How to send POST dvidenie mouse?

There are the usual form on the page in php I want to check if move the mouse the one who filled it.
A hidden field is not suitable, because palitsya and populated with the appropriate data, cook, too, because the cookies appear after the installation just like when you reload the page, and they also can be viewed during transmission.
It is clear that 100% reliable option probably not, but at least a little bit-a little more complicated though...
July 9th 19 at 13:10
5 answers
July 9th 19 at 13:12
If it's protection from the bot - You crazy.
I half forms TAB and scroll through using space/shift-space is filled, the touchpad don't even use.
Sure I'm not the only one.

// EDIT

If it really is a protection bot - the latest version of ReCaptcha, where you have to choose a few pictures = Your cure.
July 9th 19 at 13:14
A hidden field is not suitable, because palitsya and populated with the appropriate data
well, ask in this hidden field: "You me manage to get the FAQ on the white light cannot be!"))))))))
And as the field is filled - it's a bot.

PS1: There is another option: transparent "proxying" a POST request to a URL mask (s):
1. Verification on variable captcha.
2. If the captcha is not passed - play the Turing test (direct full-page).
3. Passed - pass a POST request to the website and set variable flag that the captcha for this session - passed, and never requested it.
July 9th 19 at 13:16
Look at the event onBlur onFocus
July 9th 19 at 13:18
Do not interfere in JS to track the movement of the mouse, and using ajax to send POST to the server with the data you need.
July 9th 19 at 13:20
Maybe better websocket to merge the position of the mouse? Timer 10 times per second for example to throw json on a web socket and a server has to listen and do with this info what you wish
Interesting option but as in the simplest case, the websocket know about the position of the mouse? - Clovis_Gulgowski43 commented on July 9th 19 at 13:23
:
document.onmousemove=function(e){window.x=e;console.log(e.screenX);};<code></code>
- camryn27 commented on July 9th 19 at 13:26
console.log() remove, take the timer value from the document.x of the values of the screenX and screenY and send to the websocket. - camryn27 commented on July 9th 19 at 13:29
: how to take coordinato I know, the question was how exactly to send to the websocket? - Clovis_Gulgowski43 commented on July 9th 19 at 13:32
: well, it's already too much). In Google the first link upon request js websoket - camryn27 commented on July 9th 19 at 13:35

Find more questions by tags JavaScriptPHP