在JPA中将实体列表保存在具有ManyToOne关系的数据库中

Z时代
2024-01-10
分类：问答

我有2类：Game和User

User 有游戏列表（ManyToOne关系）。

如何向已保存在数据库中的用户添加其他游戏。

这是我的方法：

@RequestMapping(value = "/game/play", method = RequestMethod.POST)
@ResponseBody
public User indexRequestPlay(@RequestParam String username, @RequestParam String password) {
    User user = userRepository.findByUsernameAndPassword(username, password);
    Random random = new Random();
    int userScore = random.nextInt(5) + 1;
    int npcScore = random.nextInt(5) + 1;
    Date date = new Date();
    List<Date> startSessions = user.getStartSessions();
    startSessions.add(date);
    user.setStartSessions(startSessions);
    Game game = new Game(userScore, npcScore, date);
    List<Game> games = new ArrayList<Game>();
    games.add(game);
    games.addAll(user.getGames());
    user.setGames(games);
    userRepository.save(user);
    return user;
}

回答：

答案是：

@RequestMapping(value = "/game/play", method = RequestMethod.POST)
@ResponseBody
public User indexRequestPlay(@RequestParam String username, @RequestParam String password) {
    User user = userRepository.findByUsernameAndPassword(username, password);
    Random random = new Random();
    int userScore = random.nextInt(5) + 1;
    int npcScore = random.nextInt(5) + 1;
    /////////////////////////////////////////////////////////////////////////////////
    // aici adaug un camp in user.
    user.getStartSessions().add(new Date());
    /////////////////////////////////////////////////////////////////////////////////
    Game game = new Game(userScore, npcScore, new Date());
    game.setUser(user);
    user.getGames().add(game);
    userRepository.save(user);
    return user;
}