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Process substitution inside a subshell to set a variable

Bad substitution inside xargs subshell?Creating temp file vs process substitution vs variable expansion?Assign Subshell background process pid to variableScope of variables in a process substitutionbash array with variable in the nameIs the command in a process substitution invoked in a subshell?Subshell and process substitutionSubshell inside quotesSet environment variable for subshellSyntax error near unexpected token 'else'

1

I'm trying to run a script remotely and use its standard output to populate a variable. I'm doing this to avoid temporary files.

Here's the pattern I'm trying:

var=$(bash <(curl -fsSkL http://remote/file.sh))

echo "var=${var}"

I'm testing this pattern without curl using cat:

var=$(bash <(cat ./local/file.sh))

echo "var=${var}"

This should be the same as far as syntax is concerned. ./local/file.sh contains echo hello, so I would expect var to contain the value hello, but alas, executing the above results in the following:

test.sh: command substitution: line 4: syntax error near unexpected token `('

test.sh: command substitution: line 4: `bash <(cat ./local/file.sh)'

var=

How can I accomplish my goal without using temporary files?

bash variable subshell process-substitution

share|improve this question

asked 1 hour ago

Sean AllredSean Allred

8143816

add a comment | 

1

I'm trying to run a script remotely and use its standard output to populate a variable. I'm doing this to avoid temporary files.

Here's the pattern I'm trying:

var=$(bash <(curl -fsSkL http://remote/file.sh))

echo "var=${var}"

I'm testing this pattern without curl using cat:

var=$(bash <(cat ./local/file.sh))

echo "var=${var}"

This should be the same as far as syntax is concerned. ./local/file.sh contains echo hello, so I would expect var to contain the value hello, but alas, executing the above results in the following:

test.sh: command substitution: line 4: syntax error near unexpected token `('

test.sh: command substitution: line 4: `bash <(cat ./local/file.sh)'

var=

How can I accomplish my goal without using temporary files?

bash variable subshell process-substitution

share|improve this question

asked 1 hour ago

Sean AllredSean Allred

8143816

add a comment | 

I'm trying to run a script remotely and use its standard output to populate a variable. I'm doing this to avoid temporary files.

Here's the pattern I'm trying:

var=$(bash <(curl -fsSkL http://remote/file.sh))

echo "var=${var}"

I'm testing this pattern without curl using cat:

var=$(bash <(cat ./local/file.sh))

echo "var=${var}"

This should be the same as far as syntax is concerned. ./local/file.sh contains echo hello, so I would expect var to contain the value hello, but alas, executing the above results in the following:

test.sh: command substitution: line 4: syntax error near unexpected token `('

test.sh: command substitution: line 4: `bash <(cat ./local/file.sh)'

var=

How can I accomplish my goal without using temporary files?

bash variable subshell process-substitution

share|improve this question

asked 1 hour ago

Sean AllredSean Allred

8143816

var=$(bash <(curl -fsSkL http://remote/file.sh))

echo "var=${var}"

var=$(bash <(cat ./local/file.sh))

echo "var=${var}"

test.sh: command substitution: line 4: syntax error near unexpected token `('

test.sh: command substitution: line 4: `bash <(cat ./local/file.sh)'

var=

share|improve this question

asked 1 hour ago

Sean AllredSean Allred

8143816

share|improve this question

asked 1 hour ago

Sean AllredSean Allred

8143816

asked 1 hour ago

Sean AllredSean Allred

8143816

asked 1 hour ago

Sean AllredSean Allred

8143816

1 Answer
1

active

oldest

votes

7

Those are the errors you get when trying to perform a process substitution in bash when the shell is running in POSIX mode. The bash shell does not support process substitutions in POSIX mode.

bash will run in POSIX mode when either

1. 
   set -o posix has been used, or


2. the shell is being invoked as sh.

My hunch is that you have a script, test.sh, that you are running with sh test.sh or that has a #!/bin/sh hashbang line, and that your sh happens to be bash. Another possibility is that the script does not have #!-line at all, and it is being invoked by bash-as-sh in some other way.

Instead, see to that your test.sh script is being invoked by bash.

