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How do I add a php variable to a onclick function argument?

Beginner JS here.
I am trying to add a PHP variable to a Javascript onclick function. I converted the PHP variable to a JS variable just fine. However, when adding the JS variable to the function I'm not receiving the desired output. What am I doing wrong?
<script>
js_logo_number = "<?php echo $logo_number; ?>";
</script>
<img src='<?php echo $image1[0]; ?>' onclick="openModal();currentSlide(js_logo_number)">

You can do like this :
<img src='<?php echo $image1[0]; ?>' onclick="openModal();currentSlide(<?php echo $logo_number; ?>)">

This code works for me. Be sure you use a let or const (or the outdated var) prefix to declare js_logo_number a new variable. Also make sure that $logo_number is set like this:
<script>
let js_logo_number = <?php isset($logo_number)?$logo_number:null; ?>;
</script>
Then make sure you convert it to an integer in Javascript, if that's how you want to use it:
function currentSlide(num){
console.log('currentSlide fired', parseInt(num));
}
let js_logo_number = "5";
function openModal(){
console.log('openModal fired');
}
function currentSlide(num){
console.log('currentSlide fired', parseInt(num));
}
<img src='https://placekitten.com/200/200' onclick="openModal();currentSlide(js_logo_number)">

When PHP parses a file, it looks for opening and closing tags, which are which tell PHP to start and stop interpreting the code between them. Parsing in this manner allows PHP to be embedded in all sorts of different documents, as everything outside of a pair of opening and closing tags is ignored by the PHP parser.
See: PHP tags
1) echo
<?php echo 'if you want to serve PHP code in XHTML or XML documents, use these tags'; ?>
Solution:
<img src='<?php echo $image_src ?>'
onclick="openModal();currentSlide(<?php echo $logo_number ?>)">
2) Short echo
You can use the short echo tag to <?= 'print this string' ?>.
It's equivalent to <?php echo 'print this string' ?>.
Solution:
<img src='<?= $image_src ?>'
onclick="openModal();currentSlide(<?= $logo_number ?>)">
3) If short_open_tag is enabled
<? echo 'this code is within short tags, but will only work '
.'if short_open_tag is enabled'; ?>
Solution:
<img src='<? echo $image_src ?>'
onclick="openModal();currentSlide(<? echo $logo_number ?>)">
See: PHP tags

Cannot echo .load function

For some reason this code doesn't work. Adding \ to the front of the $ doesn't change anything. Can anyone see the problem?
$commentid = 2; //for example
echo "<script>";
echo "<div id = 'commentinput".$commentid."'>";
echo "</div>";
echo "$('#commentinput".$commentid."').load('editcomments2.php')";
echo "</script>";

Do it like this:
echo '$("#commentinput'.$commentid.'").load("editcomments2.php")';

I'd suggest not using echo to output HTML (in most cases anyway), but closing the PHP tag instead.
This becomes (edited with a loop example, see comments below):
<?php while ($commentid = $dbObject->dbMethod()): ?>
<div id="commentinput<?= $commentid ?>">
</div>
<script>
$('#commentinput<?= $commentid ?>').load('editcomments2.php');
</script>
<?php endwhile ?>
Note that in this case, the <script> portion should probably be outside of the loop, with a jQuery selector matching all the comment inputs.

So besides all of the other answers, I found another solution to!
When echoing out js code, like
echo '$("#commentinput'.$commentid.'").load("editcomments2.php")';
//and
echo "alert('stuff')";
it might then print it out like: $("#commentinput2").load("editcomments2.php")alert('stuff') and the code will get confused. So lastly I need to added a semicolon to prevent the program from reading all my code as just 1 line:
echo '$("#commentinput'.$commentid.'").load("editcomments2.php")';

How to change a href when using multiple variables in PHP?

So I didn't found what I was looking for in last two days.
To be clear I have a table "tracks" in my database.
So I can show from "tracks" the "demo" field wich is an audio file "url"..
<?= $tracks['demo']; ?>
I have multiple url's but then I need to replace the a href
<a href="<?php echo $tracks['demo'];?>"
in a logical way.
In simple terms it's like I want to click on button 1 that loads
url 1 into this a href with the ID, title field from that track.
When you press button 2 you need to load another url and replace url 1 with url2.
I tried many ways including javascript but it won't work.
Problem now is that when I list 4 items it always loads the latest "URL" i have posted in my source code. :)
When I echo out <?php echo $tracks['demo]; ?> on all items I can see the correct url's so the functionality to replace the url is my issue.
* I basically want to load many mp3 url's in a player I made in the footer of my website. It works by hardcoding the urls in the url field but this is not what it should be... *
Code:
<?php while($tracks = mysqli_fetch_assoc($muziek)) : ?>
<div class="col-md-2 viny" id="muziek">
<h4><?= $tracks['title']; ?></h4>
<img src="<?= $tracks['img']; ?>"
alt=" <?= $tracks['title']; ?> />
<p class="artist">Artist: <?= $tracks['artist']; ?></p>
</div>
<?= $tracks['demo'] = $audiourl ; ?>
<button type="button" onclick="play">Play</button>
</div>
<?php endwhile; ?>
*THIS LOOPS PERFECT FOR ALL ITEMS TRACKS ARE LISTED PERFECT *
In my player I have
<a href="<?php echo $audiourl;?>" ><?= $tracks['artist']; ?></b> - <?= $tracks['title']; ?></a>
function play(){
I'm not good at JS... And this is my issue because it now loads the latest output from the php loop...
}

