我有如下的 Java 實體@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); }}我在實體內(nèi)部有很多孩子，我從 DAO 調(diào)用中獲取對象User user = userRepository.findById(userId);現(xiàn)在我想根據(jù)一些引用 Map 的 if 條件獲取一些用戶詳細(xì)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();}我需要創(chuàng)建一個通用方法來根據(jù)地圖獲取用戶對象。我怎樣才能做到這一點。
1 回答
牧羊人nacy
TA貢獻(xiàn)1862條經(jīng)驗 獲得超7個贊
不確定我是否完全理解你的問題，但你可以這樣做：
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Getter像這樣聲明一個接口：
public interface Getter<T,R> {
public R get(T obj);
}
然后做這樣的事情：
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);
}
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