The bug seems rather simple. Using command-line PHP, the following fragment should work:

#!/usr/bin/php -q
<?php

   $fd = fopen('php://stdout','w');
   if (posix_isatty($fd))
      print "yes, it is a terminal\n";
   else
      print "no, it is not a terminal\n";

?>

However, it appears that the function definition requires a long passed, instead of a resource. The error message reported is:

dns:root-/usr/bin> ./test.php
PHP Warning:  posix_isatty() expects parameter 1 to be long, resource given in /usr/bin/test.php on line 4
<br />
<b>Warning</b>:  posix_isatty() expects parameter 1 to be long, resource given in <b>/usr/bin/test.php</b> on line <b>4</b
><br />
no, it is not a terminal

A workaround appears to be the following:

#!/usr/bin/php -q
<?php

   $fd = fopen('php://stdout','w');
   settype($fd,'int');
   if (posix_isatty($fd))
      print "yes, it is a terminal\n";
   else
      print "no, it is not a terminal\n";

?>

The explicit type-cast fixes the problem. Maybe I have settings in the php.ini file wrong, but I don't think this function should behave like this. Perhaps it requires just a change to the source. This is also broke with posix_ttyname(). Am I suppose to acquire a file descriptor another way?

Regards,

Patrick O'Lone