為什么會出現此錯誤？：“字段列表”中的未知列“名字”

Question

  if(isset($_POST["submit"]))
  {

    // Details for inserting into the database

        $id = htmlentities($_POST["id"]);
        $firstname = htmlspecialchars($_POST["firstname"]);
        $lastname = htmlspecialchars($_POST["lastname"]);
        $username = htmlspecialchars($_POST["username"]);
        $password = htmlspecialchars($_POST["password"]);

    // Dealing with inserting 

        $query = "INSERT INTO `myDatabaseForAll`.`users` (`id`, `firstname`, `lastname`, `username`, `password`)
                 VALUES (NULL, $firstname, $lastname,$username,$password)";
        $result = mysqli_query($connection,$query);


        if(!$result) {
            die('There were some errors '.mysqli_error($connection));
        } else {
            redirect_to('index.php');
        }

}

Answer 1

$query = "INSERT INTO `myDatabaseForAll`.`users` (`id`, `firstname`, `lastname`, `username`, `password`)
         VALUES (NULL, $firstname, $lastname,$username,$password)";

您需要在SQL查詢中的文本字段周圍加上單引號

將以上查詢更改為

$query = "INSERT INTO `myDatabaseForAll`.`users` (`id`, `firstname`, `lastname`, `username`, `password`)
         VALUES (NULL, '$firstname', '$lastname','$username','$password')";

Answer 2

由於表列名稱與您在查詢中給定的名稱不匹配，可能會發生此錯誤。 如果不是這種情況，請嘗試以下代碼。

$query = "INSERT INTO `myDatabaseForAll`.`users` (id, firstname, lastname, username, password)
         VALUES (NULL, '".$firstname."', '".$lastname."','".$username."','".$password."')";

在查詢中，您將一些字符串部分與php變量連接在一起。 因此，應使用連接運算符以上述方式完成此連接。