显示来自数据库的图像，当我尝试访问图像时给出 about:blank#blocked

Question

我正在尝试从数据库中显示图像，但我不知道如何显示它，这就是我上传图像的方式

//image upload + validation
    $file          = $_FILES['image'];
    $file_name     = $_FILES['image']['name'];//file name
    $file_location = $_FILES['image']['tmp_name']; //temporary location
    $file_size     = $_FILES['image']['size'];// size
    $file_error    = $_FILES['image']['error'];// error 0 if no error or 1 if there is an error

    // $file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));

    $temp_extension = explode('.',$file_name);//explode from . file extension (here we have file name and extension)
    $file_extension = strtolower(end($temp_extension)); // extension name (ex: .jpg)
    $allowed_extensions = array('jpg', 'jpeg', 'png', 'pdf');

    if (in_array($file_extension, $allowed_extensions)) {
        if ($file_error === 0) {
            if ($file_size < 31457280) { //31457280b(bytes) in 30mb 
                $new_file_name = uniqid('',true).".".$file_extension;
                var_dump($new_file_name);
                $file_destination = dirname(__FILE__, 2)."/images/".$new_file_name;
                move_uploaded_file($file_location, $file_destination);
            }else {
                echo "Sorry your file size it's too big!";
            }
        }else {
            echo "Sorry, there was an error, try again";
        }
    }else {
        echo "Sorry, your file type is not accepted";
    }

    $sql = "SELECT * FROM `product` WHERE `id` = '{$_POST['id']}'";
    $result = mysqli_query(get_connection(), $sql);
    $row = $result->fetch_assoc();

    if($error_message == ""){
    
        if(is_array($row)){

            $sql = "UPDATE `product` SET `category_id` = '$category_id', `name` = '$name', `description` = '$description', `price` = '$price', `quantity` = '$quantity', `type` = '$type', `image` = '$file', `modified_on` = NOW(), `color_id` = '$color_id', `session_id` = '$session_id' WHERE `id` = '{$_POST['id']}'";
            $result = mysqli_query(get_connection(), $sql);
            // print_r($sql);
            // var_dump($sql);
        }else{

            $sql = "INSERT INTO `product` (`category_id`, `name`, `description`, `price`, `quantity`, `type`, `image`, `added_on`, `color_id`, `session_id`) VALUES ('$category_id', '$name', '$description', '$price', '$quantity', '$type', '$new_file_name', NOW(), '$color_id', '$session_id')";
            $result = mysqli_query(get_connection(), $sql);
            // var_dump($sql);
            // var_dump($sql);
        }//end elseif
        header("Location: admin.php?page=product_list");
    }//end if

这就是我尝试从数据库中显示图像的方式

<!-- TABLE BODY-->
        <tbody>
            <?php 
                foreach ($result as $row) {
                    
                    $sql_category = "SELECT * FROM `category` WHERE `id` = '{$row['category_id']}'";
                    $result_category = mysqli_query(get_connection(), $sql_category);
                    $row_category = $result_category->fetch_assoc();

                    $sql_color = "SELECT * FROM `color` WHERE `id` = '{$row['color_id']}'";
                    $result_color = mysqli_query(get_connection(), $sql_color);
                    $row_color = $result_color->fetch_assoc();
            ?>
            <tr>

                
                <td><p><?=$nr++?></p></td>
                <td><p><?=$row_category['name']?></p></td>
                <td><p><?=$row['name']?></p></td>
                <td><p><?=$row['description']?></p></td>
                <td><p><?=$row['price']?></p></td>
                <td><p><?=$row['quantity']?></p></td>
                <td><p><?=$row['type']?></p></td>
                <?php var_dump(dirname(__FILE__, 2)."/images/".$row['image']); ?>
                <td><p><img src="<?=dirname(__FILE__, 2)."/images/"?><?=$row['image'];?>" alt="" style="width: 100px; height: 100px;"></p></td>
                <td><p><?=$row['added_on']?></p></td>
                <td><p><?=$row['modified_on']?></p></td>
                <td><p><?=$row_color['name']?></p></td>
                <td>
                <button type="button" class="btn btn-outline-primary"><a href="admin.php?page=product_add_edit&id=<?=$row['id']?>">Edit</a></button>
                &nbsp;<button type="button" class="btn btn-outline-primary"><a href="admin.php?page=product_list&action=delete&id=<?=$row['id']?>" onclick="return confirm('Are you sure ?')">Delete</a></button>
                &nbsp;<button type="button" class="btn btn-outline-primary"><a href="admin.php?page=cart&action=add_to_cart&id=<?=$row['id']?>">Add to cart</a></button>
                </td>
            </tr>
            <?php }//end foreach ?>
        </tbody>
        <!-- END OF THE TABLE BODY -->
    </table>

Var_dump 显示 C:\\wamp64\\www\\Project6Bootstrap4-V1\\modules\\product_list.php:101:string 'C:\\wamp64\\www\\Project6Bootstrap4-V1/images/6103fb6fe3da77.64812338.jpg'
如果我将该地址粘贴到浏览器中，它会显示该图像，但是如果我尝试复制图像地址，它会给我关于：blank#blocked
同样在数据库图像类型是 varchar(200)

Answer 1

这不是有效的 HTTP URL：

C:\wamp64\www\Project6Bootstrap4-V1/images/6103fb6fe3da77.64812338.jpg

您正试图让网页访问本地文件，但浏览器出于安全原因阻止了该文件。

去掉dirname(__FILE__, 2)部分，只使用与正在查看的页面相关的图像的 URL：

/images/6103fb6fe3da77.64812338.jpg

请注意，我不知道这是否是图像的正确URL。 你必须确定那是什么。 但据推测，您要显示的图像是您网站的一部分，您只需要使用该图像的 URL。