Bash script loop on python variable

Question

I'm trying to do a simple script, but I don't get how to pass a variable to the command that I need:

#!/bin/bash 
for i in 1 2 3
do
python -c 'print "a"*' $i
done

score 1 · Accepted Answer · answered Feb 27 '19 at 08:57

1

If you really want to go on with your solution using python -c, you would have to remove the space:

#!/bin/bash 
for i in 1 2 3
do
  python -c 'print "a"*'$i
done

But the approach suggested by Asmox makes more sense to me (and it has nothing to do with the question where you pipe the standard output to).

answered Feb 27 '19 at 08:57

user1934428

You absolutely *do not* have to remove the space. You just have to quote the whole argument. – Konrad Rudolph Feb 27 '19 at 10:28
This is **also** one possibility. I suggested to remove the space, because this requires the least changes to the original code, compared to `python -c "print 'a'* $i"`. – user1934428 Feb 28 '19 at 10:45

score 0 · Answer 2 · answered Feb 27 '19 at 08:47

0

Have you considered making a whole script? Like this one:

import sys

def f(a):
    print a

if __name__ == "__main__":
    a = int(sys.argv[1])
    f(a)

This example might seem overcomplicated, but I wanted to show you how to pass parameter from console to function

answered Feb 27 '19 at 08:47

Asmoox

i need this to pipe to another file, the whole line would be something like "python -c 'print "a"* $i ' | ./example " – neorus Feb 27 '19 at 08:52

score 0 · Answer 3 · answered Aug 06 '19 at 07:52

I have tried the solution provided above and had to modify for python3+

due to lack of sufficient points I am not allowed to make a comment, thats why I posted my solution separately

#!/bin/bash
for i in {1..3}
do
  python -c "print ('a'* ${i})"
done

output~>
a
aa
aaa

3 Answers3