I'm maiking simple form with image upload option. I have problem with getting access to my uploaded file through jQuery in my PHP code.
How I can check uploaded file name, size in PHP?
HTML:
    <form id="conversion_form" method="post" action="#" enctype="multipart/form-data">
            <input type="text" id="prize_name" size="25" value="">
<textarea rows="8" cols="75" name="lista_losowanie" id="lista_losowanie">
            <input id="file_img" type="file" name="image">
            <div id="uploadPreview"></div>
            </div>
            <p><input id="button_draw" class="btn btn-lg btn-success" type="submit" value="Losowanie" /></p>
            </form>
This is my Ajax, sending prize_name works good, but I have problem with my uploaded image. How I can send it in Ajax and receive in my PHP?
AJAX:
$('#button_draw').click(function(e) {
e.preventDefault();
    var formData = new FormData();
    // let's just take the first image
    var file = $('#file_img')[0].files[0];
    // but you also need to add the other fields to formData
    var lista = $(this).closest('form').find('#lista_losowanie').val();
    var prize_name = $(this).closest('form').find('#prizename').val();
    formData.append('prize_img', file);
    formData.append('lista', lista);
    formData.append('prize_name', prize_name);
    console.log(formData);
    alert (formData);
            $.ajax({
            context: this,
            url: './inc/new_draw.php',
            data: formData,
            type: 'POST',
            success: function(response){
            //odpowiedź serwera
            //Trzeba sparsować zwrot
            var json = JSON.parse(response);
            if(json.odp == 1){
            }
            if(response == 2){
            }
            },
            error: function(x,n,o){
            }
           });
  });
PHP:
<?php
include ("config.php");
$prize_name = mysqli_real_escape_string($con, $prize_name);
$prize_img = mysqli_real_escape_string($con, $prize_img);
//HOW I CAN GET ACCESS TO UPLOADED IMAGE? HOW I CAN CHECK FILE SIZE ETC. THIS BELOW DOESNT WORK.
        $file_size = $prize_img['size'];
        $file_name = $prize_img['name'];
        $file_type = $prize_img['type'];
        echo $file_size;
        if ($prize_name == NULL) {
            $problem = TRUE;
            $problem_code = 1;
            echo json_encode($dodano);
        }
?>
 
    