执行shell脚本代码取决于php返回值

Question

I haven't that much of knowledge in PHP, and I need to run a script that depends upon the PHP return value.

script

presentDir=`pwd`
success=$(php -f $presentDir/optimize.php $file)
echo echo "value:".$success

PHP code

<?php
try{        
    //code  
} catch(Exception $e){
    return 1
}
return 0

When executing the above shell script, success is empty.

Answer 1

PHP's return statement doesn't send any output to STDOUT, which is what you'd be catching with the $() construct. To see the exit value from a command you use the exit statement in PHP and the shell's $? variable . Maybe this is what you're looking for?

<?php
try {
    //code
} catch(Exception $e){
    exit(1);
}
exit(0);
?>

Followed by:

#!/bin/sh
php -f optimize.php "$file"
printf "PHP script returned %d" $?

# or you could just use if:
if php -f optimize.php "$file"; then
    do_stuff
fi

There's no need to include the current working directory in a script call. (Further to that, you could make the script executable and call it directly!)