52

I have a script that process a folder, and count the files in the mean time.

i=1
find tmp -type f | while read x
do 
   i=$(($i + 1))
   echo $i
done
echo $i

However, $i is always 1, how do I resolve this?

1

2 Answers 2

62

In your example the while-loop is executed in a subshell, so changes to the variable inside the while-loop won't affect the external variable. This is because you're using the loop with a pipe, which automatically causes it to run in a subshell.

Here is an alternative solution using a while loop:

i=1
while read x; do
   i=$(($i + 1))
   echo $i
done <<<$(find tmp -type f)
echo $i

And here is the same approach using a for-loop:

i=1
for x in $(find tmp -type f);
do 
   i=$(($i + 1))
   echo $i
done
echo $i

For more information see the following posts:

Also look at the following chapter from the Advanced Bash Scripting Guide:

4
  • 2
    regarding the first solution, isn't it process substitution like so < <(find ...) instead of like so <<<(find ...) .... yours is the latter and that seems incorrect. Jun 22, 2018 at 3:47
  • @AlexanderMills The <<< operator is called a here-string.
    – igal
    Feb 17, 2019 at 22:11
  • solution using for-loop doent work if the output contains spaces; but while-loops works Sep 23, 2019 at 10:54
  • The "for" loop is the answer I've been looking for! :)
    – fig
    Aug 31, 2021 at 10:38
9
#!/bin/bash
i=1
while read x
do
   i=$((i+1))
   echo $i
done < <(find . -type f)
echo $i

https://stackoverflow.com/questions/7390497/bash-propagate-value-of-variable-to-outside-of-the-loop

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