ID: 26814
Updated by: [EMAIL PROTECTED]
Reported By: mccarthy36 at earthlink dot net
Status: Verified
Bug Type: *General Issues
Operating System: *
PHP Version: 5CVS, 4CVS
New Comment:
one more comment: using require() doesn't matter either, still the main
script is run to the end..
Previous Comments:
------------------------------------------------------------------------
[2004-01-06 17:35:55] [EMAIL PROTECTED]
Parse error -> execution should stop ?
(but it doesn't, that's the bug here, IMO)
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[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().
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--
Edit this bug report at http://bugs.php.net/?id=26814&edit=1
