16

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?

marked as duplicate by muru, Stephen Rauch, peterh, G-Man, taliezin Nov 6 '17 at 7:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

11

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:

  • 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. – Alexander Mills Jun 22 '18 at 3:47
  • @AlexanderMills The <<< operator is called a here-string. – igal Feb 17 at 22:11
1
#!/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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