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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
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sheehanmedia commented Mar 13, 2018

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
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Helpmetoday commented Apr 15, 2018

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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