Question

I’m writing a routine that will identify if a process stops running and will do something once the processes targeted is gone.

I came up with this code (as a test for my future code):

#!/bin/bash
value="aaa"
ls | grep $value
while [ $? = 0 ]; 
    do
    sleep 5
    ls | grep $value
    echo $?
done;
echo DONE

My problem is that for some reason, the loop never stops and echoes 1 after I delete the file “aaa”.

0
0    >>> I delete the file at that point (in another terminal)
1
1
1
1
.
.
    .

I would expect the output to be “DONE” as soon as I delete the file…

What’s the problem?

SOLUTION:

#!/bin/bash
value="aaa"
ls | grep $value
while [ $? = 0 ];
    do
    sleep 5
    ls | grep $value
done;
echo DONE

Answer 1

The value of $? changes very easily. In the current version of your code, this line:

echo $?

prints the status of the previous command (grep) — but then it sets $? to 0, the status of the echo command.

Save the value of $? in another variable, one that won’t be clobbered next time you execute a command:

#!/bin/bash
value="aaa"
ls | grep $value
status=$?
while [ $status = 0 ]; 
    do
    sleep 5
    ls | grep $value
    status=$?
    echo $status
done;
echo DONE

If the ls | grep aaa is intended to check whether a file named aaa exists, this:

while [ ! -f aaa ] ; ...

is a cleaner way to do it.