PHP count of array returns more than it should?

Question

This is my code:

$dir = "img/";
$files = scandir($dir);

for ($i=0; $i <= count($files); $i++) { 
    echo $files[$i]."<br/>";
}
echo count($files);

count of array returns value of 2 on empty array, I cheched for hidden files, zero resault. So what can cause this? var_dump resault

array(7) { 
[0]=> string(1) "." 
[1]=> string(2) ".." 
[2]=> string(8) "img1.gif" 
[3]=> string(8) "img2.gif" 
[4]=> string(8) "img3.jpg" 
[5]=> string(8) "img4.png" 
[6]=> string(8) "img5.png" 
}

Answer 1

Its because your array contain '.' & '..' two file names.

You can get rid of it by using below code

$files = array_diff(scandir($dir), array('..', '.'));

Answer 2

You should be checking for return value of scandir first. See if $files is really an array?

<?php
$dir = "t";
$files = scandir($dir);
if(is_array($files))
{
for ($i=0; $i <= count($files); $i++) { 
    echo $files[$i]."<br/>";
}
echo count($files);
}
?>

Answer 3

This is normal behaviour of scandir() and also true.

This because every directory contain two logical entry

1) "." reference to current directory.
2) ".. reference to parent directory.

So for empty directory also u will get at list 2 thing.

See : PHP scandir() Function

Answer 4

The problem is that scandir() also return "." and ".." referring to the parent directory. If you var_dump($files) you will see what i am talking about.