The following is my code.I am unable to send the image file to the php page.I'm using formData and i'm not clear about the concept of it.How do i send it to php page and how do i retrieve The image in php page?
JAVASCRIPT CODE
function UpdateUserDetails() {
   var title = $("#title1").val();
   var store = $(".search-box").val();
   var category= $("#category").val();
   var descp=$("#descp1").val();
   var url=$("#url1").val();
   var id = $("#hidden_user_id").val();
   var form = $('#image1')[0]; 
   var formData = new FormData(form);
   $.post("update.php", {
        id: id,
        title:title,
        store:store,
        category:category,
        descp:descp,
        data:formData,  
        url:url
    },
    function (data, status) {
        $("#update_user_modal").modal("hide");
        readRecords();
    }
);
}
update.php
<?php
include("db_connection.php");
if(isset($_POST))
{
   $id = $_POST['id'];
   $title=$_POST['title'];
   $desc=$_POST['descp'];
   $pname=$_POST['store'];
   $category=$_POST['category'];
   $url=$_POST['url'];
   $path = $_FILES['tmp_name'];
   $name = $_FILES['name'];
   $size = $_FILES['size'];
   $type = $_FILES['type'];
   $content = file_get_contents($path);
   $content = base64_encode($content);
   $sql1 = "update products set title='$title',url='$url',store='$pname', product_catagory='$category', image='$content',size='$size',type='$type',descp='$desc' where id=".$id."";
        if(mysql_query($sql1))
        {
            echo"updated";
        }
        else   
            echo "Not Updated";       
 }
 ?>
 
    