点击提交按钮后如何关闭灯箱表单？

Question

I have a light box form (and have a page in the background). I want to close the light box and stay on that page after clicking on the submit button. But it's redirecting to another page (PHP database connection page) after clicking the submit button.

I want to know how to close light box and stay on that page.

Answer 1

Add an event listener to the form that handles the form request and then return false from that handler and the page won't refresh. Because we don't know which lightbox you are using we can't exactly tell you how to remove it... but if you're using jQuery you could use something like:

$("#lightbox-id").hide();

or

$("#lightbox-id").remove();

Answer 2

If you haven't found your answer i believe this is it... check in your extension from chrome;

1. go to settings in your chrome browser

2. then extensions [chrome://extensions/].

It should have the words "form" "promo"

here is the link after uninstalling

https://www.geteasyforms.com/exit-survey?ti=7BCNr3xaTXfKQfvyuDBX

Answer 3

Use JavaScript to submit the form within the lightbox & close the lightbox.

Answer 4

function send_form ( )
{
ajax = new XMLHttpRequest;
if (!ajax) return false;

ajax.open('POST', '/index/send_form/'); //your post url should be here
ajax.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');

var field0= document.getElementById('submit').value;
var field1 = document.getElementById('message').value; //you should modify these fields according to your form 

ajax.send('sub=' + submit + '&message=' + content);
//then post url becomes something like: www.example.com/index/send_form/?sub=submit&message=this

//then close the lightbox in here
}

and for the html part remove action attribute and call this function in onsubmit event.