shell question

convenientstore · #1 (**permalink**) 06-03-2007

Can someone please explain the PS1="${x##*/}\$ " ??

how does that lop off leading components?

I don't understand it.. thanks

cd ( ) {

command cd "$@" Actually change directory

x=$(pwd) Get current directory name into variable x

PS1="${x##*/}\$ " Lop off leading components, assign to PS1

}

aigles · #2 (**permalink**) 06-03-2007

Extract from bash man pages :

Quote:

${parameter##word}
The word is expanded to produce a pattern just as in pathname expansion. If
the pattern matches the beginning of the value of parameter, then the result
of the expansion is the expanded value of parameter with the shortest match-
ing pattern (the ``#'' case) or the longest matching pattern (the ``##''
case) deleted. If parameter is @ or *, the pattern removal operation is
applied to each positional parameter in turn, and the expansion is the
resultant list. If parameter is an array variable subscripted with @ or *,
the pattern removal operation is applied to each member of the array in
turn, and the expansion is the resultant list.

An example :

Code:

$ var=/a/b/c/d
$ echo ${var##*/}
d
$

convenientstore · #3 (**permalink**) 06-03-2007

how do u find these things?

that's great.. thank you so much

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