java - 如何让泛型函数根据Java中的输入获取对象属性

Question

我有如下 Java 实体

@Entity
@Table(name = "user", schema = "xxxx")
public class User extends AuditableBase<Long> {

  /** The Constant serialVersionUID. */
  private static final long serialVersionUID = -5827659402853250569L;

  public static final String FTL = "FTL";

  /** The refer number. */
  @Column(name = "reference_number")
  private String referenceNumber;

  /** The customer id. */
  @OneToOne(fetch = FetchType.LAZY)
  @JoinColumn(name = "customer_id")
  @Audited(targetAuditMode = RelationTargetAuditMode.NOT_AUDITED)
  private Customer customer;

  /** The load Notes. */
  @OneToMany(cascade = CascadeType.ALL, mappedBy = "load")
  @JsonBackReference
  private Set<Notes> notes = new HashSet<>();

  /** The customer location. */
  @OneToOne(fetch = FetchType.LAZY)
  @JoinColumn(name = "customer_location_id")
  @Audited(targetAuditMode = RelationTargetAuditMode.NOT_AUDITED)
  private CustomerLocation customerLocation;

  /** The invoice number. */
  @Column(name = "invoice_number")
  private String invoiceNumber;

  // Lot going

  /**
   * Instantiates a new User.
   */
  public User() {
    super();
  }

  /**
   * Instantiates a new User.
   *
   * @param id
   *          the id
   */
  public User(Long id) {
    super(id);
  }

}

我在实体中有很多子对象，我从 DAO 调用获取对象

User user = userRepository.findById(userId);

现在我想根据引用 Map 的一些 if 条件获取一些用户详细信息。

Map<Integer, String> cc = new HashMap<Integer, String>();
cc.put(1, "getCustomer()");
cc.put(2, "getNotes()");
cc.put(3, "getCustomerLocation()");
cc.put(4, "getReferenceNumber()");


for (Entry<Integer, String> map : cc.entrySet()) {

    user.map.getvalue();

}

我需要创建一个通用方法来根据 map 获取用户对象。

我怎样才能做到这一点。

Answer 1

不确定我完全理解你的问题，但你可以这样做:

        Map<Integer,Method> cc = new HashMap<>();
        cc.put(1, User.class.getDeclaredMethod("getCustomer"));
        cc.put(2, User.class.getDeclaredMethod("getNotes"));
        cc.put(3, User.class.getDeclaredMethod("getCustomerLocation"));
        cc.put(4, User.class.getDeclaredMethod("getReferenceNumber"));

        for (Entry<Integer, Method> map : cc.entrySet()) {
            Integer index = map.getKey();
            Method getter = map.getValue();
            Object value = getter.invoke(user);
            doSomethingUsefulWith(index, value);
        }

更新:

你可以像这样声明一个接口(interface)Getter:

public interface Getter<T,R> {
    public R get(T obj);
}

然后做这样的事情:

    Map<Integer,Getter<User,?>> cc = new HashMap<>();
    cc.put(1, (u) -> u.getCustomer().getName());
    cc.put(2, (u) -> u.getNotes());
    cc.put(3, (u) -> u.getCustomerLocation());
    cc.put(4, (u) -> u.getReferenceNumber());
    for (Entry<Integer,Getter<User,?>> map : cc.entrySet()) {
        Integer index = map.getKey();
        Getter<User,?> getter = map.getValue();
        Object value = getter.get(user);
        doSomethingUsefulWith(index, value);
    }