Assign value to a variable in unix dynamically - variable indirection

Question

HOST_NAME_="localhost:8080" 
HOST_NAME_stg="stg.com:8080"
HOST_NAME_qa="qa.com:8080"
HOST_NAME=${!"HOST_NAME_$1"}

echo -n ${HOST_NAME}

I get the error bad substitution.But I want stg.com:8080 to printed if the argument passed is stg.How do I do that

mklement0 · Accepted Answer · 2017-09-19 00:59:19Z

6

Note: Given your use of ${!...} to refer to a variable indirectly, I assume that you're using bash - POSIX-features-only shells and many others do NOT support this feature (at least with this syntax).

Try the following:

HOST_NAME_="localhost:8080" 
HOST_NAME_stg="stg.com:8080"
HOST_NAME_qa="qa.com:8080"

HOST_NAME_INDIRECT=HOST_NAME_$1

HOST_NAME=${!HOST_NAME_INDIRECT}

echo -n "${HOST_NAME}"

When referencing a variable indirectly with ${!...}, the argument containing the target variable name must itself be a variable, not a string literal, hence the intermediate variable $HOST_NAME_INDIRECT.

Also consider avoiding all-uppercase variable names to avoid potential conflicts with standard environment variables (such as $HOSTNAME) and special shell variables (such as $SECONDS).

Here's how various popular shells do variable indirection:

Assume the following two variables:

a='value of a'
nameOfA='a'

bash

 echo "${!nameOfA}" # -> 'value of a'

 # bash 4.3+
 declare -n aliasOfA='a'  # in functions: `local -n`
 echo "$aliasOfA"         # same as `echo "$a"` -> 'value of a'

zsh

echo "${(P)nameOfA}"  # -> 'value of a'

ksh 93+:

 nameref aliasOfA='a'  # same as `typeset -n`
 echo "$aliasOfA"      # same as `echo "$a"` -> 'value of a'

sh (POSIX-features-only shells)

 eval echo \"'$'$nameOfA\"  # -> 'value of a'

Caveat: Using eval is potentially unsafe. If $nameOfA contained something like 'a; $(rm *)', the rm command would execute too.

edited Sep 19, 2017 at 0:59

answered Apr 19, 2014 at 15:20

mklement0

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2 Comments

Abhijeet Kushe Over a year ago

Thanks for providing such a detailed answer.This does solve my problem but I would like to know more details about why eval is unsafe.I looked at some links of stackoverflow which talk about eval but they just add to the ambiguity. If you know some good sources which talk about eval can you post it here. Thanks

mklement0 Over a year ago

@AbhijeetKushe: In a nutshell: a string passed to eval can contain any command. If you don't have full control over that string - e.g., if it is passed from a calling script - it may contain commands that either inadvertently or maliciously do things you do not want or expect. Say someone passed qa; $(rm ...) instead of qa as $1 to your script. It's a contrived example, but shows you the principle of how unwanted things can happen. This is not to say that eval must never be used - just be aware of its dangers and limit use to carefully controlled circumstances.

Abhijeet Kushe · Accepted Answer · 2014-04-19 15:53:27Z

0

Thanks for the reply.Yeah using a temporary variable does solve the problem.But I figured out if to do this it serves the purpose but it is slightly complicated

eval "HOST_NAME=\$HOST_NAME_$2"

answered Apr 19, 2014 at 15:53

Abhijeet Kushe

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1 Comment

mklement0 Over a year ago

While your version has the advantage of working in all Bourne-like shells, it's potentially unsafe, because it uses eval. Thus, in bash I suggest you use the ${!...} approach.

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