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

score 6 · Accepted Answer · edited Nov 14 '14 at 20:10

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.

As an example, for this specific case:

echo "${!threshold}"

edited Nov 14 '14 at 20:10

Charles Duffy

answered Nov 14 '14 at 18:55

Etan Reisner

score -1 · Answer 2 · edited May 23 '17 at 10:26

-1

Use eval command :

eval echo \${$threshold}

More details about this command can be found here:

edited May 23 '17 at 10:26

Community

answered Nov 14 '14 at 18:58

RBH

1

Boo, hiss -- `eval` introduces room for bugs (including security bugs), which more conventional indirection approaches would not. See BashFAQ #48: http://mywiki.wooledge.org/BashFAQ/048 – Charles Duffy Nov 14 '14 at 20:08
...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. – Charles Duffy Nov 14 '14 at 20:09
Downvoting for recommending `eval` to accomplish a task for which `bash` provides explicit support. – chepner Nov 14 '14 at 20:25

2 Answers2