我没有找到适合我的stackoverflow的答案,所以当我使用PHP将数据保存到mySQL时,我添加了2个参数。保存imgfile参数时,只保存参数名称。
参数:
$stmt->bindParam(':imgfile', $newfilename);
更新后在数据库中:
---------------------------
ID imgfile
---------------------------
1 :imgfile
---------------------------
期待看到的内容:
---------------------------
ID imgfile
---------------------------
1 45678_profile.png
---------------------------
我的整个代码(我将id设置为1,存在的用户,如果PHP无法获取ID):
if (move_uploaded_file($_FILES["file"]["tmp_name"], "/home1/bubos/public_html/homework/images/pfimages/" . $newfilename)) {
$stmt = $conn->prepare("UPDATE `users` SET `profileimg` = ':imgfile' WHERE `users`.`id` = 1");
$stmt->bindParam(':id', $_SESSION['user_id']);
$stmt->bindParam(':imgfile', $newfilename );
if ( $stmt = $conn->prepare("UPDATE `users` SET `profileimg` = ':imgfile' WHERE `users`.`id` = 1")) {
$stmt->execute();
$message = "Image was uploaded to server and set successfully!";
} else {
$message = "Image uploaded to server, but error occurred on image setting!";
}
} else {
$message = "Error on image uploading to server!";
}
有谁能说我该如何解决?