在数据库中插入图像文件，然后用php显示

// 图像文件上传处理
$uploadDir = 'path/to/upload/directory/';
$uploadedFile = $uploadDir . $_FILES['image']['name'];
move_uploaded_file($_FILES['image']['tmp_name'], $uploadedFile);

// 数据库插入
$dbHost = 'localhost';
$dbUser = 'username';
$dbPass = 'password';
$dbName = 'database_name';

$conn = new mysqli($dbHost, $dbUser, $dbPass, $dbName);
if ($conn->connect_error) {
    die('数据库连接失败：' . $conn->connect_error);
}

$imageData = file_get_contents($uploadedFile);
$imageName = $_FILES['image']['name'];

$sql = "INSERT INTO images (image_data, image_name) VALUES (?, ?)";
$stmt = $conn->prepare($sql);
$stmt->bind_param("ss", $imageData, $imageName);
$stmt->execute();
$stmt->close();

// 图像文件显示
$sql = "SELECT image_data, image_name FROM images WHERE image_name = ?";
$stmt = $conn->prepare($sql);
$stmt->bind_param("s", $imageName);
$stmt->execute();
$stmt->bind_result($imageData, $imageName);
$stmt->fetch();
$stmt->close();

header("Content-Type: image/jpeg"); // 根据实际图像类型设置正确的Content-Type
echo $imageData;

$conn->close();