php-將記錄插入數據庫

Question

我對使用 PHP 插入到我的數據庫中的以下代碼感到有些困惑。 幾天前我使用了這段代碼，它運行良好，我現在嘗試了它，但它不起作用。

這是PHP代碼：

<?php


session_start();
 require_once('dbconn.php');
$user = $_SESSION['who'];
if (!$_SESSION['who'])
{
    header("location: login.php");
}


if (isset($_POST['submit']))
{ 
    if (isset($_POST['flights'])){
         $flightID = $_POST["flights"];
    }
    date_default_timezone_set('Australia/Sydney');
    $date = date("Y-m-d h:i:s");

    if (isset($_POST['baggage']))
    {
         $baggage = $_POST['baggage'];
    }

    $sql = "INSERT into booking (flight_id,customer_id,booking_datetime,baggage)
            values ('$flightID','$user','$date','$baggage')";
    $result = mysqli_query($dbConn, $sql) or trigger_error("Query Failed! SQL: $sql - Error: ".mysqli_error($dbConn), E_USER_ERROR);

    }

?>

調試代碼后，我的 flightID 變量似乎不包含任何內容。 這是錯誤

“查詢失敗！SQL：插入預訂（flight_id，customer_id，booking_datetime，baggage）值（''，'10'，'2020-01-28 01:28:13','121'） - 錯誤：不正確的整數值: '' 列 'flight_id'"

這是我的 FlightID 字段的 html

 <div>
     <label for="flights"> choose a flight:</label> <br>
     <select name = "flights ">
        <option id = "flights" name = "flights" value = "1"> QF1769: cape york -> darwin </option>
        <option id = "flights" name = "flights" value = "2"> QF1988: cape york -> QLd </option>
        <option id = "flights" name = "flights" value = "3"> QF1654: cape york -> melbourne </option>
      </select>

      <span class="error" id="flightRequired">flight required</span>

 </div>

Answer 1

 <div>
  <label for="flights"> choose a flight:</label> <br>
  <select name="flights">
    <option id="flights1" value="1"> QF1769: cape york -> darwin </option>
    <option id="flights2" value="2"> QF1988: cape york -> QLd </option>
    <option id="flights3" value="3"> QF1654: cape york -> melbourne 
    </option>
  </select>

  <span class="error" id="flightRequired">flight required</span>

 </div>