David C. Rankin wrote:
> Guys,
>
> How do I call 'ls' from within a script without it also returning the
> contents of the present working directory? Here is the line from my script:
>
> ls -al /usr/lib/libGL.so*
>
> Here is the output:
>
> # ./linux/scripts/showLibConfig
> 250sata.pdf 7857.pdf Bannykh-ArizMedBoard.pdf Bannykh-TennMedBoard.pdf
> bin broadway.pdf david.asc Desktop Documents linux log Pictures
> public_html westlaw-renewal_20071129.pdf /usr/lib/libGL.so Config
>
> lrwxrwxrwx 1 root root 10 2007-11-09 16:19 /usr/lib/libGL.so ->
> libGL.so.1
> lrwxrwxrwx 1 root root 12 2007-12-16 16:25 /usr/lib/libGL.so.1 ->
> libGL.so.1.2
> lrwxrwxrwx 1 root root 27 2007-12-16 16:25 /usr/lib/libGL.so.1.2 ->
> /usr/X11R6/lib/libGL.so.1.2
> -rwxr-xr-x 1 root root 391344 2007-09-21 20:34 /usr/lib/libGL.so.1.2.sav
>
> as you can see, it looks like 'ls' is evaluated before 'ls -al
> /usr/lib/libGL.so*' gets evaluated. How do I fix this?
>
>
That doesn't make any sense. Could you post the actual code of the
script? If the code is as you show it above there is no way that it
lists the current directory, unless you have some kind of wierd alias
for ls. Type "which ls" without the quotes to see where ls is being run
from.
--
kr
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