How to display images from a folder using php - PHP

Question

It would be great if someone could help me figure out why the browser cannot load the images (error 404). The code works, and the image source is correct, but I cannot figure out what is wrong. (using localhost)

$dir          = '/home/user/Pictures';
$file_display = array(
    'jpg',
    'jpeg',
    'png',
    'gif'
);

if (file_exists($dir) == false) {
    echo 'Directory \'', $dir, '\' not found!';
} else {
    $dir_contents = scandir($dir);

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

        if ($file !== '.' && $file !== '..' && in_array($file_type, $file_display) == true) {
            echo '<img src="', $dir, '/', $file, '" alt="', $file, '" />';
        }
    }
}

there are syntax errors on the code above like in_array( does not have the closing parenthesis. the $ char in the echo — jerjer
– jerjer, Commented Oct 2, 2013 at 2:33

Bill · Accepted Answer · 2013-10-02 03:34:03Z

11

You had a mistake on the statement below. Use . not ,

echo '<img src="', $dir, '/', $file, '" alt="', $file, $

to

echo '<img src="'. $dir. '/'. $file. '" alt="'. $file. $

and

echo 'Directory \'', $dir, '\' not found!';

to

echo 'Directory \''. $dir. '\' not found!';

edited Oct 2, 2013 at 3:34

answered Oct 2, 2013 at 2:30

Bill

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11 Comments

jerjer Over a year ago

actually for echo statement comma can be used instead of the concat operator

jerjer Over a year ago

the comma actually has better performance compared to . on echo statements

jerjer Over a year ago

string concatenation is memory intensive

jerjer Over a year ago

@user2837048, have you corrected the syntax errors from your code?

user2837048 Over a year ago

Yes, I did, as you can tell.

|

jerjer · Accepted Answer · 2013-10-02 03:05:46Z

Here is a possible solution the solution #3 on my comments to blubill's answer:

yourscript.php
========================
<?php
    $dir = '/home/user/Pictures';
    $file_display = array('jpg', 'jpeg', 'png', 'gif');

    if (file_exists($dir) == false) 
    {
        echo 'Directory "', $dir, '" not found!';
    } 
    else 
    {
        $dir_contents = scandir($dir);

        foreach ($dir_contents as $file) 
        {
            $file_type = strtolower(end(explode('.', $file)));
            if ($file !== '.' && $file !== '..' && in_array($file_type, $file_display) == true)     
            {
                $name = basename($file);
                echo "<img src='img.php?name={$name}' />";
            }
        }
    }
?>


img.php
========================
<?php
    $name = $_GET['name'];
    $mimes = array
    (
        'jpg' => 'image/jpg',
        'jpeg' => 'image/jpg',
        'gif' => 'image/gif',
        'png' => 'image/png'
    );

    $ext = strtolower(end(explode('.', $name)));

    $file = '/home/users/Pictures/'.$name;
    header('content-type: '. $mimes[$ext]);
    header('content-disposition: inline; filename="'.$name.'";');
    readfile($file);
?>

Rully · Accepted Answer · 2013-10-02 02:53:22Z

You have two ways to do that:

METHOD 1. The secure way.

Put the images on /www/htdocs/

<?php
    $www_root = 'http://localhost/images';
    $dir = '/var/www/images';
    $file_display = array('jpg', 'jpeg', 'png', 'gif');

    if ( file_exists( $dir ) == false ) {
       echo 'Directory \'', $dir, '\' not found!';
    } else {
       $dir_contents = scandir( $dir );

        foreach ( $dir_contents as $file ) {
           $file_type = strtolower( end( explode('.', $file ) ) );
           if ( ($file !== '.') && ($file !== '..') && (in_array( $file_type, $file_display)) ) {
              echo '<img src="', $www_root, '/', $file, '" alt="', $file, '"/>';
           break;
           }
        }
    }
?>

METHOD 2. Unsecure but more flexible.

Put the images on any directory (apache must have permission to read the file).

<?php
    $dir = '/home/user/Pictures';
    $file_display = array('jpg', 'jpeg', 'png', 'gif');

    if ( file_exists( $dir ) == false ) {
       echo 'Directory \'', $dir, '\' not found!';
    } else {
       $dir_contents = scandir( $dir );

        foreach ( $dir_contents as $file ) {
           $file_type = strtolower( end( explode('.', $file ) ) );
           if ( ($file !== '.') && ($file !== '..') && (in_array( $file_type, $file_display)) ) {
              echo '<img src="file_viewer.php?file=', base64_encode($dir . '/' . $file), '" alt="', $file, '"/>';
           break;
           }
        }
    }
?>

And create another script to read the image file.

<?php
    $filename = base64_decode($_GET['file']);
    // Check the folder location to avoid exploit
    if (dirname($filename) == '/home/user/Pictures')
        echo file_get_contents($filename);
?>

considering the first method I think the "break" statement and checking if the ending of the file is "." or ".." are unnecessary unless I missed something, then would like to know what.

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