How to get a variable value if variable name is stored as string

Question

How can I retrieve a bash variable value if I have the variable name as string  var1  quot this is the real value quot  a  quot var1 quot  Do something to get value of var1 just using variable a   Context  I have some AMI s  Amazon Machine Image  and I want to fire up a few instances of each AMI  As soon as they finish booting  I want to setup each instance according to its AMI type  I don t want to bake lots of scripts or secret keys inside any AMI so I prepared a generalized startup script and I put it on S3 with a publicly accessible link  In rc local I put small piece of code which fetches the startup script and executes it  This is all I have in the AMIs  Then each AMI accesses a common configuration script which is applicable to all AMIs and special setup scripts for each  These scripts are private and require a signed URL to access them  So now  when I fire an instance of an AMI  my private ami 1   I pass a signed URL for one more file presented on S3 which contains signed URL for all private scripts in terms of key value pair config url  quot http   s3 amazo    config signature quot  my private ami 1  quot http   s3 amazo    ami 1 signature quot       When the startup script runs  it downloads the above file and source s it  Then it checks for its AMI type and picks the correct setup script for itself  ami  type GET AMI TYPE  ex  sets ami  type to my  private  ami  1 setup  url GET THE SETUP FILE URL BASED ON AMI  TYPE   this is where this problem arises  So now I can have a generic code which can fire instances irrespective of their AMI types and instances can take care of themselves

User · Answer

For my fellow zsh users  the way to accomplish the same thing as the accepted answer is to use     P a  It is appropriately called Parameter name replacement  This forces the value of the parameter name to be interpreted as a further parameter name  whose value will be used where appropriate  Note that flags set with one of the typeset family of commands  in particular case transformations  are not applied to the value of name used in this fashion  If used with a nested parameter or command substitution  the result of that will be taken as a parameter name in the same way  For example  if you have    foo bar    and    bar baz     the strings    P foo      P   foo    and    P   echo bar   will be expanded to    baz     Likewise  if the reference is itself nested  the expression with the flag is treated as if it were directly replaced by the parameter name  It is an error if this nested substitution produces an array with more than one word  For example  if    name assoc    where the parameter assoc is an associative array  then         P name  elt      refers to the element of the associative subscripted    elt

User · Answer

Had the same issue with arrays  here is how to do it if you re manipulating arrays too    array name  ARRAY NAME  ARRAY NAME   Val0   Val1   Val2    ARRAY  array name    echo  ARRAY   ARRAY   ARRAY      ARRAY    echo  ARRAY   ARRAY      echo  ARRAY 0    ARRAY 0    echo  ARRAY 1    ARRAY 1    echo  ARRAY 2    ARRAY 2      This will output    ARRAY ARRAY NAME    ARRAY Val0 Val1 Val2 ARRAY 0  Val0 ARRAY 1  Val1 ARRAY 2  Val2

User · Answer

X foo Y X eval  Z    Y    sets Z to  foo   Take care using eval since this may allow accidential excution of code through values in   Y   This may cause harm through code injection   For example  Y    touch  tmp eval-is-evil      would create  tmp eval-is-evil  This could also be some rm -rf    of course

User · Answer

Modified my search keywords and Got it     eval a    a Thanks for your time

User · Answer

In bash 4 3  the  -v  test for set variables was introduced  At the same time   nameref  declaration was added  These two features together with the indirection operator     enable a simplified version of the previous example  get value       declare -n var name  1   if    -v var name      then     echo  quot   var name  quot    else     echo  quot variable with name  lt    var name  gt  is not set quot    fi    test 123 get value test 123  test  quot    echo   quot something nasty  quot   quot  get value test   echo  quot something nasty quot    unset test get value test variable with name  lt test gt  is not set  As this approach eliminates the need for  eval   it is safer  This code checked under bash 5 0 3 1

User · Answer

Based on the answer  https   unix stackexchange com a 111627                                                                                    Summary  Returns the value of a variable given it s name as a string    Required Positional Argument       variable name - The name of the variable to return the value of   Returns  The value if variable exists  otherwise  empty string                                                                                       get value of         variable name  1     variable value        if set   grep -q    variable name    then         eval variable value     variable name      fi     echo   variable value     test 123 get value of test   123 test     echo   something nasty     get value of test     echo  something nasty

User · Answer

You can use    a    var1  this is the real value  a  var1  echo     a     outputs  this is the real value    This is an example of indirect parameter expansion      The basic form of parameter expansion is   parameter   The value of    parameter is substituted       If the first character of parameter is an exclamation point      it   introduces a level of variable indirection  Bash uses the value of the   variable formed from the rest of parameter as the name of the   variable  this variable is then expanded and that value is used in the   rest of the substitution  rather than the value of parameter itself

User · Answer

modern shells already support arrays  and even associative arrays   So please do use them  and use less of eval   var1  this is the real value  array    var1     or array 0    var1    then when you want to call it   echo   array 0

[string] How to get a variable value if variable name is stored as string?

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