Script Fails
 

Hi, Am new to Linux. Here is what am trying.
#!/usr/bin/bash
export D1_DIR=/my/dir1
export D2_DIR=/my/dir2
for i in 1 2
do
echo $D$i_DIR ### Errors from here
mkdir -p $D$i_DIR ##Errors
ln -s $D$i_DIR L$i #Error
done

--- give a clue or help me in pointing out my mistake. I also tried eval but there is no use.....

First of all, please use [code][/code] tags around your code, to preserve formatting and to improve readability.

The string $D$i_DIR has two variables, $D and $i. Since you didn't define $D, it fails. I think you want echo "D$i_DIR" instead. (See addendum below)

It's also generally safer to double-quote a variable, as I demonstrated.

Edit: Actually there's a second problem. Since the underscore can also be part of a variable name, the $i variable needs to be differentiated from it also.

So you have to use echo "D${i}_DIR".

By the way, is your bash exec really in /usr/bin/bash? Most systems have it in /bin/bash.

Nesting variables appears pretty tricky. You will need to use eval and the back slash to escape the variable.

Something like
Either should work

More references can be found at google, using "bash nesting variables"

Hope this helps.

spazticclown is right. I forgot about the indirect referencing needed to get the actual values.

However you don't really need to use exec here. Bash has an indirect reference pattern. Try this:

One more thing. You can avoid all this indirect stuff completely if you simply use an array variable instead.

Damn, that's how you do that. It's been a while since I had to do anything like that, and I wanted to reply so bad before, but couldn't think for the life of me. Thanks dude!

Thanks
 

Thanks a lot spazticclown, David...

hopefully you saved my life and my time.

once Again Thanks

I like David's option using the array, very clean and simple... mine is just hobbled together because I was trying to do the same thing and failing to find a good answer, bookmarked this thread.