dossier en cours d'analyse en utilisant PHP

Question

J'ai une structure de dossiers comme celui-ci:

/articles 
    .index.php 
    .second.php 
    .third.php 
    .fourth.php 

Si j'écris mon code dans second.php, comment puis-je analyser le dossier en cours (articles)?

Merci

Answer 1

$files = glob(dirname(__FILE__) . "/*.php"); 

http://php.net/manual/en/function.glob.php

Answer 2

Cela dépend de ce que vous entendez par 'scan' Je suppose que vous voulez faire quelque chose comme ceci:

$dir_handle = opendir("."); 

while($file = readdir($dir_handle)){ 
     //do stuff with $file 
} 

Answer 3

De PHP manual :

$dir = new DirectoryIterator(dirname($path)); 
foreach ($dir as $fileinfo) { 
    if (!$fileinfo->isDot()) { 
     var_dump($fileinfo->getFilename()); 
    } 
} 

Answer 4

<?php 

$path = new DirectoryIterator('/articles'); 

foreach ($path as $file) { 
    echo $file->getFilename() . "\t"; 
    echo $file->getSize() . "\t"; 
    echo $file->getOwner() . "\t"; 
    echo $file->getMTime() . "\n"; 
} 

?> 

From The Standard PHP Library (SPL)

Answer 5

foreach (scandir('.') as $file) 
    echo $file . "\n"; 

Answer 6

dossier en cours d'analyse

$zip = new ZipArchive(); 
$x = $zip->open($filepath); 
if ($x === true) { 
    $zip->extractTo($uploadPath); // place in the directory 
    $zip->close(); 

    $fileArray = scandir($uploadPath); 
    unlink($filepath); 
} 
foreach ($fileArray as $file) { 
    if ('.' === $file || '..' === $file) 
    continue; 
    if (!is_dir("$file")){ 
     //do stuff with $file 
    } 
} 

Answer 7

essayer

$dir = glob(dirname(__FILE__)); 
    $directory = array_diff(scandir($dir[0]), array('..', '.')); 
    print_r($directory); 

Answer 8

Liste toutes les images dans un dossier

$dir = glob(dirname(__FILE__)); 

$path = $dir[0].'\\images'; 

$imagePaths = array_diff(scandir($path), array('.', '..', 'Thumbs.db')); 
?> 
<ul style="overflow-y: auto; max-height: 80vh;"> 
<?php 

    foreach($imagePaths as $imagePath) 
    { 
?> 
    <li><?php echo '<img class="pagina" src="images/'.$imagePath.'" />'; ?></li> 
<?php 
    } 
     ?> 
</ul>