ID:               26814
 Updated by:       [EMAIL PROTECTED]
-Summary:          Misleading data returned by get_included_files
 Reported By:      mccarthy36 at earthlink dot net
-Status:           Open
+Status:           Verified
-Bug Type:         PHP options/info functions
+Bug Type:         *General Issues
-Operating System: Linux
+Operating System: *
-PHP Version:      4.3.4
+PHP Version:      5CVS, 4CVS
 New Comment:

Parse error -> execution should stop ?
(but it doesn't, that's the bug here, IMO)



Previous Comments:
------------------------------------------------------------------------

[2004-01-06 11:10:02] mccarthy36 at earthlink dot net

Description:
------------
I don't know if this is considered a bug, but in my opinion it's
undesirable behavior.  I'm finding that if I try to include a file that
has a parse error, the file is not included -- the include function
used returns false -- but the "included" file name is in the array
returned by get_included_files().

Reproduce code:
---------------
(file 1)
<?php

$worked = "NO";

echo "*", include_once( 'included.php' ), "*";

echo "<pre>"; var_dump( get_included_files() ); echo "</pre>";

echo "#{$worked}#";

?>

(file 2)
<?php

$worked = "YES";

$name = "blah "whatever";

?>

Expected result:
----------------
Since there is a parse error in the "included" file, and include_once()
returns false, I expect the name of the "included" file not to appear
in the array returned by get_included_files().

Actual result:
--------------
include_once() returns false, yet the name of the "included" file is in
the array returned by get_included_files().


------------------------------------------------------------------------


-- 
Edit this bug report at http://bugs.php.net/?id=26814&edit=1

Reply via email to