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

Yeah, either 'sed' as the others have said, or start with your template in 2 separate variables and construct on the fly. e.g.

templateprefix="my"
templatesuffix="appserver"

server=get_env

template=${templateprefix}${server}${templatesuffix}

Answer 2

Here's a bash script that wraps this up

Example:

$ search_and_replace.sh "test me out" "test me" ez
ez out

search_and_replace.sh

#!/bin/bash
function show_help()
{
  echo ""
  echo "usage: SUBJECT SEARCH_FOR REPLACE_WITH"
  echo ""
  echo "e.g. "
  echo ""
  echo "test t a                     => aesa"
  echo "'test me out' 'test me' ez   => ez out"
  echo ""

  exit
}

if [ "$1" == "help" ]
then
  show_help
  exit
fi
if [ -z "$3" ]
then
  show_help
  exit
fi

SUBJECT=$1
SEARCH_FOR=$2
REPLACE_WITH=$3

echo "$SUBJECT" | sed -e "s/$SEARCH_FOR/$REPLACE_WITH/g"