bash - Can you trigger a specific exit status on a php fatal error? -

i've tried following:

<?php function shutdown_find_exit() {     exit(50); } register_shutdown_function('shutdown_find_exit');  trigger_error("fatal error", e_user_error);  ?> 

but still exists error code of 255.

i'm trying have error code of 50 returned.

i'm checking following bash script:

php tester.php status=$? echo exit code: ${status} 

updated code based on accepted answer interested:

<?php function shutdown_find_exit() {     exit('50'); } register_shutdown_function('shutdown_find_exit');  trigger_error("fatal error", e_user_error); ?> 

and bash script

output="$(php tester.php)" status=$? echo exit code: ${status} if [[ $output == *"50"* ]]   echo "got 50"; fi 

tl;dr: use string, instead of int.

php exit( status )

if status string, function prints status before exiting.

if status integer, value used exit status , not printed. exit statuses should in range 0 254, exit status 255 reserved php , shall not used. status 0 used terminate program successfully.

note: php >= 4.2.0 not print status if integer.

execute php script in bash, assign output variable.

variable = $(/path/to/php -f $home/path/to/my.php) 

for example:

# executing php, capturing output then, prints console status = $(/usr/bin/php -f $home/path/to/tester.php) echo exit code: ${status} 

hope helps.