Качване на blob файлове/изображения в Mysql

Опитайте да инициализирате променливата. Добавих целия код. Не е необходимо да получавате $_POST['file'].

<?php 
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="clinic"; // Database name
$tbl_name="patients"; // Table name
session_start();
$con=mysqli_connect("localhost","root","root","clinic");

$id=$_REQUEST['id'];

 $name = '';
 $remarcs = '';
 $address = '';
 $test_res = '';
 $date = '';
 $phone = '';
  $new_path = '';

if (isset ($_POST['name'])) {
 $name = $_POST['name'];
 }
if (isset ($_POST['remarcs'])) {
 $remarcs = $_POST['remarcs'];
 }
if (isset ($_POST['test_res'])) {
$test_res = $_POST['test_res'];
}
if (isset ($_POST['address'])) {
  $address = $_POST['address'];
}

if (isset ($_POST['date'])) {
  $date = $_POST['date'];
}

if (isset ($_POST['phone_num'])) {
 $phone = $_POST['phone_num'];
}

if(isset($_FILES['file'])){ //Check file is uploaded or not

  $path = "../uploads/".$_FILES['file']['name'];
  if(move_uploaded_file($_FILES["file"]["tmp_name"], $path)){
  $new_path = $path;
  $sql=" update patients set 
  name = '$name',
  echo_photo = 'NULL',
  echo_file = '$new_path',
  remarcs = '$remarcs',
   test_res = '$test_res',
  date = '$date',
  address = '$address',
  phone_num = '$phone'
  WHERE id = ".$id;

  $result=mysqli_query($con,$sql) or die('Unable to execute query. '.   mysqli_error($con));

if($result){
 echo $name."<p>\n</p>";
 echo $remarcs."<p>\n</p>";
 echo $test_res."<p>\n</p>";
 echo $address."<p>\n</p>";
 echo $phone."<p>\n</p>";
 }
  echo "Uploaded";
} else {
  echo "Not uploaded";
 }
}
mysqli_close($con);

?>