CODEWITHSUNDEEP
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Roland
Roland

Reputation: 9731

Iterate through a folder?

I'm looking for the best way to iterate through a folder and put all the file names inside it into an array and in another one the count of the files.

I have found glob() as a good solution, but also a lot of alternatives for it on php.net. I'm not sure which I should use, so I'm asking here. If you're wondering for what I want to use it, it's to get all the .sql files inside a backup folder and display them as <li>thesqlfile.sql</li> and have a count of all of them too.

So I thought of having two arrays, one with their names, and one with the count of all of them. So in this case which method would be best fit to iterate ?

Method I:

<?php
    $files = array();
    foreach (glob("backup/*.txt") as $filename) {
        $files[]= $filename;
    }
    $count = sizeof($files);
?>

Method II:

function getfoldercontents($dir, $file_display = array(), $exclusions = array()) {
if(!file_exists($dir)){
    return FALSE;
}

$dir_contents = scandir($dir);
$contents = array();

foreach ($dir_contents as $file){
    $file_parts = explode('.', $file);
    $file_type = strtolower(end($file_parts));

    if ($file !== '.' && $file !== '..' && in_array($file_type, $file_display) && !in_array($file, $exclusions)) {
        $contents[] = $dir. '/'. $file;                        
    }
}

return $contents; 
}

Upvotes: 0

Views: 710

Answers (2)

Eregrith
Eregrith

Reputation: 4366

I would say the second is probably better in terms of file-control (through $file_display and $file_exclude), but maybe you should add a check to ensure the current file is not a directory named something.typeyouwishtodisplay

Upvotes: 0

Michael Berkowski
Michael Berkowski

Reputation: 270785

Since glob() already returns an array, you don't need to iterate over it to append to an array at all. Your first method is a little over-complicated. This accomplishes the same thing:

// Just assign the array output of glob() to a variable
$files = glob("backup/*.txt");
$num_files = count($files);

Upvotes: 3

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