@Embeddablefor@ManyToMany关系和附加列是否与String? 我不使用@Generated Value for @Id
因为我的实体ApplikationUser
@Id
@Column(length = 128)
private String applikationUserId;
Run Code Online (Sandbox Code Playgroud)
按业务逻辑总是有一个 Id
这是我的代码:
@EmbeddedId
private ApplikationUserPopupMessageId applikationUserPopupMessageId;
@ManyToOne(fetch = FetchType.EAGER)
@MapsId("applikationUserId")
private ApplikationUser applikationUser;
@ManyToOne(fetch = FetchType.EAGER)
@MapsId("popupMessageId")
private PopupMessage popupMessage;
@Column(name = "gelesen")
private boolean gelesen = false;
Run Code Online (Sandbox Code Playgroud)
@Embeddable
public class ApplikationUserPopupMessageId implements Serializable {
@Column(name = "applikation_user_id")
private String applikationUserId;
@Column(name = "popup_message_id")
private Long popupMessageId;
Run Code Online (Sandbox Code Playgroud)
@Entity
@Table
public class PopupMessage {
@Id
@GeneratedValue
@Column(length = 128)
private …Run Code Online (Sandbox Code Playgroud) 我正在读这本书:“ C von A bis Z”。
有这个例子。
/* ptr14.c */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/* Fehler: Funktion gibt die Adresse
* einer lokalen Variablen zurück. */
/* [ Error: Function returns the address of a
a local variable. ] */
// ...
/* Möglichkeit2: Speicher vom Heap verwenden */
/* [ Option2: Use memory from the heap ] */
char *test3(void){
char *buffer = (char *) malloc(10);
strcpy(buffer, "testwert");
return buffer;
}
/* Möglichkeit3: Einen Zeiger …Run Code Online (Sandbox Code Playgroud) PollListenerVaadin 14 中是否有可能删除 a ?
UI.getCurrent().addPollListener(pollEvent -> { refresh(); }
Run Code Online (Sandbox Code Playgroud)
我需要开始刷新,但如果路由或 UI 发生更改,刷新必须停止。
是否有“Vaadin 可能性”,或者我必须自己构建一些东西?
谢谢你!
c ×1
java ×1
jpa ×1
many-to-many ×1
polling ×1
stack ×1
stack-memory ×1
vaadin ×1
vaadin-flow ×1