Why does PHP give a parse error if you do:
echo array_keys($arr)[0];
It makes you assign the result of the function to a var first like this:
$arr = array_keys($arr);
echo $arr[0];
I just want to grab the 1st element of the array. Why does it make you do
it in 2 lines
When in safe mode shouldn't PHP check to see if the directory that is
about to be opened with a opendir() function has the same UID as the PHP
script itself, and fail if the UIDs do not match?
Because in PHP 4.0.6 with safe_mode on, a PHP script owned by fred can
open any directory owned by
But where user fred can opendir() a directory owned by user mary
(underneath the open_basedir), that action doesn't even pass a UID check
if the UIDs are supposed to match in safe mode in order for the action to
be allowed.
How would an optional GID check help?
A.
When in safe mode
Hi again,
I believe it should disallow openning a directory in safe mode if the UID
of the directory does not match the UID of the PHP script.
That is exactly the behavior of fopen() in safe mode.
Without that behavior, users are permitted to write a PHP script that lets
them crawl around the
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