Re: bash question: reporting a variable named within a variable
On Thursday 03 April 2003 08:01 pm, Bob Proulx wrote:
> Craig Dickson wrote:
> > echo $var=$(eval echo \$$var)
>
> That works. Personally I prefer to eval the entire line. This way
> you only use one layer of processing rather than the two in the above.
>
> for var in FOO BLAH ; do
> eval echo $var = \$$var;
> done
The following works only in bash, but is smaller and faster:
for var in FOO BLAH ; do
echo $var = ${!var};
done
--
Rob
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