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, $i is always 1, how do I resolve this?
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Also see: A variable modified inside a while loop is not remembered on Stack Overflow– muruNov 6, 2017 at 4:10
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2 Answers
In your example the while-loop is executed in a subshell, so changes to the variable inside the while-loop won't affect the external variable. This is because you're using the loop with a pipe, which automatically causes it to run in a subshell.
Here is an alternative solution using a while loop:
i=1
while read x; do
i=$(($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 $i
For more information see the following posts:
Also look at the following chapter from the Advanced Bash Scripting Guide:
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2regarding the first solution, isn't it process substitution like so
< <(find ...)instead of like so<<<(find ...).... yours is the latter and that seems incorrect. Jun 22, 2018 at 3:47 -
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1solution using for-loop doent work if the output contains spaces; but while-loops works Sep 23, 2019 at 10:54
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2A simpler version of
<<<$(is< <(. Using here documents causes problems. E.g. they ignore NULL bytes, like when the commandfind … -print0is used.– anonJun 13, 2022 at 23:20
i=1
while read x
do
i=$(($i + 1))
echo $i
done < <(find tmp -type f)
echo $i
https://stackoverflow.com/questions/7390497/bash-propagate-value-of-variable-to-outside-of-the-loop
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This would solve the immediate issue of allowing the variable
ito survive past the end of the loop, but the answer does not explain why this works. Also, correcting the many other issues still present in the code would be trivial.– Kusalananda ♦Mar 25 at 14:53