Read a variable in bash with default value without -i option

Not exactly but you can do something like:

name="Alex"
read -e -p "Please enter your name [$name]: " input
name="${input:-$name}"

User will not be able to edit the name but still see it as default value. Not exactly what you wanted. I guess you maybe already thought about that..

I'm going to guess that the reason you're using a 12+ year old version of Bash is that you must be trying to get this to work in modern OSX, which for some reason still ships Bash 3.2.57. For reference, Ubuntu 10.04 (released 2010-04-29) had Bash 4.1.5(1) ^cite.

The answer is probably going to be that -- for a lot of shell features -- your Bash version is simply too old to support this. You'll have to either rely on an external program, a workaround, or be content with what the old version supports.

Your Options

Note: I added -r to read (i.e. don't treat \ specially) because shellcheck complains about this.

1. Use #!/usr/bin/env bash as your script's bangline to bypass the default OSX shell (requires homebrew).

2. Do without it:

name=Alex
read -r -e -p "Please enter your name [$name]: " input
name=${input:-$name}

3. Test Bash version before using -i:

name=Alex
if [[ ${BASH_VERSINFO[0]} -ge 4 ]]; then
    # We have a version of Bash that supports "read -i"
    read -r -e -i "$name" -p 'Please enter your name: ' input
else
    # Older version of Bash
    read -r -e -p "Please enter your name [$name]: " input
fi
# Proceed as normal
name=${input:-$name}

In BASH <4 you can do this:

name="Alex"
read -p "Please enter your name: " input && [[ -z "$input" ]] && input="$name"

This basically check after read if input var is empty then sets it to $name.