我有一个小问题,从数据库表中获取图像,也许有人可以帮助我。我有一个名为:
的数据库 -RESPONSES with next rows->
id,
user_avatar,
first_name,
last_name,
msg_user,
user_date,
AND
if (isset($_POST['r_sub'])) {
$image = $_FILES['r_image'];
$user_name = $_POST['r_username'];
$last_u_name = $_POST['r_lastname'];
$user_msg = $_POST['r_usermsg'];
$u_date = $_POST['r_date'];
$q_r = "INSERT INTO responses (user_avatar, first_name, last_name, msg_user, user_date) VALUES (:image, :user_name, :last_u_name, :user_msg, :u_date);";
$query_r = $pdo->prepare($q_r);
$results = $query_r->execute(array(
":image" => $image,
":user_name" => $user_name,
":last_u_name" => $last_u_name,
":user_msg" => $user_msg,
":u_date" => $u_date));
header("Location: index.php");
我想知道如何从表中获得BLOB(在我的情况下:user_avatar
);也许有人有现成的解决方案?我真的需要一个代码示例!
try this
echo '<img src="data:image/jpeg;base64,'.base64_encode($image->load()) .'" />';
确保
$image->load()
是从数据库中选择的图像