java - 按特定值分组列出数据

Question

我有一个包含帐单 ID 和电子邮件数据的列表。账单 ID 和电子邮件可以在列表中重复。

这是我的列表数据:

List<Bill> billings = new ArrayList<Bill>();

        Bill bill1 = new Bill("90008489", "demo@gmail.com");
        Bill bill2 = new Bill("90008489", "oke@sample.com");
        Bill bill3 = new Bill("90008489", "welcom@gmail.com");
        Bill bill4 = new Bill("90008490", "hore@yahoo.com");
        Bill bill5 = new Bill("90008490", "fix.it@demo.co.id");
        Bill bill6 = new Bill("90008491", "yuhuuu@demo.co.id");

        billings.add(bill1);
        billings.add(bill2);
        billings.add(bill3);
        billings.add(bill4);
        billings.add(bill5);
        billings.add(bill6);

这是我为 Bill 准备的 Java 类:

public class Bill {
    private String id;
    private String email;
    private List<String> emails;

    public Bill(String id, String email) {
        super();
        this.id = id;
        this.email = email;
    }

    public Bill(String id, List<String> emails) {
        super();
        this.id = id;
        this.emails = emails;
    }

... Getter and Setter

我想按帐单 ID 对该列表数据进行分组。如果找到相同的帐单 ID，我想合并电子邮件数据。

我不知道何时构建它。这是我的代码。

List<Bill> newBillings = new ArrayList<Bill>();
        for (int i = 0; i < (billings.size() - 1); i++) {

            List<String> emails = new ArrayList<String>();
            emails.add(billings.get(i).getEmail());

            //System.out.println(billings.get(i+1).getId());

            if (billings.get(i).getId() == billings.get(i + 1).getId()) {
                emails.add(billings.get(i+1).getEmail());


            }
        }

        for (Bill bill : newBillings) {
            System.out.println(bill.getId());
            for (String email : bill.getEmails()) {
                System.out.print(email  + ",");
            }

            System.out.println("\n-----");
        }

我的预期结果是:

90008489 - [demo@gmail.com, oke@sample.com, welcome@gmail.com]
90008490 - [hore@yahoo.com, fix.it@demo.co.id]
90008491 - [yuhuuu@demo.co.id]

Answer 1

我想说你为此使用了错误的数据结构。

我会完全重写它并使用 Map<String, List<String>>因为它会更合适，因为您需要将每个 id 映射到邮件列表。

Map<String, List<String>> map = new HashMap<>(); 
for(Bill b : billings) {
    List<String> list = map.get(b.getId()); //try to get the list with the specified id
    if(list == null) { //if the list is null, it means that there is no mapping yet in the map, so create one
        list = new ArrayList<>();
        map.put(b.getId(), list);
     }
     list.add(b.getEmail()); //add the corresponding email to the list
}

可以翻译成Java 8风格

Map<String, List<String>> map = billings.stream()
                    .collect(groupingBy(Bill::getId, mapping(Bill::getEmail, toList())));