CODEWITHSUNDEEP
bashshellwhile-loop
Rgonomike
Rgonomike

Reputation: 338

Shell while loop no stopping

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

Upvotes: 0

Views: 4968

Answers (5)

groovyspaceman
groovyspaceman

Reputation: 2804

Presumably your objective is to wait until a filename containing the string $value is present in the local directory and not necessarily a single filename.

try:

#!/bin/bash
value="aaa"
while ! ls *$value*; do
  sleep 5
done
echo DONE

Your original code failed because $?is filled with the return code of the echo command upon every iteration following the first.

BTW, if you intend to use ps instead of ls in the future, you will pick up your own grep unless you are clever. Use ps -ef | grep [m]ysqlplus.

Upvotes: 0

William Pursell
William Pursell

Reputation: 212178

You ought not duplicate the command to be tested. You can always write:

while cmd; do ...; done

instead of

cmd
while [ $? = 0 ]; do ...; cmd; done

In your case, you mention in a comment that the command you are testing is parsing the output of ps. Although there are very good arguments that you ought not do that, and that the followon processing should be done by the parent of the command for which you are waiting, we'll ignore that issue at the moment. You can simply write:

while ps -ef | grep -v "grep mysqldump" |
    grep mysqldump > /dev/null; do sleep 1200; done

Note that I changed the order of your pipe, since grep -v will return true if it matches anything. In this case, I think it is not necessary, but I believe is more readable. I've also discarded the output to clean things up a bit.

Upvotes: 1

anubhava
anubhava

Reputation: 784898

You can rewrite that loop in much simpler way like this:

while [ -f aaa ]; do
   sleep 5;
   echo "sleeping...";
done

Upvotes: 1

Keith Thompson
Keith Thompson

Reputation: 263177

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.

Upvotes: 6

Ulrich Schwarz
Ulrich Schwarz

Reputation: 7727

$? is the return code of the last command, in this case your sleep.

Upvotes: 2

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