Bashscripts.org

Hello,

Need some help here!



I Think you see what i wanna do, but why doesent it work within a variable?

Thanks in advance!

It's not the variable that is causing the problem, but the backticks. The backticks cause an extra evaluation round, which in turn 'destroys' your escaped \$. So you can either escape the escaped $ or you can forget about backticks altogether and use $().

Thanks Patsie!

You are a doll!

Can you explain why i need the eval command, i've googled it but i doesent realy understand?

//Barty

Normally your shell goes through an evaluation step to substitute variables ($var) with their actual contents.
The eval command forces an extra evaluation step. Take the 3 echo's below:

The first echo will be evaluated as if $hello_ and $var are 2 separate variables, because they both start with a $-sign. so the echo will evaluate to nothing, because $hello_ isn't defined anywhere, and to 'guys', because that's what $var is filled with.
If you escape the $-sign before 'hello', the evaluation will just print the $-sign. Therefor 'hello' is no variable anymore and thus the string hello_ is printed. $var isn't excaped, so the eval step will substitute it with it's contents guys. This results in the string $hello_guys.
But you want the contents of $hello_guys so you have to force another eval step.

ps: I didn't know you are into guys, 'doll'. Unfortunately, I'm not.