database - 如何将数据库条目加载到我的 Scala 模板中？

Question

我正在使用 play 2.3.8 并构建一个程序，您可以在其中提出问题并回答问题。我有几个不同的用户，所以我使用他们唯一的电子邮件作为数据库中的 ID。

问题/答案有一个 ownerID，以显示是谁写的。

我的用户是这样存储在数据库中的:

Email           Name Password
bob@mail.com    Bob  secret

我的问题是这样存储在数据库中的:

QUESTION_ID                             QUESTION_TEXT       VOTE_SCORE      OWNER_ID        PAGE  
77b7f88a-41df-4d68-9f89-de508fce8f71    How tall is tall?   1228            bob@mail.com    1

我的 Controller 类，它从数据库中收集问题/答案并将列表发送到 View 类:

public static List<Question> questionListAll = new ArrayList<Question>();
public static List<Answer> answerListAll = new ArrayList<Answer>();

public static Result index() {
        questionListAll.clear();
        answerListAll.clear();

        // Get all questions from DB
        for (Question questionItem : Question.find.all()) {
            questionListAll.add(questionItem);
        }

        // Get all answers from DB
        for (Answer answerItem : Answer.find.all()) {
            answerListAll.add(answerItem);
        }

        Collections.sort(questionListAll, Collections.reverseOrder());
        Collections.sort(answerListAll, Collections.reverseOrder());

        return ok(views.html.index.render(questionListAll, answerListAll));
    }

用户.java:

// FIXME Dont really save the password as String...
@Id
public String email;
public String name;
public String password;
(...)

问题.java:

@Entity
public class Question extends Model implements Comparable<Question> {
    @Id
    public String questionID;
    public String questionText;
    public Integer voteScore;
    public String ownerID;
    public Integer page;
(...)
}

在我的 View 类中，我通过使用 @answer.ownerID 向用户显示问题和问题背后的用户......但是当我使用电子邮件字段作为 ID 时，条目类似于:

“有多高？” - bob@mail.com

我想要的是:

“有多高？” - Bob Ross(使用他们的名字而不是他们的电子邮件)

我知道我可以简单地在我的 Controller 中找到所有问题及其所有者，然后将用户放入另一个列表，从那里我可以使用 @users.username，但这意味着我总是必须查找所有用户，将他们放入列表中，...

那么是否可以查看数据库，使用 ownerID 并在 View 类中获取用户名？

或更笼统:您能否从 View 类本身查找 View 类的数据库条目？

Answer 1

您可以使用 JPA 2.0 添加映射以在您的 View 中实现您想要的，然后您可以使用 @question.owner.name 或 @answer.owner.name .

User.java

[...]
@Id
@Column(name="user_id")
public Long userId;

@OneToMany(fetch = FetchType.LAZY, mappedBy = "owner", cascade = CascadeType.ALL)
public List<Question> questionList;

@OneToMany(fetch = FetchType.LAZY, mappedBy = "owner", cascade = CascadeType.ALL)
public List<Answer> answerList;
[...]

Question.java

[...]
@Id
@Column(name="question_id")
public Long questionId;

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name="user_id", nullable = false)
public User owner;

// @OneToMany: ORM join relationship, one Question to many Answers
// FetchType.LAZY: ORM will not fetch from db until you use it
// CascadeType.ALL: any changes to question will be propagated
//   onQuestionDelete: all answers associated with the question will be deleted
//   onQuestionUpdateAnswerList: any updates made will reflect on db
// mappedBy: Question object variable name on Answer object for ORM to make a connection
@OneToMany(fetch = FetchType.LAZY, mappedBy = "question", cascade = CascadeType.ALL)
public List<Answer> answerList;
[...]

Answer.java

[...]
@Id
@Column(name="answer_id")
public Long answerId;

// FetchType.EAGER: ORM will fetch this from db when Answer is fetch
@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name="user_id", nullable = false)
public User owner;

// @ManyToOne: ORM join relationship, many Question to one Answer
// @JoinColumn: name, pk column name of Question; nullable false, required constraint;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name="question_id", nullable = false)
public Question question;
[...]