使用unix运行php脚本作为后台进程

Question

I am trying to run the php script (below) named daemon.php as a back ground process using unix.

#!/usr/bin/php
<?php
$count = 0;
while(true){
    $count = $count + 1;
    file_put_contents('daemon1.log', $count, FILE_APPEND);
    sleep(1);  
}
?>

If I run it in the foreground using the command below

php daemon.php

The file deamon1.log starts getting written to. Also if I enter the command:

ps | grep php

I get the out put

10573 ttys000    0:00.20 php daemon.php

If I try to run it as a background process using the command

php daemon.php &

I get the console out put

[1] 10584

And the command

ps | grep php

Returns

10584 ttys000    0:00.02 php daemon.php

But nothing is written to deamon1.log. Can anyone tell me what I'm doing wrong?

Answer 1

Try a different approach.

$file_handler = fopen( 'daemon1tempname.log', 'a+') or die ('Can not work with the File!');

$count = 0;
while(true){
    $count = $count + 1;
    fwrite($file_handler, "Test {$count}
" );
    sleep(1);  
}
fclose($file_handler);

Notice the different name of the file. Just make sure you work with a new file. If that works, test it with the file you want. If that does not work, then you know it's a permission issue.