Cron Job Confusion

[TheServant]

Hey guys, here is my PHP code I want to run in a cron job:

<?php
mysql_connect ("localhost", "username", "userpass") or die ('I cannot connect to MySQL because: ' . mysql_error());
echo('Connected to MySQL<br />');

mysql_select_db ("dbname") or die('I cannot connect to the database because: ' . mysql_error());
echo('Connected to Database<br />');
    
mysql_query( "
    UPDATE users...
    " ) or die(mysql_error());
?>

I get the following error in the output being sent to my email:

/home/user/public_html/.../file.php: line 1: ?php
  : No such file or directory
  /home/user/public_html/.../file.php: line 2: syntax error near unexpected token `"localhost",'
  /home/user/public_html/.../file.php: line 2: `mysql_connect ("localhost", "username", "userpass") or die ('I cannot connect to MySQL because: ' . mysql_error()); '

I know it's running, but I don't know why it's not working? There is no update so it's not getting to the function, but the path is all correct? Does that first erro mean I should not have <?php in my cron job file?

[Atli]

Hi.

I'm guessing the problem is that the code is not being executed as PHP, but rather as shell commands.

You need tell the shell that the code is meant to be executed using the PHP parser.
That can be done by adding the path the the PHP parser as the first line of the file:
[code=php]
#!/usr/bin/php

<?php
// Do something.
?>[/code]
If you don't know the path the the PHP executable, try doing:

whereis php

That should show you where it is.

[TheServant]

Thanks Atli. I spoke with support and they said a very similar thing to what you said. Although I cannot try it right now, it does make sense what you (and they) said, so thanks for your reply.