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
