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="this is the real value"
a="var1"
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="http://s3.amazo.../config?signature"
my_private_ami_1="http://s3.amazo.../ami_1?signature"
...
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.

1
119
7/9/2014 1:48:55 PM

Accepted 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.

210
3/22/2019 12:41:29 PM

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.


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