1
我想更新狀態值-tinyint(1) - 激活和取消激活用戶。每當我嘗試更新時,我都會收到下面的消息,設置爲「Attendant update failed」。任何幫助都很感激它。由於更新TINYINT(1)
if (empty($errors)) {
// Perform Update
$id = $attendant["id"];
$status = mysql_prep($_POST["status"]);
$query = "UPDATE attendant SET ";
$query .= "status = '{$status}', ";
$query .= "WHERE id = {$id} ";
$query .= "LIMIT 1";
$result = mysqli_query($connection, $query);
if ($result && mysqli_affected_rows($connection) == 1) {
// Success
$_SESSION["message"] = "Attendant updated.";
redirect_to("activate_attendant.php");
} else {
// Failure
$_SESSION["message"] = "Attendant update failed.";
}
}
} else {
// This is probably a GET request
}
刪除''{status}'中的引號' – Rupam
刪除'status ='{$ status}'中的逗號,'<= ---執行'$ result = mysqli_query ($連接,$查詢)或死亡(mysqli_error($連接));' –
@弗雷德-II-我做到了,它的工作原理,只是當我更新的價值,它說:「數據庫查詢失敗」,但它更新。謝謝。你有什麼想法可能是什麼? – Ash23