Bash shell variable inside variables

Question

I'm pretty sure this can be done but I'm not quite sure what the correct syntax is:

dev=$(someprog1 -flag -flag);
tpcli=$(someprog2 -flag);
if [[ $tpcli = $(someprog3 -flag $dev -flag | grep stuff | cut -c 25-55) ]]; then
blah;
blah;

Basically I want to create the variable and add it inside a variable in side the IF statement. Any help would be greatly appreaciated.

bos · Accepted Answer · 2011-11-11 09:52:52Z

1

I'm not quite sure what you're asking, but there's nothing wrong with what you've done (except that I would never have a subshell within an if-clause).

bos@bos:$ foo=42
bos@bos:$ [ $foo = $(echo 42) ] && echo yes || echo no
yes
bos@bos:$ [ $foo = $(echo 242) ] && echo yes || echo no
no

answered Nov 11, 2011 at 9:52

bos

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Comments

Dirk-Willem van Gulik · Accepted Answer · 2011-11-11 12:58:44Z

0

That generally works; e.g. try

P=$(echo $PATH)

when in doubt - use the ${PATH} braces and/or escaped "'s - as to control a single argument with spaces or something which is expanded; and has its spaces seen as separators between arguments. (Which is by the way not needed in your example if we assume that we have no filenames and what not with spaces in them).

As a larger example - consider:

  #!/bin/sh
  dev=$(echo /dev/null)
  if [ $(cat $dev | wc -l) = "0" ]; then
      echo Fine
    exit 0
  else
      echo Now that is a surprize
  fi
  exit 1

which when ran gives us

  beeb-2:~ dirkx$ sh xx.sh
  Fine

or more elaborate:

  beeb-2:~ dirkx$ sh -x xx.sh
  ++ echo /dev/null
  + dev=/dev/null
  ++ cat /dev/null
  ++ wc -l
  + '[' 0 = 0 ']'
  + echo Fine
  Fine
  + exit 0

so that should help you find the issue. You sure there is not some space or transposition ?

edited Nov 11, 2011 at 12:58

answered Nov 11, 2011 at 9:23

Dirk-Willem van Gulik

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5 Comments

bikerben Over a year ago

Hmmm this stillo isn't working for me. If I had multiple IF ELSE statements each with a slightly different $dev varaible in them, would I need to name each of the vaiables differntly? Or could I get away with calling them the same name.

Dirk-Willem van Gulik Over a year ago

Eh no - they get evaluated there and then. See the scoping rules around ", { and (.

bikerben Over a year ago

Nope I'm still having no luck. Its really puzzling me, when the variable executes in the shell it just shows: someprog3 -flag $dev -flag. Do you have any suggestions to what I may of missed here?

Dirk-Willem van Gulik Over a year ago

You'll have to run sh with the -x flag. As this does not quite make sense from both the sample and the man-pages. Updated the example. But it has got to be in there - you are making some sort of scope/space/transposition error which you are not showing in your summary.

bikerben Over a year ago

Thank you for your help. I traced my steps back and found out that I had inadvertinly changed my grep scope when doing a search and replace. This is just a snippit of a larger script and when I ran it appeared as if the variable was not been called. Once again thank you for your time.

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