[Discuss] What the use of .bashrc
Joe Polcari
joe at polcari.com
Wed Oct 31 08:12:14 EDT 2012
If bash can do it, it's in this guide, my bash bible:
http://www.tldp.org/LDP/abs/abs-guide.pdf
Sent from my iPad
On Oct 30, 2012, at 12:08 PM, Jerry Feldman <gaf at blu.org> wrote:
> I generally use "Learning the BASH Shell" as a reference, but here is
> the definition:
> http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion
>
>
>
> On 10/30/2012 11:46 AM, John Abreau wrote:
>> I just looked for that in the bash manpage, and i can't find anything
>> describing
>> that behavior. Can you highlight where you discovered that?
>>
>>
>>
>> On Tue, Oct 30, 2012 at 11:07 AM, Jerry Feldman <gaf at blu.org> wrote:
>>> On 10/30/2012 10:58 AM, joe at polcari.com wrote:
>>>> Looks to me like the first test only tests if $1 is not at the end of $PATHor am I missing something? ----- Original Message -----From: "Jerry Feldman" >;gaf at blu.org
>>>>
>>> No, it tests is $1 exists in $PATH.
>>> I really hate bash pattern matching because I have to read the manual
>>> every time I use them.
>>> in this case '*:"$1":*' looks for $1 anywhere in $PATH.
>>
>>
> Look at expressions. A path is delimited by colons. So, this means look
> for $1 anywhere in a path. You can easily test it. I have not looked at
> some of the boundary cases, but they appear to work since I've been
> using this for years.
>
> case ":${PATH}:" in
> *:"$1":*)
> ;;
> Note that $PATH is prepended and appended by ':'. So, assume a PATH is
> $HOME/bin/usr/bin, the pattern is ":$HOME/bin:/usr/bin:"
> So, it will look for $1 anywhere between 2 colons.
> http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion
>
>
> --
> Jerry Feldman <gaf at blu.org>
> Boston Linux and Unix
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>
>
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