The output files from the folder?

Question

The output files from the folder?

Hi!
Recently asked me how to get the list of files for the slider, but the topic is stalled.
How list the files in a folder?

there is a code tag

spoiler

<?php
 $fienames = glob('/img/*.png');
 foreach ($filenames as $filename) {
 echo "<div-->$filename

";
}
?>

The files themselves
But it glows white in the editor and does not work including the hosting.
In the folder only 5 images.
How to fix errors in this code?
Thanks for the replies!

PHP

Lisette asked June 10th 19 at 14:22

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More answers about "The output files from the folder?"

2 answers

Baylee_Kulas answered on · Accepted Answer · 2019-06-10T12:24:32.330Z

Baylee_Kulas answered on June 10th 19 at 14:24

Solution

Ochepyatka is the worst enemy of the encoder:
$fienames !== $filenames

After correcting two ochepyatka:

Oh, how much is nerves are lost because of a simple comma =) - Lisette commented on June 10th 19 at 14:27

even the closing tag of the PHP looked like this:
/?> - Baylee_Kulas commented on June 10th 19 at 14:30

Caesar.Keeli answered on · Accepted Answer · 2019-06-10T12:26:32.330Z

Caesar.Keeli answered on June 10th 19 at 14:26

Solution

<?php
 $dir = 'images/'; 
 $cols = 3; 
 $files = scandir($dir); 
 echo "<table-->";
 $k = 0; 
 for ($i = 0; $i < count($files); $i++) { 
 if (($files[$i] != ".") && ($files[$i] != "..")) { 
 if ($k % $cols == 0) echo ""; 
 echo ""; 
 $path = $dir.$files[$i]; 
 echo "<a href="$path">"; 
 echo "<img src="$path" alt width="100">"; 
 echo "</a>"; 
 echo ""; 


 if ((($k + 1) % $cols == 0) || (($i + 1) == count($files))) echo "";
$k++;
}
}
 echo ""; 
?>

tested it personally

Thank you!
But as a picture not a table issue and inside div ?
List type

the <div>img.png</div>
the <div>img.png</div>
the <div>img.png</div>
 the <div>img.png</div>

- Lisette commented on June 10th 19 at 14:29

you can't get there div and insert the same img is also not necessary ...
just write echo and how your code is written - Baylee_Kulas commented on June 10th 19 at 14:32

I've just got a slider is. his slides are laid out divами. I was wondering how to make the formation of these DIVS from the folder. that is to say to minimize the content of slider without code. - Lisette commented on June 10th 19 at 14:35

have reduced the code to this:

<?php
 $dir = 'img/'; 
 $cols = 3; 
 $files = scandir($dir); 
 $k = 0; 
 for ($i = 0; $i < count($files); $i++) { 
 if (($files[$i] != ".") && ($files[$i] != "..")) { 
 echo "<div-->"; 
 $path = $dir.$files[$i]; 
 echo "<a href="$path">"; 
 echo "<img src="$path" alt width="100">"; 
 echo "</a>"; 
 echo ""; 
}
}
?>

Thank you!
It works!! - Caesar.Keeli commented on June 10th 19 at 14:38

I left - I'm glad that helped - Lisette commented on June 10th 19 at 14:41