Someone help with better explanations on code below.
I want to test uploading picture in database using the script below. The code is not showing success or error and I'm unable to figure out what to do.
[code]
<html>
<head>
<?php
$servername = "server";
$username = "username";
$password = "passme";
$dbname = "mydb";
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST["submit"])) {
$name = $_FILES["fileupload"]["name"];
$username = $_POST["username"];
$text = $_POST["textdata"];
$target_dir = "upload/";
$target_file = $target_dir . basename($_FILES["fileupload"]["temp_name"]);
// Select file type
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Valid file extensions
$extensions_arr = array("jpg","jpeg","png","gif");
// Check extension
if( in_array($imageFileType,$extensions_arr) ){
// Upload file
if(move_uploaded_file($_FILES['fileupload']["name"],$target_dir.$name)){
// Convert to base64
$image_base64 = base64_encode(file_get_contents('upload/'.$name) );
$image = 'data:image/'.$imageFileType.';base64,'.$image_base64;
// prepare and bind
$stmt = $conn->prepare("INSERT INTO messagebox (image, username, reg_date, text) VALUES (?, ?, ?, ?)");
$stmt->bind_param("ssss", $image, $username, NOW(), $text);
$stmt->execute();
echo "<>Success!";
$stmt->close();
$conn->close();
}
}
}
?>
</body>
</html>
[/code]