'echo foo')

share|improve this answer

answered 2 days ago

Jesse_bJesse_b

14k23572

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, this works. An additional question; if i replace echo with node, which was my initial goal, calling a node-script. I get command not found: node despite node being inside my $PATH variable. Which I checked by echoing path from inside the script. Thoughts?
  
  – Audun Olsen
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Without seeing the command you're attempting to execute and your path I can't really speculate.
  
  – Jesse_b
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  node script.js succeeds, output=$(node script.js) fails. Falls perfectly in line with the question title, actually.
  
  – Audun Olsen
  2 days ago
  

  
  
  
  

add a comment | 

1

It seems like you found the issue. You should quote only the string passed to echo and not the entire command itself:

output=$("echo foo")

Should instead be:

output=$(echo "foo")

If you want the output variable to contain the literal string echo foo then you should just set it as:

output='echo foo'

but could also do:

output=$(echo 'echo foo')

share|improve this answer

answered 2 days ago

Jesse_bJesse_b

14k23572

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, this works. An additional question; if i replace echo with node, which was my initial goal, calling a node-script. I get command not found: node despite node being inside my $PATH variable. Which I checked by echoing path from inside the script. Thoughts?
  
  – Audun Olsen
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Without seeing the command you're attempting to execute and your path I can't really speculate.
  
  – Jesse_b
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  node script.js succeeds, output=$(node script.js) fails. Falls perfectly in line with the question title, actually.
  
  – Audun Olsen
  2 days ago
  

  
  
  
  

add a comment | 

It seems like you found the issue. You should quote only the string passed to echo and not the entire command itself:

output=$("echo foo")

Should instead be:

output=$(echo "foo")

If you want the output variable to contain the literal string echo foo then you should just set it as:

output='echo foo'

but could also do:

output=$(echo 'echo foo')

share|improve this answer

answered 2 days ago

Jesse_bJesse_b

14k23572

output=$("echo foo")

output=$(echo "foo")

output='echo foo'

output=$(echo 'echo foo')

share|improve this answer

answered 2 days ago

Jesse_bJesse_b

14k23572

answered 2 days ago

Jesse_bJesse_b

14k23572

answered 2 days ago

Jesse_bJesse_b

14k23572

0

As one might expect commands inside parentheses creates a subshell. But $() is not what you might think. It does not return the output of a command as a string it substitutes the output of a command into the shell. If you want proof try this: $(echo "echo 123")

Instead you want to use double quotes to capture it into a string instead of the array (the output split by spaces) being run on the shell.

output="$(echo foo)"

Note this isn't an issue specific to zsh it applies to bash as well.

share|improve this answer

answered 2 days ago

jdwolfjdwolf

2,833217

add a comment | 

0

As one might expect commands inside parentheses creates a subshell. But $() is not what you might think. It does not return the output of a command as a string it substitutes the output of a command into the shell. If you want proof try this: $(echo "echo 123")

Instead you want to use double quotes to capture it into a string instead of the array (the output split by spaces) being run on the shell.

output="$(echo foo)"

Note this isn't an issue specific to zsh it applies to bash as well.

share|improve this answer

answered 2 days ago

jdwolfjdwolf

2,833217

add a comment | 

As one might expect commands inside parentheses creates a subshell. But $() is not what you might think. It does not return the output of a command as a string it substitutes the output of a command into the shell. If you want proof try this: $(echo "echo 123")

Instead you want to use double quotes to capture it into a string instead of the array (the output split by spaces) being run on the shell.

output="$(echo foo)"

Note this isn't an issue specific to zsh it applies to bash as well.

share|improve this answer

answered 2 days ago

jdwolfjdwolf

2,833217

share|improve this answer

answered 2 days ago

jdwolfjdwolf

2,833217

answered 2 days ago

jdwolfjdwolf

2,833217

answered 2 days ago

jdwolfjdwolf

2,833217