Re: [PHP] is_dir on WIndows
if (is_dir($file) ($file!=.) ($file!=..)) {
$users[]=$file;
}
It's only adding it to the array if all 3 of those are true. and since $file can't be
both '.' and '..' at the same time, it's never matching.
I'd do something like:
if (is_dir($file) || ($file!=.) || ($file!=..)) continue;
Mark wrote:
I'm having a problem I haven't encountered before. I'm using PHP
4.3.4 (upgrading the 4.3.9, but humor me) on Windows.
I'm using the following code snippet to try to get all the
subdirectories of a directory into an array.
$maildir=MERCURY./MAIL;
$handle=opendir($maildir);
// echo $handle;
while ($file = readdir($handle)) {
// echo $file;
// echo is_dir($file).BR/;
if (is_dir($file) ($file!=.) ($file!=..)) {
$users[]=$file;
}
}
What this should do (if I'm not a complete idiot), is put all the
true subdirectories (excluding . and ..) into the array $users. There
are two additional subdirectories under the $maildir directory.
However the array is empty. As you can see, I tried echoing out $file
and whether it is a directory, but it came up as not recognizing the
directories as such.
With the echos above in place, I get the resource handle, and it
echos everything you would expect except that is_dir() fails to
recognize the directories.
It DOES recognize . and .. as directories, for some bizarre reason.
And no, I can't switch to linux.
Any ideas?
Mark
=
Mark Weinstock
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Re: [PHP] is_dir on WIndows
Chris wrote:
if (is_dir($file) ($file!=.) ($file!=..)) {
$users[]=$file;
}
It's only adding it to the array if all 3 of those are true. and since
$file can't be both '.' and '..' at the same time, it's never matching.
I'd do something like:
if (is_dir($file) || ($file!=.) || ($file!=..)) continue;
That will match everything and anything, files and dot directories...
The first statement by the OP should work fine. All three will eval to
true if it is a directory, and the directory isn't . or ..
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John C. Nichel
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KegWorks.com
716.856.9675
[EMAIL PROTECTED]
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Re: [PHP] is_dir on WIndows
On Tuesday 28 September 2004 03:54, Mark wrote: With the echos above in place, I get the resource handle, and it echos everything you would expect except that is_dir() fails to recognize the directories. Because is_dir() expects a path to the file as well, otherwise it would be trying to look at $file in it's current working directory (cwd) wherever that may be (most likely not $maildir). It DOES recognize . and .. as directories, for some bizarre reason. Because all directories contain '.' and '..' and hence the cwd of is_dir() would have those. (BTW '.' and '..' ARE directories). -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-general -- /* QOTD: I looked out my window, and saw Kyle Pettys' car upside down, then I thought 'One of us is in real trouble'. -- Davey Allison, on a 150 m.p.h. crash */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] is_dir on WIndows
--- John Nichel [EMAIL PROTECTED] wrote:
Chris wrote:
if (is_dir($file) ($file!=.) ($file!=..)) {
$users[]=$file;
}
It's only adding it to the array if all 3 of those are true. and
since
$file can't be both '.' and '..' at the same time, it's never
matching.
I'd do something like:
if (is_dir($file) || ($file!=.) || ($file!=..)) continue;
That will match everything and anything, files and dot
directories...
The first statement by the OP should work fine. All three will
eval to
true if it is a directory, and the directory isn't . or ..
Which is the problem I'm having. It's not evaluating anything as a
directory, except . and ..
=
Mark Weinstock
[EMAIL PROTECTED]
***
You can't demand something as a right unless you are willing to fight to death to
defend everyone else's right to the same thing.
***
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Re: [PHP] is_dir on WIndows
--- Jason Wong [EMAIL PROTECTED] wrote:
On Tuesday 28 September 2004 03:54, Mark wrote:
With the echos above in place, I get the resource handle, and it
echos everything you would expect except that is_dir() fails to
recognize the directories.
Because is_dir() expects a path to the file as well, otherwise it
would be
trying to look at $file in it's current working directory (cwd)
wherever that
may be (most likely not $maildir).
