Java：通過一個屬性值過濾不同對象的2 collections

Question

給定包含 2 種不同對象的 2 個列表（就像您從 API 獲取一個集合以創建客戶端或更新（如果存在））：

public static void main(String[] args) {
        List<ClientA> clientsA = new ArrayList<>();
        List<ClientB> clientsB = new ArrayList<>();
        for (int i = 1; i <=5; i++) {
            clientsA.add(new ClientA("JohnA-" + i, "DoeA-" + i, "A-" + i));
            clientsB.add(new ClientB("JohnB-" + i, "DoeB-" + i, "B-" + i));
        }
    }

    @Getter
    @Setter
    @AllArgsConstructor
    static class ClientA {
        private String firstName;
        private String lastName;
        private String ssNumber;

    }

    @Getter
    @Setter
    @AllArgsConstructor
    static class ClientB {
        private String firstName;
        private String lastName;
        private String security;
    }

目的是建立一個新的ClientA對象列表：

• 如果clientsA列表中有一個條目，其ssNumber值等於ClientB列表中客戶端的security值，則更新找到的條目firstName和lastName屬性；
• 否則，創建一個具有相同屬性/值的新ClientA clientsB ，分配firstName -> firstName ， lastName -> lastName ， securityNumber -> ssNumber 。

我打算使用contains或retainAll方法，但它需要覆蓋上述類的equals和hashCode ，這是我做不到的。

我希望有這樣的東西：

public void process() {
        List<ClientA> clientsA = new ArrayList<>();
        List<ClientB> clientsB = new ArrayList<>();
        for (int i = 1; i <=5; i++) {
            clientsA.add(new ClientA("John-" + i, "Doe-" + i, "A-" + i));
            clientsB.add(new ClientB("JohnB-" + i, "DoeB-" + i, "B-" + i));
        }

        clientsA.add(new ClientA("Samantha", "Smith", "123456789"));
        clientsB.add(new ClientB("Michael", "Smith", "123456789"));

        findExistingEClientsA(clientsA, clientsB);
        findNewClientsB(clientsA, clientsB);
    }

    private void findNewClientsB(List<ClientA> clientsA, List<ClientB> clientsB) {
        Set resultSet = new HashSet();
        for (ClientA clientA : clientsA) {
            List<ClientB> collect = clientsB.stream().filter(c -> !c.getSecurity().equals(clientA.getSsNumber())).collect(Collectors.toList());
            resultSet.addAll(collect);
        }
        System.out.println("+++++++ New clients B +++++++");
        System.out.println(resultSet);
    }

    private void findExistingEClientsA(List<ClientA> clientsA, List<ClientB> clientsB) {
        Set resultSet = new HashSet();
        for (ClientA clientA : clientsA) {
            List<ClientB> collect = clientsB.stream().filter(c -> c.getSecurity().equals(clientA.getSsNumber())).collect(Collectors.toList());
            resultSet.addAll(collect);
        }

        System.out.println("++++++ existing clients B +++++++ ");
        System.out.println(resultSet);
    }        

什么返回以下結果：

++++++ existing clients B +++++++ 
[ClientB{firstName='Michael', lastName='Smith', security='123456789'}]
+++++++ New clients B +++++++
[ClientB{firstName='JohnB-4', lastName='DoeB-4', security='B-4'}, ClientB{firstName='JohnB-2', lastName='DoeB-2', security='B-2'}, ClientB{firstName='JohnB-5', lastName='DoeB-5', security='B-5'}, ClientB{firstName='JohnB-3', lastName='DoeB-3', security='B-3'}, ClientB{firstName='JohnB-1', lastName='DoeB-1', security='B-1'}, ClientB{firstName='Michael', lastName='Smith', security='123456789'}]

這是一個好的解決方案還是有更好的解決方案？

但仍然沒有成功。

Answer 1

似乎對於這兩種情況，都應從ClientB實例創建ClientA的實例並將其收集到新列表中，因此，應將以下構造函數添加到ClientA ：

static class ClientA {
    public ClientA(ClientB b) {
        this(b.getFirstName(), b.getLastName(), b.getSecurity());
    }
}

所以轉換將是一個簡單的重新映射：

List<ClientA> newA = clientsB.stream().map(ClientA::new).collect(Collectors.toList());

---

如果“更新”意味着需要將變更提升到clientsA列表，可以先建一個ssNumber的ClientA的ClientA ：

Map<String, ClientA> map = clientsA.stream()
        .collect(Collectors.toMap(
            ClientA::getSsNumber, clientA -> clientA, 
            (c1, c2) -> c1  // select first clientB if duplicate entries are detected
        ));

然后可以像這樣創建一個新列表：

List<ClientA> newA = new ArrayList<>();

clientsB.forEach(b -> {
    ClientA a = map.getOrDefault(b.getSecurity(), new ClientA(b));
    a.setFirstName(b.getFirstName());
    a.setLastName(b.getLastName());
    newA.add(a);
});

或者應該實現一個輔助方法（可能添加到ClientA ）以從ClientB復制值：

public static ClientA copyToA(ClientA a, ClientB b) {
    Objects.requireNonNull(a);
    Objects.requireNonNull(b);
    a.setFirstName(b.getFirstName());
    a.setLastName(b.getLastName());
    return a;
}

那么新的ClientA列表可能會以更streamed化的方式構建：

List<ClientA> newClientsA = clientsB.stream()
        .map(b -> map.containsKey(b.getSecurity()) 
            ? copyToA(map.get(b.getSecurity()), b)
            : new ClientA(b)
        )
        .collect(Collectors.toList());