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Date: Sat, 23 Dec 2000 21:31:05 +0200
From: Giorgos Keramidas <keramida@ceid.upatras.gr>
To: Aaron Hill <hillaa@hotmail.com>
Cc: davidd@datasphereweb.com, questions@FreeBSD.ORG
Subject: Re: Script: Variable substition within a variable?
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References: <F157QBRbpOmK1iY1VBn00000efe@hotmail.com>
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In-Reply-To: <F157QBRbpOmK1iY1VBn00000efe@hotmail.com>; from hillaa@hotmail.com on Mon, Dec 18, 2000 at 04:10:36AM +0000
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On Mon, Dec 18, 2000 at 04:10:36AM +0000, Aaron Hill wrote:
> Thanks for your prompt reply.
> 
> >I think what you're looking for is something along the lines of:
> >echo "Your name is " $fileone
> 
> 
> I'm afraid that doesn't work. It seems that variable substitution only works 
> to one level - maybe this is for the best too.

You can evaluate expressions once more with `eval'.  If I understood the question correctly,
you have the name of a variable in $name as in:

	$ name=datavar

Then you need to get the contents of that file in $fileone variable:

	$ cat datafile
	piou
	$ datavar=`cat datafile`
	$ echo $datavar
	piou

And you need to use $name as the name of a variable, which will then be
echoed by your script:

	$ echo \$$name
	$datavar

But you need this last expression, i.e. `$datavar', including the dollar sign
to be evaluated:

	$ eval echo \$$name
	piou

The first dollar sign is escaped by the backslash, and what `eval' finally
evaluates is a dollar sign followed by `datafile', the value of $name.

I hope that this helps a bit.

- giorgos


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