How to repeat loop n times in Bash

Question

I have below scenario like:

if [file exists]; then
   exit   
elif
   recheck if file exist (max 10 times)
   if found exit else recheck again as per counter  
fi 

Answer 1

If the count is not a variable you can use brace expansion:

for i in {1..10}   # you can also use {0..9}
do
  whatever
done

If the count is a variable you can use the seq command:

count=10
for i in $(seq $count)
do
  whatever
done

Answer 2

There are many ways to do this loop.

With ksh93 syntax (also supported by zsh and bash):

for (( i=0; i<10; ++i)); do
    [ -e filename ] && break
    sleep 10
done

For any POSIX-like shell:

n=0
while [ "$n" -lt 10 ] && [ ! -e filename ]; do
    n=$(( n + 1 ))
    sleep 10
done

Both of the loops sleep 10 seconds in each iteration before testing the existence of the file again.

After the loop has finished, you will have to test for existence of the file a last time to figure out whether the loop exited due to running 10 times or due to the file appearing.

If you wish, and if you have access to inotify-tools, you may replace the sleep 10 call with

inotifywait -q -t 10 -e create ./ >/dev/null

This would wait for a file creation event to occur in the current directory, but would time out after 10 seconds. This way your loop would exit as soon as the given filename appeared (if it appeared).

The full code, with inotifywait (replace with sleep 10 if you don't want that), may look like

for (( i=0; i<10; ++i)); do
    [ -e filename ] && break
    inotifywait -q -t 10 -e create ./ >/dev/null
done

if [ -e filename ]; then
    echo 'file appeared!'
else
    echo 'file did not turn up in time'
fi

Answer 3

n=0
until [ "$((n+=1))" -gt 10 ]
do    <exists? command exit
done
echo oh noes!

though test -e file && exit is more flexible