How can I load database entries into my scala template? - database

Question

I am using play 2.3.8 and building a program where you can ask questions and answer them. I have several different users, so I use their unique email as the ID in the database.
Questions / answers have an ownerID, to show who has written them.
My users are stored in the DB like this:
Email Name Password
bob#mail.com Bob secret
My questions are stored in the DB like this:
QUESTION_ID QUESTION_TEXT VOTE_SCORE OWNER_ID PAGE
77b7f88a-41df-4d68-9f89-de508fce8f71 How tall is tall? 1228 bob#mail.com 1
My controller class, that collects the questions / answers from the DB and sends the lists to the view class:
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));
}
User.java:
// FIXME Dont really save the password as String...
#Id
public String email;
public String name;
public String password;
(...)
Question.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;
(...)
}
In my view class I show the questions and behind them the user by using #answer.ownerID ... but as I use the email-field as ID, the entry is something like:
"How tall is tall?" - bob#mail.com
What I want is:
"How tall is tall?" - Bob Ross (using their name and not their email)
I know I could simply find all questions and their owners in my controller and put the users into another list, from where I would use #users.username, but this would mean I always have to look up all users, put them into the list, ...
So is it possible to look into the DB, use the ownerID and get the username in the view class?
Or more general: Can you look up DB entries for the view class from the view class itself?

Answer 1

You may add mappings with JPA 2.0 to achieve what you want in your view, then you may use #question.owner.name or #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;
[...]