JAVA:如何根据属性值合并 arrayList 中的对象？

Question

我正在使用 Spring Boot、Java 和 MySQL 构建(或学习如何)体育 REST API。我正在构建一个方法，该方法当前从比赛集合中获取每场比赛，并返回 TeamStandings 的 ArrayList 以获取完整的比赛列表。

方法如下:

public List<TeamStanding> createStandingsTable(Match[] matches){
        List<TeamStanding> teamStandings = new ArrayList<TeamStanding>();
        for(int i = 0;i < matches.length; i++) {
            TeamStanding firstTeam = new TeamStanding();
            TeamStanding secondTeam = new TeamStanding();

            //set team ids
            firstTeam.setIdTeam(matches[i].getWcmHome());
            secondTeam.setIdTeam(matches[i].getWcmAway());


            //first team stats
            firstTeam.setTeamPlayed((long) 1);
            firstTeam.setTeamGoalsFavor(matches[i].getWcmHomeGoals());
            firstTeam.setTeamGoalsAgainst(matches[i].getWcmAwayGoals());
            firstTeam.setTeamGoalDif(firstTeam.getTeamGoalsFavor() - firstTeam.getTeamGoalsAgainst());

            //second team stats
            secondTeam.setTeamPlayed((long) 1);
            secondTeam.setTeamGoalsFavor(matches[i].getWcmAwayGoals());
            secondTeam.setTeamGoalsAgainst(matches[i].getWcmHomeGoals());
            secondTeam.setTeamGoalDif(secondTeam.getTeamGoalsFavor() - secondTeam.getTeamGoalsAgainst());

            //combined team stats

            if(firstTeam.getTeamGoalsFavor() > secondTeam.getTeamGoalsFavor()) {
                firstTeam.setTeamWins((long) 1);
                firstTeam.setTeamLoses((long) 0);
                firstTeam.setTeamDraws((long) 0);
                firstTeam.setTeamPoints((long) 3);
                secondTeam.setTeamWins((long) 0);
                secondTeam.setTeamLoses((long) 1);
                secondTeam.setTeamDraws((long) 0);
                secondTeam.setTeamPoints((long) 0);
            } else if (firstTeam.getTeamGoalsFavor() == secondTeam.getTeamGoalsFavor()) {
                firstTeam.setTeamWins((long) 0);
                firstTeam.setTeamLoses((long) 0);
                firstTeam.setTeamDraws((long) 1);
                firstTeam.setTeamPoints((long) 1);
                secondTeam.setTeamWins((long) 0);
                secondTeam.setTeamLoses((long) 0);
                secondTeam.setTeamDraws((long) 1);
                secondTeam.setTeamPoints((long) 1);
            } else {
                firstTeam.setTeamWins((long) 0);
                firstTeam.setTeamLoses((long) 1);
                firstTeam.setTeamDraws((long) 0);
                firstTeam.setTeamPoints((long) 0);
                secondTeam.setTeamWins((long) 1);
                secondTeam.setTeamLoses((long) 0);
                secondTeam.setTeamDraws((long) 0);
                secondTeam.setTeamPoints((long) 3);
            }
            teamStandings.add(firstTeam);
            teamStandings.add(secondTeam);
        }
        return teamStandings;
    }

结果是这样的:

[
    {
        "idTeam": 7,
        "teamPoints": 3,
        "teamPlayed": 1,
        "teamWins": 1,
        "teamDraws": 0,
        "teamLoses": 0,
        "teamGoalsFavor": 4,
        "teamGoalsAgainst": 1,
        "teamGoalDif": 3
    },
    {
        "idTeam": 13,
        "teamPoints": 0,
        "teamPlayed": 1,
        "teamWins": 0,
        "teamDraws": 0,
        "teamLoses": 1,
        "teamGoalsFavor": 1,
        "teamGoalsAgainst": 4,
        "teamGoalDif": -3
    },
    {
        "idTeam": 4,
        "teamPoints": 3,
        "teamPlayed": 1,
        "teamWins": 1,
        "teamDraws": 0,
        "teamLoses": 0,
        "teamGoalsFavor": 1,
        "teamGoalsAgainst": 0,
        "teamGoalDif": 1
    },
    {
        "idTeam": 7,
        "teamPoints": 0,
        "teamPlayed": 1,
        "teamWins": 0,
        "teamDraws": 0,
        "teamLoses": 1,
        "teamGoalsFavor": 0,
        "teamGoalsAgainst": 1,
        "teamGoalDif": -1
    }
]

我的问题是如何基于idTeam合并这些对象？我试图实现的结果是在 idTeam 保持不变的情况下将所有其余属性相加。在给定的示例中，预期的结果是:

[
        {
            "idTeam": 7,
            "teamPoints": 3,
            "teamPlayed": 2,
            "teamWins": 1,
            "teamDraws": 0,
            "teamLoses": 1,
            "teamGoalsFavor": 4,
            "teamGoalsAgainst": 2,
            "teamGoalDif": 2
        },
        {
            "idTeam": 13,
            "teamPoints": 0,
            "teamPlayed": 1,
            "teamWins": 0,
            "teamDraws": 0,
            "teamLoses": 1,
            "teamGoalsFavor": 1,
            "teamGoalsAgainst": 4,
            "teamGoalDif": -3
        },
        {
            "idTeam": 4,
            "teamPoints": 3,
            "teamPlayed": 1,
            "teamWins": 1,
            "teamDraws": 0,
            "teamLoses": 0,
            "teamGoalsFavor": 1,
            "teamGoalsAgainst": 0,
            "teamGoalDif": 1
        }
    ]

还有一个细节，我首先构建了 TeamStandings 的 ArrayList，现在我正在尝试合并它们，但也许我应该将它们堆叠为一个循环匹配数组，在上面相同的方法中，但我不确定。

Answer 1

迭代 TeamStanding 的列表，记下团队 ID 并执行添加。您可能希望使用 Map 将一对团队 ID 保存为键，将团队本身保存为值，以便于操作。这是截图(我还没有测试过，所以你可能需要稍微修改一下)。

List<TeamStanding> list = createStandingsTable(matches);
Map<Integer, TeamStanding> map = new HashMap<>();

for (TeamStanding team: list) {
    int id = team.getIdTeam();
    if (map.containsKey(id)) {
        TeamStanding other = map.get(id);
        other.setTeamPoints(team.getTeamPoints());
        other.setTeamPlayed(team.getTeamPlayed());
        // and so on...
    } else {
        map.put(id, team);
    }
}

List<TeamStanding> merged = new ArrayList<>(map.values());

如果您想创建合并的List<TeamStanding>直接来自Match[] ，那么你必须使用相同的想法，但是，将两个迭代组合在一起可能有点复杂。然后我建议您坚持这两个单独的迭代。简洁性、可读性和可维护性高于性能 - 而且，性能在这里并不是真正的问题。