Can someone please tell me why is the $* there in the following
example:

awk -F, '{
print $4 ", " $0
}' $*

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yusuf wrote:
> Can someone please tell me why is the $* there in the following
> example:
>
> awk -F, '{
> 	print $4 ", " $0
> 	}' $*
>

The $* expands on shell level to a list of words that has been
passed to the script containing these lines.

Say, these lines are in a script called 'mysript' and the script
is called this way...

myscript a b "c d" e

then within the shell program myscript it expands to...

awk -F, '{
print $4 ", " $0
}' a b c d e

The arguments a, b, c, d, e are then taken as the input data files
for the awk program.

The input files contain likely comma separated values (the "-F,").

Mind a couple things...
$4 and $0 are variables within awk
$* is a predefined shell variable
$* will split arguments that contain spaces (like "c d"); use the
quoted "$@" variable if you want it expanded as it has been passed


Janis

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On 15 Mar 2006 16:52:36 -0800, yusuf
<yusufm@gmail.com> wrote:
> Can someone please tell me why is the $* there in the following
> example:
>
> awk -F, '{
> 	print $4 ", " $0
> 	}' $*
>

Because the script writer didn't know to use "$@".


--
leverage, n.:
Even if someone doesn't care what the world thinks
about them, they always hope their mother doesn't find out.

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