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 (

Answer 1

As nobody mentioned it, here's a cool possibility using printf. The place-holder must be %s though, and not *.

# use %s as the place-holder
template="my%sappserver"

# replace the place-holder by 'whatever-you-like':
server="whatever-you-like"
printf -v template "$template" "$server"

Done!

If you want a function to do that (and notice how all the other solutions mentioning a function use an ugly subshell):

#!/bin/bash

# This wonderful function should be called thus:
# string_replace "replacement string" "$placeholder_string" variable_name
string_replace() {
    printf -v $3 "$2" "$1"
}

# How to use it:
template="my%sappserver"
server="whatever-you-like"

string_replace "$server" "$template" destination_variable

echo "$destination_variable"

Done (again)!

Hope you enjoyed it... now, adapt it to your needs!

Remark. It seems that this method using printf is slightly faster than bash's string substitution. And there's no subshell at all here! Wow, that's the best method in the West.

Funny. If you like funny stuff you could write the function string_replace above as

string_replace() {
    printf -v "$@"
}

# but then, use as:
string_replace destination_variable "$template" "$server"