Shell Programming and Scripting

Hi,

I'm having some trouble with the syntax in constructing a simple nested 'for' loop.

My code is as follows:


#!/bin/bash

dir1="fred flume haystack"
for dir2 in ${dir1}
do
fred="1 2 3"
flume="a b c"
haystack="x y z"

for site in ${dir2}
do
echo ${box}/${site}
done
done

What I had expected was that in the first loop, ${dir2} would take the value of 'fred' which it seems to do OK. My thought was that in the nested loop, '${site}' would take the value of each value in dir2 (fred) ie. 1, 2 then 3.
Then the next ${dir2} would be read in (flume) and the process repeated so that flume's contents would be displayed (a, b then c) and so on...

Expected/desired output:

fred/1
fred/2
fred/3
flume/a
flume/b
flume/c
haystack/x
haystack/y
haystack/z

What i got:

fred/fred
flume/flume
haystack/haystack



What am I doing wrong?

Hi,

You are making many typos here. It might be worth looking at your script. Where did you define ${box} which you are echoing later on.

Plus. You have a

for dir2 in $dir
do
..
for site in $dir2
do

This makes no sense. The only thing what you do is fill $site with the value of $dir2. $dir2 only conatains one value a time.

Please re-read your script. I realy don't understand what you want to achief here.

Regs David

sorry, 'box' should say 'dir2', and I've found a way round the problem anyway.

Thanks for your fast reply, only a few minutes after I posted. You guys are good! :P

Just a side question pertaining to the original question .....

Is there a way in shell to use the string in a variable as the name of another variable like the original poster was trying to do ?? I know you can do it in perl like this:

$xxx = "yyy";
$yyy = "zzz";
$$xxx = "aaa"; <---- now $yyy contains "aaa"

How is this done in shell ?

yeah, that's basically what I was trying to do as well, I even tried double dollar too ($$).

Can it be done in shell?

Yes, you can do this using the eval command. Use the search to find examples (there are 38 hits on eval).

syntax
VAR1=$(eval $VAR2)

Maybe you should take another look at a few those hits.

The syntax that you posted will not work in a case like this. In this case, we need:

eval $xxx=\"aaa\"