Re: bash ${parameter%%patern} does not work ?
Thanks :-) It can not get any clearer.
I must have been sleeping when reading documents. :-(
Osamu
On Mon, Jul 23, 2001 at 01:39:00PM +0200, Leonard Stiles wrote:
> Osamu Aoki <debian@aokiconsulting.com> writes:
>
> > I have been scratching my head for bash parameter substitution.
> >
> > I understand ## in following examples but I can not understand %%.
> > Is this how bash 2.0 supposed to work?
>
> [examples deleted]
>
> The examples you give do not really illustrate what exactly you don't
> understand about this form of parameter expansion. The bash
> documentation is very good in this field anyway (see "man bash", or
> the texinfo docs in the bash-doc package).
>
> Maybe this table helps demonstrate the difference between the four
> forms:
> match type
> |longest|shortest|
> ------+-------+--------+
> start | ## | # |
> Anchor match to ------+-------+--------+
> end | %% | % |
> ------+-------+--------+
>
> In each case, if there is an appropriate match for the pattern in the
> value of the variable being expanded, then the matching string is
> removed from the value of the expansion, otherwise the value of the
> variable is returned as a whole.
>
> e.g. ${PATH%%:*} returns the first directory in the path.
>
> If you still don't understand something after reading the relevant
> parts of the documentation, post examples where bash behaviour
> deviates from what you expect, and explain what you though would
> happen.
>
> --
>
> Leonard Stiles <ljs@uk2.net>
>
>
> --
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>
--
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+ Osamu Aoki <debian@aokiconsulting.com>, GnuPG-key: 1024D/D5DE453D +
+ My debian quick-reference, http://www.aokiconsulting.com/quick/ +
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