dynamically creating bash variable and accessing its value defined earlier

Question

i am trying to read the value of a bash variable defined earlier , but this variable name derived dynamically.

this is the bash script i am trying to do

  $ mythreshold=10
  $ table=my
  $ threshold="$table"threshold
  $ echo $("$threshold")
   mythreshold

but when i try to read this variable value like

    $ echo $("$threshold")
    -bash: mythreshold: command not found

however i was expecting it to print

  $ echo $("$threshold")
   10

is there a way i can get this work, it should have printed the value of mythreshold variable defined above

Charles Duffy · Accepted Answer · 2014-11-14 20:10:43Z

6

$() is Command Substitution. It runs the command inside and returns the output. A variable name is not a command.

You can $(echo "$threshold") but that will only get the mythreshold back.

You need indirection for what you want. Specifically Evaluating indirect/reference variables.

As an example, for this specific case:

echo "${!threshold}"

edited Nov 14, 2014 at 20:10

Charles Duffy

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answered Nov 14, 2014 at 18:55

Etan Reisner

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Community · Accepted Answer · 2017-05-23 10:26:26Z

-1

Use eval command :

eval echo \${$threshold}

More details about this command can be found here:

eval command in Bash and its typical uses

edited May 23, 2017 at 10:26

CommunityBot

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answered Nov 14, 2014 at 18:58

RBH

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3 Comments

Charles Duffy Over a year ago

Boo, hiss -- eval introduces room for bugs (including security bugs), which more conventional indirection approaches would not. See BashFAQ #48: mywiki.wooledge.org/BashFAQ/048

Charles Duffy Over a year ago

...in this specific case -- consider if threshold contained, for instance, 'mythreshold:-$(rm -rf /)'. A safe indirection approach would throw an error or evaluate to an empty string; eval would erase your drive.

chepner Over a year ago

Downvoting for recommending eval to accomplish a task for which bash provides explicit support.

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