[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
> PGP key id:3BC1EB90 
> PGP Key fingerprint: 49E2 C52A FC5A A31F 8D66  C0AF 7CEA 30FC 3BC1 EB90
> 
> 
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