问如何使用外部php文件查看xampp中的图像？

EN

<?php
//Upload Image
if (isset($_POST['uploadImg'])) {
    if (is_uploaded_file($_FILES['image']['tmp_name'])) {
        if (move_uploaded_file($_FILES['image']['tmp_name'], "upload/{$_FILES['image']['name']}")) {
            echo "File Uploaded";
        }else{
            echo "File not uploaded";
            $i = '';
        }
        $i = $_FILES['image']['name'];
    }else{
        $i ='';
    }

    //Update database
    $query = "UPDATE Employees SET Photo = '$i' WHERE EmployeeID = {$_SESSION['EmpId']}";
    $result = @mysqli_query($dbc, $query);//Run the query
}
$query = "SELECT EmployeeID AS id, Photo AS photo, CONCAT(FirstName, ' ', LastName) AS name, PhoneNumber AS phone, Email AS email FROM Employees WHERE EmployeeID = '{$_SESSION['EmpId']}'";
$infoResult = @mysqli_query($dbc, $query);//Run the query
if($infoResult){
    while($row = mysqli_fetch_array($infoResult, MYSQLI_ASSOC)){
?>
<div id="profile-pic">
        <?php
            echo "<img src=\"includes/view-image.inc.php?file={$row['photo']}\" alt=\"{$row['photo']}\" />";
        ?>
        <form name="uploadForm" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>?pagelet=Dashboard" method="post" enctype="multipart/form-data">
            <input type="file" name="image">

            <input type="submit" name="uploadImg" value="submit">
        </form>
    </div>
<?php
    }
}
?>

<?php
$path = "/upload/"; //directory outside public_html on the server
$file = ($_GET['file']);
$image = $path.$file;
header("Content-Type: image/jpg");
@readfile($image);
?>