I don't know how you are replacing it but
Why not pass the audio in the function. Here like this
<button type="button" onclick="play(<?= $tracks['demo'] = $audiourl ; ?>)">Play</button>
assign a id to anchor tag
<a id="someId" href="<?php echo $audiourl;?>" ><?= $tracks['artist']; ?></b> - <?= $tracks['title']; ?></a>
and in the play function, you receive it like this
function play(audioUrl){
}
and change the href like this in the play function
If you are using jquery
$('#someId').attr("href", audioUrl);
sometimes the above does not works. Can't remember why but if that does not works try this
$('#someId').href = audioUrl
If you are using javascript
document.getElementById('abcd').href = audioUrl

So I tried many things but I had many issues with the player it's js file using the same id's and so on.
The new logic I tried out seems to work:
<form action="" method="POST">
<input type="submit" value="Play" name="Play">
</form>
<?php
if (isset($_POST["Play"]))
{
echo $tracks['demo'];
}else{
echo ' ';
}
?>
There is more going on but I tried to write it down as simple as this to show the logic. All Track url's are hidden and when the page loads there is an empty $audiourl by default loaded in the a href from my audioplayer. By clicking the button play I show the url in the source but not on the site.
Then the latest $audiourl variable changes to this value.

include variable in javascript?

I need some help with the following.
I have a script that close down the web page using the following:
<a href=\"Javascript:void(0);\" onclick=\"parent.location.reload();window.parent.Shadowbox.close();\">
It works as it should. My problem is that I need to change parent to another page. For this I used the following:
<a href=\"Javascript:void(0);\" onclick=\"parent.location='http://www.google.com';window.parent.Shadowbox.close();\">
It also works, but now I do not want it to go to google but one of my pages where it should include a variable. Normally I would use:
<?=$array["id"]?>
But since it is a javascript then I cant use PHP and the following solution does not work:
<a href=\"Javascript:void(0);\" onclick=\"parent.location='new-new.php?$array[‘id’];window.parent.Shadowbox.close();\">
So I tried this:
function getNewURL() {
var root="http://minside.net";
var id="topic/#entry667119";
var nURL=root+"/"+id;
return nURL;
}
With the javascript:
<a href=\"Javascript:void(0);\" onclick=\"parent.location.replace(getNewURL);window.parent.Shadowbox.close();\">
For in this way to get my variables with me, but it does not work.
Someone who can help me? Or have another idea for how to resolve this?

you can assign a php variable to javascript variable.
var a=<?php echo $array[‘id’]?>;

you are using echo for that right since you use \ for ", so it should be like this:
<a href=\"Javascript:void(0);\" onclick=\"parent.location='new-new.php?'.$array[‘id’]'.window.parent.Shadowbox.close();\">

The best way to assign a variable in Javascript is
var id = <?php echo json_encode($array["id"]); ?>

You need to use echo:
<? echo $array["id"]; ?>
Edit
var id = <?php echo $array["id"]; ?>;
var link = "<a href=\"Javascript:void(0);\" onclick=\"parent.location='new-new.php?id="+id+";window.parent.Shadowbox.close();\">";
assuming you use $_GET['id'] in new-new.php (you omitted it in the URL)
Edit
if(strlen($large_photo_exists) > 0 && strlen($thumb_photo_exists) > 0)
{
echo $thumb_photo_exists;
echo "<p>Use this photo as my profile photo</p>";
echo "<br> OR <br>";
echo "<p>Upload and use another photo</p>";
}

Preload next couple images in a dynamic gallery

I've searched and found out solutions to preload images IF I have their file names. However my gallery and images are dynamic so it runs a php echo image url code to call my images.
<a href="<?php echo $image->imageURL ?>" <?php echo $image->thumbcode ?> >
<?php if ( !$image->hidden ) { ?>
<img title="<?php echo $image->alttext ?>" alt="
<?php echo $image->alttext ?>" src="<?php echo $image->thumbnailURL ?>"
<?php echo $image->size ?> />
<?php } ?></a>
What can I do to preload the next 2 or 3 images in the gallery since I wont be able to give the exact image file for each of my galleries. The only solution I found was to make all the images preload which I don't want.

You can probably hide your images when page load and you can run your preload image function
$('#gallery img').hide();
$('#gallery img').each(function(e) {
$(this).delay(500*e).fadeIn('slow');
})

I'm not clear as to what you're asking, but you can't load a resource without knowing it's path.
Also, I assume $image->size returns the string width="X" height="Y". You'll probably be better off having a height and width property, or store the dimensions in an array.
if($image->width) {
echo 'width="' . $image->width . '"';
}