php - 同时更新两个表

Question

有没有办法同时更新两个表？我有一张 table food 和一张 table food_r

这是我插入食物的代码

$rest_id = null;
if ( !empty($_GET['rest_id'])) 
{
    $rest_id = $_REQUEST['rest_id'];
}
if ( null==$rest_id ) 
{
    echo "null==$rest_id";
}
if(isSet($_POST['submit'])) 
{
    // keep track post values
    $food_name = $_POST['food_name'];
    $food_description = $_POST['food_description'];
    $food_menu = $rest_id;
    $usertype = $_SESSION['usertype'];

     // update data
     $pdo = Database::connect();
     $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
     $sql = $pdo->prepare("INSERT INTO food ( food_name, food_description, food_menu, usertype ) 
              VALUES (:food_name, :food_description, :food_menu, :usertype)");

                    $sql->execute(array(
                    ':food_name'    => $food_name,
                    ':food_description' => $food_description,
                    ':food_menu'    => $food_id,
                    ':usertype'     => $_SESSION['usertype']                        
                    ));
                    Database::disconnect();
                    echo "Product added!";
            }

现在，如果我想让插入的产品可见，我必须在表 food_r 中插入 food_menu 的值和 usertype 的值。我怎样才能做到这一点？

更新:它就是这样工作的。感谢@JonathonWisnoski 向我指出交易。。

  $pdo->beginTransaction();
    $sql = $pdo->prepare("INSERT INTO food ( food_name, food_description, food_menu, usertype ) 
                VALUES (:food_name, :food_description, :food_menu, :usertype)");

        $sql->execute(array(
        ':food_name'    => $food_name,
        ':food_description' => $food_description,
        ':food_menu'    => $rest_id,
        ':usertype'     => $_SESSION['usertype']                        
        ));
        $lastInsertID = $pdo->lastInsertId();
    $sql = $pdo->prepare("INSERT INTO food_r (food_id, usertype) 
                    VALUES (:rest_id, :usertype)");
        $sql->execute(array(
        ':rest_id'      => $lastInsertID,
        ':usertype' => $rest_id
        ));
        $pdo->commit();

Answer 1

更新:它就是这样工作的。感谢@JonathonWisnoski 向我指出交易。。

我还为任何错误添加了 try{}catch{} block 。

 try 
 {
   $pdo->beginTransaction();
     $sql = $pdo->prepare("INSERT INTO food ( food_name, food_description, food_menu, usertype ) 
                VALUES (:food_name, :food_description, :food_menu, :usertype)");

        $sql->execute(array(
        ':food_name'    => $food_name,
        ':food_description' => $food_description,
        ':food_menu'    => $rest_id,
        ':usertype'     => $_SESSION['usertype']                        
        ));
        $lastInsertID = $pdo->lastInsertId();
    $sql = $pdo->prepare("INSERT INTO food_r (food_id, usertype) 
                    VALUES (:rest_id, :usertype)");
        $sql->execute(array(
        ':rest_id'      => $lastInsertID,
        ':usertype' => $rest_id
        ));
        $pdo->commit();
   }                
   // any errors from the above database queries will be catched
   catch (PDOException $e)
   {
        // roll back transaction
        $pdo->rollback();
        // log any errors to file
        ExceptionErrorHandler($e);
        exit;
    }