Ytukyg

I apologise for the pretty terrible title - and the poor quality post - but what I basically want to do is this:

for I in 1 2 3 4

    echo $VAR$I # echo the contents of $VAR1, $VAR2, $VAR3, etc.

Obviously the above does not work - it will (I think) try and echo the variable called $VAR$I Is this possible in Bash?

Yes, but don't do that. Use an array instead.

If you still insist on doing it that way...

$ foo1=123

$ bar=foo1

$ echo "${!bar}"

123

for I in 1 2 3 4 5; do

    TMP="VAR$I"

    echo ${!TMP}

done

I have a general rule that if I need indirect access to variables (ditto arrays), then it is time to convert the script from shell into Perl/Python/etc. Advanced coding in shell though possible quickly becomes a mess.

You should think about using bash arrays for this sort of work:

pax> set arr=(9 8 7 6)

pax> set idx=2

pax> echo ${arr[!idx]}

7

For the case where you don't want to refactor your variables into arrays...

One way...

$ for I in 1 2 3 4; do TEMP=VAR$I ; echo ${!TEMP} ; done

Another way...

$ for I in 1 2 3 4; do eval echo $$(eval echo VAR$I) ; done

I haven't found a simpler way that works. For example, this does not work...

$ for I in 1 2 3 4; do echo ${!VAR$I} ; done

bash: ${!VAR$I}: bad substitution

for I in {1..5}; do

    echo $((VAR$I))

done

Yes. See Advanced Bash-Scripting Guide

This definately looks like an array type of situation. Here's a link that has a very nice discussion (with many examples) of how to use arrays in bash: Arrays