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Cómo obtener los nombres de los archivos de otra carpeta en mi servidor XAMPP con JS

Estoy creando una aplicación de álbum de fotos en la que, cuando cargas una imagen, se guarda en un directorio de fotos de mi servidor XAMPP. Ahora necesito acceder a los nombres de los archivos de esas fotos desde el directorio de fotos para poder mostrarlas nuevamente. ¿Cómo hacer esto con javascript?

 <?php if(isset($_FILES['image'])){ $error = arraY(); $file_name = $_FILES['image']['name']; $file_size = $_FILES['image']['size']; $file_tmp = $_FILES['image']['tmp_name']; $file_type = $_FILES['image']['type']; $hello = explode('.',$_FILES['image']['name']); $hello2 = end($hello); $file_ext = strtolower($hello2); $extensions = array('jpeg','jpg','png'); if(in_array($file_ext,$extensions)==false){ $error[] = "Please choose the image which has the extension as jpeg,jpg,png"; } if(empty($error) == true){ move_uploaded_file($file_tmp,"images/".$file_name); } } ?> <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <meta http-equiv="X-UA-Compatible" content="IE=edge"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>Document</title> </head> <body> <h3>Image Upload</h3> <form action="album.php" method="POST" enctype="multipart/form-data" id="form"> <input type="file" name="image" id="image"> <input type="submit" value="Upload"> </form> <ul class="photoalbum"></ul> </body> </html>
about 4 years ago · Juan Pablo Isaza
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