>men der kommer intet input.
skal det forståes sådan at der ikke kommer noget i database ??
kig på dette som inspiration, coden skal sikres før brug i production.
for at gemme i databasen
<?php
//
https://www.youtube.com/watch?v=-tldcpFO6ds//
https://www.webslesson.info/2016/10/insert-and-fetch-images-from-mysql-database-in-php.htmlif(isset($_POST['insert']) ){
require_once "serverconnect.php";
$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
$name = $_FILES['image']['name'];
$query = "INSERT INTO store (name, image) VALUES('$name','$image')";
$qry = mysqli_query($conn, $query);
}
?>
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<br /><br />
<div>
<h3 align="center">Insert and Display Images From Mysql Database in PHP</h3>
<br />
<form method="post" enctype="multipart/form-data" action="">
<input type="file" name="image" id="image" />
<br />
<input type="submit" name="insert" id="insert" value="Insert">
</form>
</div>
</body>
</html>
for at vise indholdet fra databasen
<?php require_once "serverconnect.php"; ?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<table border="1">
<?php
$sql = "SELECT * FROM store ORDER BY id DESC ";
$result = $conn->query($sql);
while($data = $result->fetch_assoc())
{
// header("Content-type: image/jpeg");
echo '<tr><td>'.$data["id"] . '</td><td>'.$data["name"]. '</td><td><img src=data:image/jpeg;base64,'.base64_encode($data["image"] ).'></td></tr>';
}
?>
</table>
</body>
</html>
men du bør overveje at ligge billederne i en seperat folder og ikke i databasen, og i din database bare have et felt hvor du henvisser til deres url.
at ligge billederne som blob får din database til at svumle voldsomt op.
