通过php中的scandir函数在html文件中以透视图类型显示文件

Question

I am trying to show all images in a directory through scandir function but I am getting just the first letter of every file name instead of full name to show a picture as :

<?php
$path    = 'images/';
$files = scandir($path);
foreach ($files as $file) {
$x = 0;
$x_new = $x++;
echo "<img src='images/".$file[$x_new]."' >";
} 
?>

Here I am getting the result as :

<img src='images/.' >
<img src='images/.' >
<img src='images/J' >
<img src='images/J' >
<img src='images/i' >

doing a var dump gives me this as :

array(5) { [0]=> string(1) "." [1]=> string(2) ".." [2]=> string(17) "Job Board New.png" [3]=> string(13) "Job Board.png" [4]=> string(9) "index.jpg" }

Plus there are two . and .. elements in the array I don't have any files in the directory with these names so what these are really?

Will be great if you people please guide me in the right way..!

Plus How can I get the extension of files and then do different things using functions..!

Answer 1

Use trim() to remove . , also use $file instead of $file[$x_new] . Example here...

foreach ($files as $file) {
    $name = trim($file, ' .');
    if($name){
        echo "<img src='images/".$name."' >";
    }
}

Answer 2

The '.' directory means the current directory. The '..' directory means the parent directory of the current directory. This is common under UNIX, LINUX and Windows.

Use pathinfo to read the info about the file and its types. The array index extension gives the extension of the file.

$path_parts = pathinfo('/www/htdocs/inc/lib.inc.php');
echo $path_parts['basename'], "\n";
echo $path_parts['extension'], "\n";