Bash: How to set a variable from argument, and with a default value
bash, shell, unix
Solution
I see several questions here.
“Can I write something that actually reflects this logic”
Yes. There are a few ways you can do it. Here's one:
if [[ "$1" != "" ]]; then
DIR="$1"
else
DIR=.
fi
“What is the difference between this and `DIR=${1-.}`?”
The syntax `${1-.}` expands to `.` if `$1` is unset, but expands like `$1` if `$1` is set—even if `$1` is set to the empty string.
The syntax `${1:-.}` expands to `.` if `$1` is unset or is set to the empty string. It expands like `$1` only if `$1` is set to something other than the empty string.
“Why can't I do this? `DIR="$1" || '.'`”
Because this is bash, not perl or ruby or some other language. (Pardon my snideness.)
In bash, `||` separates entire commands (technically it separates pipelines). It doesn't separate expressions.
So `DIR="$1" || '.'` means “execute `DIR="$1"`, and if that exits with a non-zero exit code, execute `'.'`”.
Problem
It is pretty clear that with shell scripting this sort of thing can be accomplished in a huge number of ways (more than most programming languages) because of all the different variable expansion methods and programs like `test` and `[` and `[[`, etc. Right now I'm just looking for ``` DIR=$1 or . ``` Meaning, my DIR variable should contain either what is specified in the first arg or the current directory. What is the difference between this and `DIR=${1-.}`? I find the hyphen syntax confusing, and seek more readable syntax. Why can't I do this? ``` DIR="$1" || '.' ``` I'm guessing this means "if $1 is empty, the assignment still works (DIR becomes empty), so the invalid command '.' never gets executed."