Как принять post с файлом на php?

Question

[fuckingawesomenigga]

var ajaxPicture = function(){
    	var picture = new FormData();
    	picture.append('picture', $('input[name=picture]').prop('files')[0]);
    	$.ajax({
    		url: 'functions.php',
    		type: 'POST',
    		dataType: 'html',
    		processData: false,
    		data: picture
    	})
    	.done(function(data) {
    		alert(data);
    		console.log("img done");
    	});
    }

Answer 1

[Исрапил Ахмедов]

$_FILES

Answer 2

[irishmann]

<?php
$file = $_FILES['picture'];