Iterating over variable name in bash script

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?

Oh, I could python something.py each_filename, but later I need to run some other scripts over the same files, that's why i put it on those variables — rafa
– rafa, Commented Jun 26, 2013 at 11:46

glenn jackman · Accepted Answer · 2013-06-26 18:47:00Z

9

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

edited Jun 26, 2013 at 18:47

answered Jun 26, 2013 at 18:37

glenn jackman

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Comments

Nikolai Popov · Accepted Answer · 2013-06-26 11:52:03Z

4

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

edited Jun 26, 2013 at 11:52

answered Jun 26, 2013 at 11:46

Nikolai Popov

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3 Comments

rafa Over a year ago

Thank you! Is this structure with square brackets some kind of dictionary?

Nikolai Popov Over a year ago

@rafa It's the way get reference to array element. Some kind of short manual.

l0b0 Over a year ago

Simpler is for train in "${trains[@]}". Of course, the array variable name should be plural :)

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