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";
}
?>