Read a variable in bash with a default value

Question

I need to read a value from the terminal in a bash script. I would like to be able to provide a default value that the user can change.

# Please enter your name: Ricardo^

In this script the prompt is "Please enter your name: " the default value is "Ricardo" and the cursor would be after the default value. Is there a way to do this in a bash script?

Answer 1

name=Ricardo
echo "Please enter your name: $name \c"
read newname
[ -n "$newname" ] && name=$newname

Set the default; print it; read a new value; if there is a new value, use it in place of the default. There is (or was) some variations between shells and systems on how to suppress a newline at the end of a prompt. The '\c' notation seems to work on MacOS X 10.6.3 with a 3.x bash, and works on most variants of Unix derived from System V, using Bourne or Korn shells.

Also note that the user would probably not realize what is going on behind the scenes; their new data would be entered after the name already on the screen. It might be better to format it:

echo "Please enter your name ($name): \c"

Answer 2

you can use parameter expansion eg

read -p "Enter: " name
name=${name:-Richard}
echo $name

Answer 3

In Bash 4:

name="Ricardo"
read -e -i "$name" -p "Please enter your name: " input
name="${input:-$name}"

This displays the name after the prompt like this:

Please enter your name: Ricardo

with the cursor at the end of the name and allows the user to edit it. The last line is optional and forces the name to be the original default if the user erases the input or default (submitting a null).