How to send page title and url in feedback form?

Question

PHP

How to send a feedback form with the data from the form, also the title of the form and the url of the page from which it was sent

Answer 1

you can add any hidden fields to the form using the html input tag with the hidden type

<input type="hidden" name="url" value="тут-url-страницы">
<input type="hidden" name="form-title" value="тут-название-формы">

the fields themselves can either be filled in advance through php, or catch the event of the "on submit" form through js and fill them before submitting the form

Answer 2

Create a hidden field on a form

<input class="some-class" type="hidden" name="url">

Next, in the script file we write:

// jQuery
$(function ($) {
 $('.some-class').val(window.location.href);
});

or

// js
document.addEventListener("DOMContentLoaded", () => { 
 let urlInput = document.querySelector('.some-class');
 urlInput.value = window.location.href;
});

Don't forget to wrap it in a script tag if you add it to html.

Answer 3

Figured it out:
$(function ($) {
$('.set-url').val(window.location.href);
$('.set-tit').val(window.document.title);
});
THANK!