php - 更新 tinyint(1)

Question

我想更新状态值 -tinyint(1)- 以激活和停用用户。每当我尝试更新时，我都会收到下面设置为“助理更新失败”的消息。任何帮助都很感激。谢谢

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}', <=

中的尾随逗号

MySQL 会抛出一个错误:

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

---

• 我还想指出，您目前的密码是开放给 SQL injection 的.
  使用 prepared statements , 或 PDO with prepared statements ，它们更安全。