Hi I'm a newbie with PHP and this site, so please be nice :)
I'm currently having trouble working the below PHP foreach code out as I'm trying to echo all images in a HTML table 3 column but it echo's with 2 only.
UPDATE: I've managed to fix some issues thanks to the comments guy's, thank you. However, I', now experiencing another issue which is confusing.
Basically, If I have one picture in a folder, it will echo that one picture, but If I put two pictures there, it echo's out with 4, 1 first picture echo's with 2 and the second is with 2 as well. Basically showing 4 images even though I have 2 images in that folder. I can't seem to fix this..
Here's the code:
<?php
// get images
$images = glob($imagedir.'/' . "*.png");
$i = 0;
echo'<table><tr>';
foreach($images as $image)
{
$i++;
echo '<td><img src="'.$image.'" height="200"></td>';
if($i == 3)
{
echo '</tr><tr>';
$i = 0;
}
}
echo '</tr></table>';
?>
Thanks in Advance
You're code for rendering the HTML is just fine. If you have duplicates, the content of your imagedir must be wrong.
A few remarks:
glob($imagedir.'/' . "*.png");
also include directories which names end as .png
.?>
at the end a the php file.I've altered you code to avoid above to problems. I'm sure there more/other ways to do this, but this came in mind first.
<?php
$imagedir = 'images';
//Get *.png files only
$images = array_filter(glob("$imagedir/*.png"), 'is_file');
//Make image array divisble by 3 (columns)
while (count($images) % 3 != 0) {
$images[] = '';
}
echo'<table><tr>';
for ($i = 1; $i <= count($images); $i++) {
//Render TD if $image is not empty
if ($images[$i-1] != '' != '') {
echo '<td><img src="' . $images[$i-1] . '">', "<br>Image $i</td>";
}
//Close table row after 3 TD's
if($i % 3 == 0)
{
echo '</tr><tr>';
}
}
echo '</tr></table>';