Question

如果我的交易类别为：

public class Transaction {
    @Id
    @Column(name = "id")
    private Long id;
    
    private String customerName;
    
    private String phoneNumber;
    
    //...getters and setters
    
    }

然后我通过Spring存储库按customerName查找交易：

    List<Transaction> findByCustomerName(String customerName);

然后我想使用以下代码将“事务列表”转换为地图，而不是特定值，有没有办法使地图的键成为对象属性？

Map<Long, Transaction> transactionMap = transactionList.stream()
         .collect(Collectors.toMap(Transaction::getId, Function.identity()));

因此，我只希望它成为字符串属性“ Id”，而不是该Transaction对象的特定ID，但我希望它成为列表中的任何属性。。因此，Transaction的下一个属性将是customerName，因此它不会显示“ Id”，而应显示“ customerName”

Answer 1

我不确定您想要的显示如下，我假设类Transaction中有一个默认构造函数及其字段。

代码段

Transaction t1 = new Transaction(10L, "AAA", "0987654321");
Transaction t2 = new Transaction(20L, "BBB", "0901234567");

List<Transaction> transactionList = new ArrayList<>();
transactionList.add(t1);
transactionList.add(t2);

// Use id as the key
Map<Long, Transaction> transactionMap1 = transactionList.stream()
         .collect(Collectors.toMap(Transaction::getId, Function.identity()));
System.out.println(transactionMap1.toString());

// Use customerName as the key
Map<String, Transaction> transactionMap2 = transactionList.stream()
         .collect(Collectors.toMap(Transaction::getCustomerName, Function.identity()));
System.out.println(transactionMap2.toString());

控制台输出

{20 =交易[id = 20，customerName = BBB，电话号码= 0901234567]，10 =交易[id = 10，customerName = AAA，电话号码= 0987654321]}
{AAA =交易[id = 10，customerName = AAA，电话号码= 0987654321]，BBB =交易[id = 20，customerName = BBB，电话号码= 0901234567]}

如何将List转换为Map并使Map键成为对象的属性

1 个答案: