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@projectxcappe
Created September 15, 2011 23:22
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Display Images From A Folder with PHP
//Display Images From A Folder with PHP
<?php
$files = glob("images/*.*");
for ($i=1; $i<count($files); $i++)
{
$num = $files[$i];
echo '<img src="'.$num.'" alt="random image">'."&nbsp;&nbsp;";
}
?>
//Display Images With Image Name From A Folder
<?php
$files = glob("images/*.*");
for ($i=1; $i<count($files); $i++)
{
$num = $files[$i];
print $num."<br />";
echo '<img src="'.$num.'" alt="random image" />'."<br /><br />";
}
?>
//Using lightbox
<!-- Start VisualLightBox.com HEAD section -->
<link rel="stylesheet" href="engine/css/vlightbox1.css" type="text/css" />
<link rel="stylesheet" href="engine/css/visuallightbox.css" type="text/css" media="screen" />
<script src="engine/js/visuallightbox.js" type="text/javascript"></script>
<script src="engine/js/vlbdata.js" type="text/javascript"></script>
<!-- End VisualLightBox.com HEAD section -->
</head>
<body>
....
<div id="vlightbox1">
<?php
$thumbs = glob("data/facepainting_thumbs/*.*");
$images = glob("data/facepainting_images/*.*");
for ($i=1; $i<count($thumbs); $i++)
{
$numT = $thumbs[$i];
$numI = $images[$i];
echo '<a class="vlightbox1" href="'.$numI.'" title="'.$i.'"><img src="'.$numT.'"/></a>';
}
?>
</div>
....
@sheehanmedia

sheehanmedia commented Mar 13, 2018

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Awesome snippets, but there's an error in the fromfolder.php example. Because an array index starts at zero, so should the for loop. Should update that line to:

for ($i=0; $i<count($files); $i++)

Hope this is helpful!

@Helpmetoday

Helpmetoday commented Apr 15, 2018

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I would like to do something very specific, and I think you can help me...

I want to display one random png image at a time from 250 images total. The images must not be displayed until 12-midnight and stay on display until 12-midnight the next day before the next image replaces the previous image. Can someone make that script? I want to use php + mysql for the script. This script is what used to be called a picture of the day script.

@sasijarvis

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thanks man!! this worked well

@philg2018

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This helped a LOT!! Thank you so much. I have a question though, if anyone can help? I want to display the name of the image but without the path. How can I do that?

@riki1972

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Change this
print $num ."
";
with this
print basename($num) ."
";

@abaaskills

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Amazing, thanks.

I wish to see how to display img_name also 👍

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