It DOES recognize . and .. as directories, for some bizarre
reason.
Because all directories contain '.' and '..' and hence the cwd of
is_dir()
would have those. (BTW '.' and '..' ARE directories).
Perhaps I should have been more precise (I'm sure I should have...)
When I echo $file, it echos what I would expect (the names of the
files and directories. I guess I need to prefix $file in the is_dir
statement with the full path to the file?
So like this?
$maildir=MERCURY./MAIL;
$handle=opendir($maildir);
while ($file = readdir($handle)) {
if (is_dir($maildir/$file) ($file!=.) ($file!=..)) {
$users[]=$file;
}
}
It seems contrary to what's in the help, but I'll try it out. Thanks!
Mark
=
Mark Weinstock
[EMAIL PROTECTED]
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Re: [PHP] is_dir on WIndows
Mark wrote:
--- John Nichel [EMAIL PROTECTED] wrote:
Chris wrote:
if (is_dir($file) ($file!=.) ($file!=..)) {
$users[]=$file;
}
It's only adding it to the array if all 3 of those are true. and
since
$file can't be both '.' and '..' at the same time, it's never
matching.
I'd do something like:
if (is_dir($file) || ($file!=.) || ($file!=..)) continue;
That will match everything and anything, files and dot
directories...
The first statement by the OP should work fine. All three will
eval to
true if it is a directory, and the directory isn't . or ..
Which is the problem I'm having. It's not evaluating anything as a
directory, except . and ..
Are you giving is_dir() the full path from root to the directory?
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KegWorks.com
716.856.9675
[EMAIL PROTECTED]
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Re: [PHP] is_dir on WIndows
--- Jason Wong [EMAIL PROTECTED] wrote: On Tuesday 28 September 2004 03:54, Mark wrote: With the echos above in place, I get the resource handle, and it echos everything you would expect except that is_dir() fails to recognize the directories. Because is_dir() expects a path to the file as well, otherwise it would be trying to look at $file in it's current working directory (cwd) wherever that may be (most likely not $maildir). It DOES recognize . and .. as directories, for some bizarre reason. Because all directories contain '.' and '..' and hence the cwd of is_dir() would have those. (BTW '.' and '..' ARE directories). Thanks. That worked. Now that I think about it, it makse sense. I should have thought about it first... = Mark Weinstock [EMAIL PROTECTED] *** You can't demand something as a right unless you are willing to fight to death to defend everyone else's right to the same thing. *** ___ Do you Yahoo!? Declare Yourself - Register online to vote today! http://vote.yahoo.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] is_dir on WIndows
* Thus wrote Mark:
I'm having a problem I haven't encountered before. I'm using PHP
4.3.4 (upgrading the 4.3.9, but humor me) on Windows.
I'm using the following code snippet to try to get all the
subdirectories of a directory into an array.
$maildir=MERCURY./MAIL;
$handle=opendir($maildir);
// echo $handle;
while ($file = readdir($handle)) {
// echo $file;
// echo is_dir($file).BR/;
if (is_dir($file) ($file!=.) ($file!=..)) {
You need to provide the full path to is_dir():
is_dir($maildir/$file);
Otherwise it will simply look for $file relative to where the
script was called from.
A another way you could do this is:
foreach(glob($maildir/*, GLOB_ONLYDIR) as $dirname) {
// only use the directory name not the full path.
$users[] = substr($dirname, strrpos($dirname, '/')+1);
}
Curt
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Re: [PHP] is_dir on WIndows
John Nichel wrote:
Chris wrote:
if (is_dir($file) || ($file!=.) || ($file!=..)) continue;
That will match everything and anything, files and dot directories...
The first statement by the OP should work fine. All three will eval
to true if it is a directory, and the directory isn't . or ..
Heh, you're right. I knew I didn't spend enough time on that answer...
What I * should* have said was something like:
while($file = readdir($handle))
{
if ('.' == $file || '..' == $file) continue;
if(is_dir($handle.'\\'.$file)) $users[] = $file;
}
That will work, but it appears you've already sorted the issue out,
despite my help.
Chris
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