How to intercept the AJAX action to send form?

Button code
<button class="btn btn--full btn__primary btn--md letter2-act" type="submit">Send</button>

When you press the button and completing the form I popup
Code:
<div class="white-popup" id="letter"><a class="popup-close popup-modal-dismiss" href="#"><i class="is-icon is-icon--close"></i></a>
<div class="white-popup-header popup-body text--center">
 <h3 class="light">Thank you. Your message has been sent.</h3>
</div>
<div class="white-popup-notice popup-bg grey--body text--center"><span class="light">We carefully review Your message and will contact You.</span></div></div>

How to catch the correct form submission? I used jquery
$(".sign-act").submit(function(){
checked++;
 })

But the checked value 0 Oates as standard. A reference to the button $(".sign-act") one hundred percent working, how to intercept?
June 10th 19 at 15:00
1 answer
June 10th 19 at 15:02
Reframe the question. Very confused, or not enough code is not clear on what to say
I need to press the button in the PopUp form to send target data to the system analysts of Yandex, test the possibilities on the variable, the variable is checked when submitting ajax forms are not functions submit click and other. - Darby_Bernier commented on June 10th 19 at 15:05
Works. Don't forget to open the console
link - tiffany_Wunsch commented on June 10th 19 at 15:08

Find more questions by tags AJAXjQuery