更新tinyint（1）

Question

我想更新状态值-tinyint（1） - 激活和停用用户。每当我尝试更新时，我都会收到下面的消息，该消息设置为“Attendant update failed”。任何帮助都很感激。感谢

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

}

Answer 1

删除status = '{$status}',＆lt; =

中的尾随逗号

MySQL会通过执行以下操作向您抛出错误：

$result = mysqli_query($connection, $query) or die(mysqli_error($connection));

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• 我还要注意，您目前的代码对SQL injection开放 使用prepared statements或PDO with prepared statements，他们更安全。