Example:

$ cat script.sh

echo hello

$ cat test.sh

var=$(bash <( cat script.sh ))

printf 'var="%s"n' "$var"

$ bash -o posix test.sh

test.sh: command substitution: line 2: syntax error near unexpected token `('

test.sh: command substitution: line 2: `bash <( cat script.sh ))'

var=""

$ bash test.sh

var="hello"

share|improve this answer

edited 1 hour ago

answered 1 hour ago

KusalanandaKusalananda

132k17250411

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yep, I was invoking as sh – I didn't realize POSIX didn't support these features. Thanks! As a short follow-up, is there a way to ensure bash-as-sh (if invoked incorrectly) will not run in POSIX mode? In other words, what is the converse of set -o posix?
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SeanAllred Since sh does not need to be bash and indeed, sh is often not bash, you should rather make sure that your scripts are executed by bash. A simple way to do this is to make the script executable and add #!/bin/bash as the first line in the script (adapt that to the correct path to the bash executable on your system). Then run it without specifying an explicit interpreter.
  
  – Kusalananda
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes I can ensure that my tools use bash, via shebang or otherwise, but I'm concerned that those using the script I'm making will not invoke it correctly. Is there no converse to option posix? (Working under the assumption now that there's no means by which to accomplish my goal – without temp files – with only POSIX features.)
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SeanAllred If you are sharing your script with others, it extra important to have a #!/bin/bash line! The sh shell on many Linux machines is dash, and that shell would not understand process substitutions however hard you tried. On my machine (OpenBSD), sh is pdksh, and that also can't be made to run process substitutions. What you are asking is the equivalent to how to make someone invoke a Python script with the correct Python interpreter even when they run it with a Ruby interpreter. If the user picks the wrong interpreter, then the documentation was not good enough.
  
  – Kusalananda
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well of course there will be a shebang line :-) but fair enough – I guess there's a limit to how much you can protect others from themselves. Thanks for your help!
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

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7

Those are the errors you get when trying to perform a process substitution in bash when the shell is running in POSIX mode. The bash shell does not support process substitutions in POSIX mode.

bash will run in POSIX mode when either

1. 
   set -o posix has been used, or


2. the shell is being invoked as sh.

My hunch is that you have a script, test.sh, that you are running with sh test.sh or that has a #!/bin/sh hashbang line, and that your sh happens to be bash. Another possibility is that the script does not have #!-line at all, and it is being invoked by bash-as-sh in some other way.

Instead, see to that your test.sh script is being invoked by bash.

Example:

$ cat script.sh

echo hello

$ cat test.sh

var=$(bash <( cat script.sh ))

printf 'var="%s"n' "$var"

$ bash -o posix test.sh

test.sh: command substitution: line 2: syntax error near unexpected token `('

test.sh: command substitution: line 2: `bash <( cat script.sh ))'

var=""

$ bash test.sh

var="hello"

share|improve this answer

edited 1 hour ago

answered 1 hour ago

KusalanandaKusalananda

132k17250411

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yep, I was invoking as sh – I didn't realize POSIX didn't support these features. Thanks! As a short follow-up, is there a way to ensure bash-as-sh (if invoked incorrectly) will not run in POSIX mode? In other words, what is the converse of set -o posix?
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SeanAllred Since sh does not need to be bash and indeed, sh is often not bash, you should rather make sure that your scripts are executed by bash. A simple way to do this is to make the script executable and add #!/bin/bash as the first line in the script (adapt that to the correct path to the bash executable on your system). Then run it without specifying an explicit interpreter.
  
  – Kusalananda
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes I can ensure that my tools use bash, via shebang or otherwise, but I'm concerned that those using the script I'm making will not invoke it correctly. Is there no converse to option posix? (Working under the assumption now that there's no means by which to accomplish my goal – without temp files – with only POSIX features.)
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SeanAllred If you are sharing your script with others, it extra important to have a #!/bin/bash line! The sh shell on many Linux machines is dash, and that shell would not understand process substitutions however hard you tried. On my machine (OpenBSD), sh is pdksh, and that also can't be made to run process substitutions. What you are asking is the equivalent to how to make someone invoke a Python script with the correct Python interpreter even when they run it with a Ruby interpreter. If the user picks the wrong interpreter, then the documentation was not good enough.
  
