在yii2中,我如何将多个相关记录保存到数据库中,以进行单个保存调用和单个事务。我有两张桌子:
User - id, name
UserAddress - id , user_id , city
用户表与UserAdress表有一对多关系
我想做的是:
UserAddress ua = new UserAddress();
ua.city = "fff"
User u = new User();
u.name = "test";
u.userAddress = new Array(ua);
u.save();
在用户上调用save应该保存user_id设置为user.id的用户和useraddress以及
// You need create hasMany relation 'userAddress' (look guide relations)
$transaction = Yii::$app->db->beginTransaction();
try {
$user = new User();
$user->name = 'Name';
$user->save();
$ua = new UserAddress();
$ua->city = 'City';
$user->link('userAddress', $ua); // <-- it creates new record in UserAddress table with ua.user_id = user.id
$transaction->commit();
} catch (Exception $e) {
$transaction->rollBack();
}
除了前面的答案之外,我还提出了一种变体,它在没有预先定义关系的情况下工作,并且可以显式处理验证错误。
Yii::$app->db->transaction(function(){
$user = new User();
$user->name = 'Name';
if( !$user->save() ){
throw new Exception('Can't be saved user model. Errors: '. join(', ', $user->getFirstErrors()));
}
$userAddress = new UserAddress();
$userAddress->city = 'City';
$userAddress->user_id = $user->id;
if( !$userAddress->save() ){
throw new Exception('Can't be saved user address model. Errors: '. join(', ', $userAddress->getFirstErrors()));
}
});
此代码严格确保两个记录都将被保存。如果其中一个模型无法保存,将引发带有验证错误的异常。