some help using glob and foreach loop

[webdeveloper123]

Hi Guys,

I am creating an image gallery. I already have working code to select and upload a image into a directory. I have successfully used glob and a foreach loop to display the images from that folder on to a page. I found some code on "W3C How to" which had ready made image gallery with css and JS included. Here is the link: https://www.w3schools.com/howto/howto_js_slideshow.asp

What I am trying to do is echo the html in a foreach loop. I am doing ok so far but I can't get an html attribute into my echo code. Here is what I am trying to echo:

<div class="mySlides fade">
    <div class="numbertext">1 / 3</div>
    <img src="img1.jpg" style="width:100%">  
  </div>

Here is what I have:

$files = glob("images/*.{jpg,png,gif,svg,jpeg,bmp,webp}", GLOB_BRACE);


foreach ($files as $filename) {
           echo '<div class="mySlides fade">';
           echo '<div class="numbertext">1 / 3</div>';
        printf("<img src='images/%s'/>" , basename($filename));

}

The above code does display all images from the folder "images" onto the page so that's fine.

Now the printf replaces the 

           <img src="img1.jpg" style="width:100%">;

But I can't seem to get the 

style="width:100%"

Into the printf statement. I have been trying for hours, escaping characters, going in and out php etc etc but Can't figure it out

Can someone help please?

[ginerjm]

Don't know why the printf is needed.  Here is how I would tackle it, no testing of course.  I'll leave that to you.

$files = glob("images/*.{jpg,png,gif,svg,jpeg,bmp,webp}", GLOB_BRACE);
$cnt = count($files);
$i = 0;
foreach ($files as $filename) 
{
	$i++;
	echo "
		<div class='mySlides fade'>
		<div class='numbertext'>$i / $cnt</div>
		<img src='$filename' style='width:100%;'>
		</div>
		";
}

Not sure what $filename actually contains but you can easily append anything needed to the src= attribute clause if it's not already there

Edited August 6, 2022 by ginerjm

[mac_gyver]

the format specifier character is a % and the format specifier for a literal % is %, so the result is %%. this is similar to the %s you are also using. the % is the format specifier character. the s is the format specifier for a string argument, resulting in %s.

[webdeveloper123]

thanks ginerjm i'll give that a go

Sorry mac_gyver I don't know what your getting at. I read about the format specifier on php.net

are you saying the %s is clashing with width:100%?

[webdeveloper123]

12 minutes ago, ginerjm said:

Not sure what $filename actually contains

it's passed to basename in my original code

[ginerjm]

I did a test of my own and my code works just fine with no need to modify the use of $filename.  Try it.  Don't know why you thought you had to play with the glob output.

[webdeveloper123]

8 minutes ago, ginerjm said:

I did a test of my own and my code works just fine

haha clever guy. Thanks, and thanks for the numbertext, I wasn't sure how to do that one!

[ginerjm]

Can I see your final code?

[Barand]

17 minutes ago, webdeveloper123 said:

are you saying the %s is clashing with width:100%?

He is saying that to output a "%" in a format string you need to have "%%".

printf("<img src='images/%s' style='width: 100%%' />" , basename($filename));

 

[webdeveloper123]

3 minutes ago, ginerjm said:

Can I see your final code?

<?php


$files = glob("images/*.{jpg,png,gif,svg,jpeg,bmp,webp}", GLOB_BRACE);
$cnt = count($files);
$i = 0;

?>

<div class="slideshow-container">

  <?php
foreach ($files as $filename) 
{
  $i++;
  echo "
    <div class='mySlides fade'>
    <div class='numbertext'>$i / $cnt</div>
    <img src='$filename' style='width:20%;'>
    </div>
    ";
}


?>

 

[webdeveloper123]

3 minutes ago, Barand said:

He is saying that to output a "%" in a format string you need to have "%%".

ahh ok. Thanks barand

[ginerjm]

So close.  You need to learn that you DO NOT NEED TO LEAVE PHP MODE just to output a line or two of html.

$files = glob("images/*.{jpg,png,gif,svg,jpeg,bmp,webp}", GLOB_BRACE);
$cnt = count($files);
$i = 0;
echo "<div class='slideshow-container'>";
foreach ($files as $filename) 
{
  $i++;
  echo "
    <div class='mySlides fade'>
    <div class='numbertext'>$i / $cnt</div>
    <img src='$filename' style='width:20%;'>
    </div>
    ";
}

Entirely in php mode!

HTH

[webdeveloper123]

2 minutes ago, ginerjm said:

You need to learn that you DO NOT NEED TO LEAVE PHP MODE just to output a line or two of html.

yes you mentioned a few times to me before

[ginerjm]

So why do you continue?

[webdeveloper123]

I will, it's just the learning process. everyone learns at their own pace, and some people get stuck in bad habits

[ginerjm]

Or is it because you aren't learning but rather continuing to copy other people's work as your own?  How hard can it be to catch on to not leaving php mode?

[Barand]

6 minutes ago, ginerjm said:

So why do you continue?

Perhaps he hasn't realized yet that ignoring your commandments can cause your wrath to descend on him in the form of seven deadly plagues.

[webdeveloper123]

Who's work did I copy as my own? I marked as your solution, I never claimed I wrote that code. That's how a forum works. You get stuck, ask a a question and sometimes you get the correct answer. 

[ginerjm]

And that is not at all what I was referring to.

Never mind.  Just wasting my time here.

[webdeveloper123]

2 minutes ago, Barand said:

Perhaps he hasn't realized yet that ignoring your commandments can cause your wrath to descend on him in the form of seven deadly plagues.

loooooooool