bash shell script conditional assignment

Question

I'm writing a bash shell script. There's a required first argument and I want to have an optional second argument.

If the second argument is omitted I want it to use the value of the first argument.

Currently I have:

SOMEVAR=${2:-Untitled}

How can I use something like basename $1 instead of Untitled?

PS: You might want to use ${2-Whatever} (without the ":") to check that $2 is undefined not empty or undefined. — l0b0
– l0b0, Commented Jul 19, 2011 at 11:26

norcalli · Accepted Answer · 2011-07-19 09:49:58Z

8

You can just do something like SOMEVAR=${2:-$(basename "$1")}. You can do any shell or variable in the optional part.

answered Jul 19, 2011 at 9:49

norcalli

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camh · Accepted Answer · 2011-07-19 09:49:48Z

3

Just use command substitution: $(basename $1), literally instead of Untitled.

However, bash also has the ability to do this without an external process: ${1##*/}

SOMEVAR=${2:-${1##*/}}

answered Jul 19, 2011 at 9:49

camh

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3 Comments

norcalli Over a year ago

The problem with using ${1##*/} shell expansion is that it is not an all encompassing replacement for basename. basename does more than just find the last '/' character.

alfonso Over a year ago

So ${1##*/} is equivalent to $(basename $1)?

norcalli Over a year ago

No, ${1##*/} isn't equivalent to $(basename $1). For example, if you had dir=/ and tried ${dir##*/}, you would get nothing, whereas $(basename "$dir") would give you /.

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