how to check if two files are existing in a directory using while loop?

Question

Hereby my code:

while [ -f  $load_extract_process ] || [ -f $Sr_details_update_process ]
do
sleep 3s
echo "mango"
done
echo "completed"

I want to check any one of the triggers is present. In that case, go to sleep mode, else print "completed".

oguz ismail · Accepted Answer · 2019-04-29 07:42:15Z

1

Your example code is missing just one character, a !. You want the loop to continue until one of the file exists, not while it exists. So the following code will work:

while ! [ -f /tmp/a1 ] || [ -f /tmp/a2 ]; do sleep 1; echo sleeping; done

Instead of the shell's ! and ||, you can also use parameters from within test ([ is just an alias for test):

while [ ! -f /tmp/a1 -a ! -f /tmp/a2 ]; do sleep 1; echo sleeping; done

Note how ! is now handled by test ([) not by the shell, and -a means and. You decide which of the two implementations is clearer to you (in my opinion, it's the second).

edited Apr 29, 2019 at 7:42

oguz ismail

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answered Apr 29, 2019 at 7:34

Nadav Har'El

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2 Comments

sreekanth rayal Over a year ago

your second option worked for me thanks :) while [ -f /tmp/a1 -o -f /tmp/a2 ]; do sleep 1; echo sleeping; done this one worked for me as my requirement thank you very much

Nadav Har'El Over a year ago

I think I misunderstood what exactly you wanted the loop condition to be, but I'm happy you were able to take pieces from my answer to do exactly what you wanted to do :-)

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