  – Kusalananda
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well of course there will be a shebang line :-) but fair enough – I guess there's a limit to how much you can protect others from themselves. Thanks for your help!
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

7

Those are the errors you get when trying to perform a process substitution in bash when the shell is running in POSIX mode. The bash shell does not support process substitutions in POSIX mode.

bash will run in POSIX mode when either

1. 
   set -o posix has been used, or


2. the shell is being invoked as sh.

My hunch is that you have a script, test.sh, that you are running with sh test.sh or that has a #!/bin/sh hashbang line, and that your sh happens to be bash. Another possibility is that the script does not have #!-line at all, and it is being invoked by bash-as-sh in some other way.

Instead, see to that your test.sh script is being invoked by bash.

Example:

$ cat script.sh

echo hello

$ cat test.sh

var=$(bash <( cat script.sh ))

printf 'var="%s"n' "$var"

$ bash -o posix test.sh

test.sh: command substitution: line 2: syntax error near unexpected token `('

test.sh: command substitution: line 2: `bash <( cat script.sh ))'

var=""

$ bash test.sh

var="hello"

share|improve this answer

edited 1 hour ago

answered 1 hour ago

KusalanandaKusalananda

132k17250411

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yep, I was invoking as sh – I didn't realize POSIX didn't support these features. Thanks! As a short follow-up, is there a way to ensure bash-as-sh (if invoked incorrectly) will not run in POSIX mode? In other words, what is the converse of set -o posix?
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SeanAllred Since sh does not need to be bash and indeed, sh is often not bash, you should rather make sure that your scripts are executed by bash. A simple way to do this is to make the script executable and add #!/bin/bash as the first line in the script (adapt that to the correct path to the bash executable on your system). Then run it without specifying an explicit interpreter.
  
  – Kusalananda
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes I can ensure that my tools use bash, via shebang or otherwise, but I'm concerned that those using the script I'm making will not invoke it correctly. Is there no converse to option posix? (Working under the assumption now that there's no means by which to accomplish my goal – without temp files – with only POSIX features.)
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SeanAllred If you are sharing your script with others, it extra important to have a #!/bin/bash line! The sh shell on many Linux machines is dash, and that shell would not understand process substitutions however hard you tried. On my machine (OpenBSD), sh is pdksh, and that also can't be made to run process substitutions. What you are asking is the equivalent to how to make someone invoke a Python script with the correct Python interpreter even when they run it with a Ruby interpreter. If the user picks the wrong interpreter, then the documentation was not good enough.
  
  – Kusalananda
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well of course there will be a shebang line :-) but fair enough – I guess there's a limit to how much you can protect others from themselves. Thanks for your help!
  
  – Sean Allred
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

Those are the errors you get when trying to perform a process substitution in bash when the shell is running in POSIX mode. The bash shell does not support process substitutions in POSIX mode.

bash will run in POSIX mode when either

1. 
   set -o posix has been used, or


2. the shell is being invoked as sh.

My hunch is that you have a script, test.sh, that you are running with sh test.sh or that has a #!/bin/sh hashbang line, and that your sh happens to be bash. Another possibility is that the script does not have #!-line at all, and it is being invoked by bash-as-sh in some other way.

Instead, see to that your test.sh script is being invoked by bash.

Example:

$ cat script.sh

echo hello

$ cat test.sh

var=$(bash <( cat script.sh ))

printf 'var="%s"n' "$var"

$ bash -o posix test.sh

test.sh: command substitution: line 2: syntax error near unexpected token `('

test.sh: command substitution: line 2: `bash <( cat script.sh ))'

var=""

$ bash test.sh

var="hello"

share|improve this answer

edited 1 hour ago

answered 1 hour ago

KusalanandaKusalananda

132k17250411

$ cat script.sh

echo hello

$ cat test.sh

var=$(bash <( cat script.sh ))

printf 'var="%s"n' "$var"

$ bash -o posix test.sh

test.sh: command substitution: line 2: syntax error near unexpected token `('

test.sh: command substitution: line 2: `bash <( cat script.sh ))'

var=""

$ bash test.sh

var="hello"

share|improve this answer

edited 1 hour ago

answered 1 hour ago

KusalanandaKusalananda

132k17250411

answered 1 hour ago

KusalanandaKusalananda

132k17250411

answered 1 hour ago

KusalanandaKusalananda

132k17250411