输入文件更新数据库中的图像

// Als de knop van de formulier is ingedrukt update de data dat van de database afkomt

if (isset($_POST['update'])) {


if (isset($_GET['id'])) {
          $chauffeurs_id = $_GET['id'];
        }


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


$sql = "SELECT * FROM chauffeurs ORDER BY 'chauffeurs_geboortedatum ASC";
        $sql = "UPDATE `chauffeurs`
        SET `Chauffeurs_foto`=$Chauffeurs_foto
        WHERE `id`='$chauffeurs_id'";


$result = $conn->query($sql);

        if ($result == TRUE){
            echo "Aanpassingen zijn voltooid.";
        }else{
            echo "Error:" . $sql . "<br>" . $conn->error;
        }
    }




if (isset($_GET['id'])) {
      $id = $_GET['id'];


$sql = "SELECT * FROM `chauffeurs` WHERE `id`='$id'";



$result = $conn->query($sql);



if ($result->num_rows > 0) {
        while ($row = $result->fetch_assoc()) {
            $id = $row['id'];
            $chauffeurs_foto = $row['chauffeurs_foto'];
}
$backId = $_SESSION['krijgid'];

<!-- Formulier waar alles in komt om te veranderen -->  
<form action="" method="post" enctype='multipart/form-data' >
          <fieldset>
           <legend>Gegevens chauffeur:</legend>
             <div class="form-group col-md-4">


//I did it like this so i could style the input file

<input type="file" name="file" id="file"  class="inputfile" value="data:image/jpeg;base64, <?php.base64_encode($row['chauffeurs_foto']).?>" />
<label class="btn btn primary"for="file">Choose a file</label>

<input type="submit" value="Chauffeur Aanpassen" name="update">
</fieldset>
</form>
</div>