Bash: Get value of variable from string of variable name (when string contains other characters than variable name)

Question

Given this code, it prints "abc_${var1}" but I want it to print "abc_def" but bash scripting has a lot of little details that I'm not aware of so hopefully someone can explain this behavior:

Shell scripts:

function main()
  var1="def"
  var2=$(get_val_sh ${another_var})

  echo ${var2}
  # I want to echo:
  #   abc_def
  # instead, I get:
  #   abc_${var1}


function get_val_sh()
  var3=call_python(get_val_py())
  echo ${var3}


Python:

def get_val_py():
  print "abc_${var1}"

The other thing is that if I directly return from a bash script (rather than calling Python), I do get the desired result (but I have to call python). Does it have something to do with print in Python returning a different kind of string than echo in bash? I don't even know how to check this.

To be clear, the following code gives the desired result though I have no idea why but can't be used because I need Python to give me the appropriate value:

function main()
  var1="def"
  var2=$(get_val_sh ${another_var})

  echo ${var2}
  # I want to echo:
  #   abc_def
  # instead, I get:
  #   abc_${var1}


function get_val_sh()
  echo "abc_${var1}"

Could you please keep the code blocks to a single language each? It's already suspicious that you use Python-style indentation in a block labeled “shell scripts”, and it gets downright confusing when we try to reason about who calls whom. — JB.
– JB., Commented Nov 14, 2013 at 7:51

anishsane · Accepted Answer · 2013-11-14 06:17:45Z

1

How are you assigning value to var2?

var1="def"
var3="var1" # var3 contains name of the variable. Note: NOT $var1
var2="abc_${var3}" # This will give abc_var1, not abc_def

Use

var2="abc_${!var3}" # This should give abc_def

answered Nov 14, 2013 at 6:17

anishsane

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