[SOLVED] bash variable substitution

Jerry Mcguire · 04-29-2010, 03:01 AM

I have a config file (not a script) that states where to find a file named key.dat. With this config, I need to operate on the key.dat, so I did the following...

Code:

$ cat myconfig
keyfile=$mypath/key.dat

$ touch $mypath/key.dat # mypath is of course defined

$ KEYFILE=`grep '^keyfile=' myconfig | cut -d '=' -f 2`

$ echo $KEYFILE
$mypath/key.dat

$ [ -f $KEYFILE ] || echo 'Not Found'
Not Found

Since '$mypath' is not expanded in the test, the key.dat is not found.

My question is, how can I make $mypath expanded in the test so that I can get hold of the file key.dat?

Thanks.

evo2 · 04-29-2010, 03:06 AM

I think you don't need to mess around with grep. Just source the file. What is the output of the following?

Code:

. myconfig
echo $keyfile

Evo2.

PS. You say that myconfig is not a script, but from what you post above the syntax is bourne shell compatible, so sourcing it should work.

Jerry Mcguire · 04-29-2010, 03:14 AM

Can't. The file is not executable. e.g. 664

and there are other stuff in the file that is not valid command.

gnashley · 04-29-2010, 03:28 AM

A file doesn't have to be executable in order to source it. Try the suggestion.

Jerry Mcguire · 04-29-2010, 03:35 AM

Right. That is true. Thank you both.

For the sake of learning, if sourcing the file doesn't work, e.g. myconfig does not contain valid bash syntax. Any work around to expand 'variable inside strings'?

colucix · 04-29-2010, 04:26 AM

Quote:

Originally Posted by Jerry Mcguire

My question is, how can I make $mypath expanded in the test so that I can get hold of the file key.dat?

The eval built-in may serve to force shell expansions. For example:

Code:

$ echo $mypath
/usr/local
$ eval $(grep '^keyfile=' myconfig)
$ echo $keyfile
/usr/local/key.dat

Jerry Mcguire · 04-29-2010, 09:33 AM

Thanks. This is what I needed.

Code:

$ KEYF=`eval echo $(grep '^keyfile=' myconfig | cut -d '=' -f 2)`

$ [ -f $KEYF ] && echo 'Found'
Found