Shell Script Segmentation Fault

Question

Hi I have a shell script which should run based on the return of php code:

x=1
while [[ "$x" != 5 ]] 
do
  echo "Welcome $x"
  php test.php
  x=$?
done

And the php code

echo "Testdfdf test".PHP_EOL;
exit(4);

So I want whenever I get 5 from php to quit the loop.

But I get sometimes:

./myshell: line 7: 20529 Segmentation fault      php test.php

Should it loop without problem?

Answer 1

It should and it does, but no clue about why php is ending with a segfault.

Answer 2

your shell while loop will loop forever, as your php script returns 4 to shell, and your while loop checks for !=5. which means the condition is not going to be met. what actually is it you are wanting to do? unless necessary, i would advise to do everything with php (or shell) , but try not to intermingle both.

Answer 3

Probably because of this error which affects both Ubuntu and Debian... https://bugs.launchpad.net/ubuntu/+source/php5/+bug/343870