通过普通表建立关系正确的方式

本文介绍了通过普通表建立关系正确的方式的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我在数据库内有三张表： interest ，订阅和 / code>。模式如下所示：I have three tables inside of a database: interest, subscription and subscriber. The schema looks like this: 订阅表对兴趣和订阅者都有一对多的关系。我想要如何工作，我不知道我想要它如何工作，它的实际工作正在排队，但它应该是这样的：一个订阅者可以拥有多个兴趣，每个兴趣可以有多个订阅者。这将让我看看订阅者的兴趣和哪些兴趣已被订阅。那么，订阅只涉及一个订阅者和一个兴趣。我认为这是如何设置的，但是我刚刚做了一些测试来查询订阅表，我回到了这里：The subscription table has a one to many relationship on both interest and subscriber. How I would like it to work, and I'm not sure if how I want it to work and how it's actually working are lining up, but it should be like this: a Subscriber can have multiple Interests, and each Interest can have multiple Subscribers. This will allow me to see which interests a subscriber has, and which interests have been subscribed to. Then, the Subscription relates only to one subscriber and one interest. I think that's how it's set up, but I just did some tests querying the subscription table and I got this back:C:\xampp\htdocs\www\public_html\playground\subscribr>php [email protected]@tester.com--Newsletter-testemail@[email protected]@[e-mail]--Promotions-testemail3@[email protected]@tester.com--Magazine-testemail4@[email protected]@[e-mail][e-mail]--Promotions-testemail9@[email protected] 我想我预期结果更像这样：I suppose I expected a result more like this:C:\xampp\htdocs\www\public_html\playground\subscribr>php [e-mail][e-mail][e-mail][e-mail][e-mail][e-mail][e-mail]--Promotions-testemail9@[email protected]如果我通过订阅者查询并获取他们的兴趣，我会得到这个结果（通过订阅表）。我正在使用ORM，所以要执行的代码如下所示：I do get this result if I query by subscriber, and grab their interests (through the subscription table). I'm using ORM, so the code to do that looks like this:$subscriberRepo = $entityManager->getRepository( 'Subscribr\Entity\Subscriber' );$subscribers = $subscriberRepo->findAll();foreach ($subscribers as $subscriber) { echo sprintf( "-%s\n", $subscriber->getEmail() ); foreach ($subscriber->getInterests() as $interest) { echo sprintf( "--%s\n", $interest->getInterest()->getName() ); }}所以我可以得到订阅者的订阅和兴趣被订阅，使用这个模式，但是有没有保留订阅表的一点？还是整个事情需要重做，做我想要做的事情？我喜欢在用户和兴趣之间建立间接关系，我希望在订阅表中添加额外的列，例如从订阅表中删除 is_subscribed ，并在订阅中添加列 subscription_status ，以这种方式，这个解决方案感觉最干净。然而，它也几乎像一个额外的字段的连接表。想法？So I can get a subscriber's subscriptions, and interests that are subscribed to, using this schema, but then is there a point in keeping the subscription table around? Or does the whole thing need to be reworked to do what I'd like it to do? I like the having an indirect relationship between subscriber and interests, incase I wanted to add extra columns to the subscription table like removing is_subscribed from the subscriber table and adding a column in subscription called subscription_status, in that way this solution feels the cleanest. However, it also almost feels like a join table with extra fields. Thoughts?推荐答案根据您的解释，您可以正确地了解需要映射的内容。简单的方法有两种：Based on your explanation a could understand properly what you need to map. In a simplified way you have two options: 选项1 订阅者 1 - > n 订阅 订阅 n 兴趣 Subscriber 1 -> n SubscriptionSubscription n <- 1 Interest这意味着间接：订阅者 n - > n 兴趣 看不到？让代码看起来更好：It means indirectly: Subscriber n -> n InterestDon't see? Lets code to see better:foreach ($subscribers as $subscriber) { echo sprintf( "-%s\n", $subscriber->getEmail() ); foreach ($subscriber->getSubscriptions() as $subscription) { echo sprintf( "--%s\n", $subscription->getInterest()->getName() ); }} 编辑 上面的代码将与您的映射中的以下更改一起使用。此代码说明：迭代订阅者，其中每个订阅者都有订阅列表，每个订阅都有兴趣。The above code will work with the below changes in your mapping. This code explanation: Iterate subscribers where each subscriber has a list of subscriptions and each subscription referes on interest. Subscriber.php /** * @var array * * @ORM\OneToMany(targetEntity="Subscription", mappedBy="subscriber", cascade={"persist", "remove"}, orphanRemoval=TRUE) */ private $subscriptions;您应该删除 private $兴趣; 订户。它不会直接访问兴趣。You should remove private $interests; from Subscriber. It won't access Interest directly. Subscription.php /** * @var Interest * * @ORM\ManyToOne(targetEntity="Interest", inversedBy="subscribers") * @ORM\JoinColumn(name="interest_id", referencedColumnName="id", nullable=FALSE) */private $interest;注意每个订阅将访问一个兴趣。这个第一个选项是你最好的。This first option is the best in your case. 选项2 您可以删除订阅类并使用其表作为参考表。并直接映射：订阅者 n - > n 兴趣。You could remove Subscription class and use its table as a reference table. And map directly: Subscriber n -> n Interest.class Subscriber { /** * @var array * * @ORM\ManyToMany(targetEntity="Interest", cascade={"persist", "remove"}, orphanRemoval=TRUE) * @ORM\JoinTable(name="subscription", * joinColumns={@ORM\JoinColumn(name="subscriber_id", referencedColumnName="id")}, * inverseJoinColumns={@ORM\JoinColumn(name="interest_id", referencedColumnName="id")} * ) */ private $interests;}此选项中的问题不能使用订阅。 interest_date 'cause subscription table现在只是一个关系表。关于这个问题。The issue in this option you cannot use subscription.interest_date 'cause subscription table is now only a relation table. More explanation about that in this question. 这篇关于通过普通表建立关系正确的方式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！