Unix bash dynamically add function using variable in its name

Question

Is it possible to define a function but use a variable to compose its name?

_internal_add_deployment_aliases(){
    environment=$1
    instanceNumber=$2
    shortName=$3
    instance="${environment}${instanceNumber}"
    ipVar="_ip_$instance"
    ip=${!ipVar}

    # lots of useful aliases


    _full_build_${instance}(){ # how do i dynamically define a function using a variable in its name
        #something useful
    }
}

Context: I'd like to add bunch of aliases and convenience functions to work with my cloud instances, defining aliases is not a problem, I can easily do

alias _ssh_into_${instance}="ssh -i \"${KEY}\" root@$ip"

and I want to have specific aliases defined when I source from this...

Now when i want to do the same for functions i have a problem, is it possible to do this?

Any help is very very much appreciated :)

Bayou · Accepted Answer · 2019-10-16 07:34:14Z

1

You need to use eval for such a problem:

$ var=tmp
$ eval "function echo_${var} {
echo 'tmp'
}"
$ echo_tmp
tmp

An example of your script:

#! /bin/bash

_internal_add_deployment_aliases(){
    environment=$1
    instanceNumber=$2
    shortName=$3
    instance="${environment}${instanceNumber}"
    ipVar="_ip_$instance"
    ip=${!ipVar}

    # lots of useful aliases


    eval "_full_build_${instance}(){
        echo _full_build_${instance}
    }"
}

_internal_add_deployment_aliases "stackoverflow" 3 so
_full_build_stackoverflow3 # Will echo _full_build_stackoverflow3
exit 0

answered Oct 16, 2019 at 7:34

Bayou

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1 Comment

Bayou Over a year ago

You're welcome :-) Just make sure you never run text from an untrusted source through eval.

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