How to replace placeholder character or word in variable with value from another variable in Bash?

Question

I'm trying to write a simple Bash script. I have a simple "template" variable:

template = "my*appserver" 

I then have a function (get_env()) that returns the values dev, qa, or live. I'd like to call get_env and then string-replace the template variable with get_env's return value and swap it out with the asterisk. So:

# Returns "dev" server = get_env  # Prints "mydevappserver" template = string_replace(server, template) 

Or:

# This time, it returns "live" server = get_env  # Prints "myliveappserver" template = string_replace(server, template) 

What should I be using in lieu of this string_replace() function to accomplish the binding?

Answer 1

Bash can do string replacement by itself:

template='my*appserver' server='live' template="${template/\*/$server}" 

See the advanced bash scripting guide for more details on string replacement.

So for a bash function:

function string_replace {     echo "${1/\*/$2}" } 

And to use:

template=$(string_replace "$template" "$server") 

Answer 2

String replacement in a bash-script can e.g. be achieved by sed:

template=$(echo $template | sed 's/old_string/new_string/g') 

This will replace old_string with new_string in the template variable.