How to understand this "bash" shell command

Question

The command is:

[ -d $x ] && echo $x | grep "${1:-.*}"

I have run it separately, and [ -d $x ] && echo $x just outputs the directory name. What does the ${1:-.*} mean?

what shell? that would be helpful since bash, c shell, etc all run the commands differently — jgr208, Nov 12 '15 at 15:30
That's [Shell Parameter Expansion](http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion). You can find it in the manual. — Etan Reisner, Nov 12 '15 at 15:32

score 0 · Accepted Answer · answered Nov 22 '15 at 02:14

In the script you refer to, grep is called. Its first argument, what it will search for, is either the first script parameter $1, if one was given, or .*, which matches anything, if no parameters were given.

"$1" or "${1}" in a bash script is replaced with the first argument the script was called with. Sometimes it's necessary to process that string a little, which can be done with shell parameter expansion, as Etan Reisner helpfully pointed out. :- is one such tool; there are several others.

"${1:-.*}" means "if parameter 1 is unset or null (i.e., no such parameter was given), then substitute the part after :; in this case, .*.

Example script pe:

#!/bin/bash
printf 'parameter count = %d\n' $#
printf 'parameter 1 is "%s"\n' "$1"
printf 'parameter 1 is "%s"\n' "${1:-(not given)}"

Output:

$ ./pe 'foo bar'
parameter count = 1
parameter 1 is "foo bar"
parameter 1 is "foo bar"

$ ./pe
parameter count = 0
parameter 1 is ""
parameter 1 is "(not given)"

How to understand this "bash" shell command

1 Answers1