Iterating over variable name in bash script - bash

Question

I needed to run a script over a bunch of files, which paths were assigned to train1, train2, ... , train20, and I thought 'why not make it automatic with a bash script?'.
So I did something like:
train1=path/to/first/file
train2=path/to/second/file
...
train20=path/to/third/file
for i in {1..20}
do
python something.py train$i
done
which didn't work because train$i echoes train1's name, but not its value.
So I tried unsuccessfully things like $(train$i) or ${train$i} or ${!train$i}.
Does anyone know how to catch the correct value of these variables?

Answer 1

Use an array.
Bash does have variable indirection, so you can say
for varname in train{1..20}
do
python something.py "${!varname}"
done
The ! introduces the indirection, so "get the value of the variable named by the value of varname"
But use an array. You can make the definition very readable:
trains=(
path/to/first/file
path/to/second/file
...
path/to/third/file
)
Note that this array's first index is at position zero, so:
for ((i=0; i<${#trains[#]}; i++)); do
echo "train $i is ${trains[$i]}"
done
or
for idx in "${!trains[#]}"; do
echo "train $idx is ${trains[$idx]}"
done

Answer 2

You can use array:
train[1]=path/to/first/file
train[2]=path/to/second/file
...
train[20]=path/to/third/file
for i in {1..20}
do
python something.py ${train[$i]}
done
Or eval, but it awfull way:
train1=path/to/first/file
train2=path/to/second/file
...
train20=path/to/third/file
for i in {1..20}
do
eval "python something.py $train$i"
done