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Show featherlightbox pop up on page load

I'm going to be totally honest. I have zero experience in wordpress php coding but I can do simple adjustments using the wordpress admin. Now I'm facing a problem. I used the feather lightbox js. I have a code below which shows the part of the footer.
<?php if ( is_front_page() ) : ?>
<script>
jQuery(document).ready(function() {
$.featherlight("#mylightbox");
//console.log("hi");
});
</script>
<?php endif; ?>
<?php wp_footer(); ?>
Open element in lightbox
<div id="mylightbox">This div will be opened in a lightbox</div>
</body>
With this the featherlightbo pops up when i clicked the anchor tag. But what i want to do is to have the feather lightbox to open it on page load. As you can see in my code there's a part there says if front_page which has also ready function. It returns an error
Uncaught TypeError: Cannot read property 'featherlight' of undefined
Please I need your help anyone.
Thanks.

<div id="mylightbox">Text to display in box</div>
For this popup use the script
$.featherlight($('#mylightbox'), {});
This will work

It seems like jQuery isn't defined using the $-sign, since jQuery.(document) works.
Change $.featherlight to jQuery.featherlight and it should work.
As you asked in your comment, you can initially hide the div by adding the css property: display: none on that div.

How do I embed Javascript in PHP code?

I have a HTML form and PHP code. In my form, I have a textbox and a textarea. Within both, I have included the "disabled" option. I want both textboxes to remain disabled until the user decides to click the "Edit" button, in which case both textboxes should be enabled so changes can be made and the output once again saved. According to the research I have done, the only way to do this is to use javascript, so I have included the following code within my PHP;
if (isset($_POST['edit']))
{
echo "<script type=\"text/javascript\">";
echo "var oldnotes = document.getElementById('oldnotes');";
echo "oldnotes.disabled = false;";
echo "var record = document.getElementById('record');";
echo "record.disabled = false;";
echo "</script>";
}
I have also tried;
if (isset($_POST['edit']))
{
echo "<script type=\"text/javascript\">";
echo "$('#oldnotes').removeAttr('disabled')";
echo "$('#record').removeAttr('disabled')";
echo "</script>";
}
But no luck :(
I am not receiving any errors, the textboxes just remain disabled after I click the Edit button. Could anyone help me with this? Thanks in advance.

This is a better approach for these kind of problems :
if (isset($_POST['edit'])){
?>
<script type="text/javascript">
var oldnotes = document.getElementById('oldnotes');
oldnotes.disabled = '';
var record = document.getElementById('record');
record.disabled = '';
</script>
<?php
}

<?
<script language='javascript'>
(javascript code here)
</script>
?>

Try use onclick on your Button
<input type="button" name="edit" id="edit" onclick="document.getElementById('oldnotes').disabled=false; document.getElementById('record').disabled=false; return false;">
I hope it helps.

The second approach seems correct, but you're missing ; there. Also, you haven't put any newline characters.
Here's what I suggest:
<?php
if (isset($_POST['edit'])) {
?>
<script type="text/javascript">
alert("Check entry"); // Use this for checking if your script is reaching here or not
$('#oldnotes, #record').removeAttr('disabled');
</script>
<?php
}
?>
You don't need to echo the whole thing. You can simply put it out of the PHP and keep it inside a if block.
Also, I've kept the alert to check if your code structure is proper. If you're getting the alert on clicking the edit button, means you're on the right path. You just need to workout on the JS.
If not, something wrong with your edit button and the form POST

Use single quotes for script type
echo "<script type='text/javascript'>\n";
//javascript goes here
echo "</script>";

use this jquery code:
<script>
$('#oldnotes,#record').attr('disabled','')
</script>

If you want to use JS with PHP you can read here:
how to write javascript code inside php
However: it seems you want the JS to be called on it own accord upon evaluation in the php logic as per the implementation provided.
Technically speaking your JS simply states it should re-enable the controls. Rather add the JS functionality to the OnClick of the submitting control and it should work fine. This is a quicker way of testing the functionality as well.
<a href="javascript:void(0);" onclick="document.getElementById('oldnotes').disabled=false;" >
Once it is clicked it will hide the appropriate control instead of trying to write new JS on the fly that is not executed. This is tried and tested and should work fine for you.
Hope it helps!

you can always make a cookie
use document.cookie()
set the onclick=“” to changing the cookie
and make a script that runs and gets the cookie when you open the site (use window.onload = function(){/*code goes here*/})

PHP: Reading DOM info of own HTML page

So, let's say I have only one file on my server called index.php :
<!DOCTYPE html>
<html>
<body style="background-color: orange;">
<!-- On/Off button's picture -->
<?php
echo ("<img id='button' src='data/button.jpg' alt='off'/>");
?>
</body>
</html>
Now, let's say I attach a JavaScript that changes the buttons ALT to ON when I click the button at some point.
How can I read the DOM elements on this page using PHP?
So far, I've found this code:
$dom = new DOMDocument;
$dom->loadHTML($html);
$main = $dom->getElementById('alt');
but I can't figure out how I would fit that in my code above, what page should I call loadHTML() for?
How do I make PHP read DOM elements it generated using echo?

First of all, your button has id button, so searching for the id alt would return no results.
Second: The piece of php you found reads a page $html (probably file_get_contents('some_url'), puts it in the variable dom and parses it, searching for the id alt. In your case it would read your own page, breaking it down in pieces looking for that button.
Thats of no use if you can read the status with javascript on your own page. You can then pass it to php trough an ajax-call or by adding it to the url like mypage.php?button=off

$main = $dom->getElementById('button');

How can I put PHP code inside JavaScript document.write

Hello can anyone please help me in fixing this code?
1st code
<script type="text/javascript">
if (document.getElementById("tester") != undefined)
{
document.write('**2nd code should be here**');
}
else
{
document.write('<img scr="./img.png" />');
}
</script>
2nd code
<?php while (have_posts()) : the_post(); get_template_part('item-video');
endwhile; ?>
What I want to happen is to insert the 2nd code into the first code. But when I try to insert it, it gives me a blank page.

the biggest thing I've heard is that PHP is server side, while Javascript is client side. Therefore, changing the html page using document.write() isn't going to do anything unless you reload the page. I think you can circumvent this problem tho, using Ajax and the load() function.

Javascript error document.getElementById('x').style.display='none'

What is wrong on this syntax:
PHP
echo '<li onMouseOver="" onMouseOut="document.getElementById(\''.$boxrow.'\').style.display=\'none\';">';
And on HTML page source code:
<li onmouseover="" onmouseout="document.getElementById('evidence').style.display='none';">
IT is working fine, but scripts after this wil causing errors, if I remove this script everything working fine.
First what is below this script and stop working is:
header("Location: index.php?module=service&no=".$_POST['id']."&"); and next after ...

PHP header redirection only works if you still didn't parse any code on your page. Try redirect by javascript instead.
Also it's a better practice not to use inline javascript. Better:
EDIT:
window.onload = function() {
li.onmouseout=function(){
document.getElementById('evidence').style.display='none';
};
};