I know I can do this...
glob(\'/dir/somewhere/*.zip\');
...to get all files ending in .zip
, but is there a way to return all f
You could always try something like this:
$all = glob('/dir/somewhere/*.*');
$zip = glob('/dir/somewhere/*.zip');
$remaining = array_diff($all, $zip);
Although, using one of the other methods Pascal mentioned might be more efficient.
I don't think glob can do a "not-wildcard"...
I see at least two other solutions :
$dir = "/path";
if (is_dir($dir)) {
if ($d = opendir($dir)) {
while (($file = readdir($d)) !== false) {
if ( substr($file, -3, 3) != "zip" ){
echo "filename: $file \n";
}
}
closedir($d);
}
}
NB: "." and ".." not taken care of. Left for OP to complete
This pattern will work:
glob('/dir/somewhere/*.{?,??,[!z][!i][!p]*}', GLOB_BRACE);
which finds everything in /dir/somewhere/ ending in a dot followed by either
?
)??
)[!z][!i][!p]*
)A quick way would be to glob()
for everything and use preg_grep() to filter out the files that you do not want.
preg_grep('#\.zip$#', glob('/dir/somewhere/*'), PREG_GREP_INVERT)
Also see Glob Patterns for File Matching in PHP