使用JPA设置一个对多个桌面的桌面

本文介绍了使用JPA设置一个对多个桌面的桌面的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我在MySQL 5数据库中有以下预先存在的表： 应用 用户 User_App_Bridge 这些关系被映射到App和User之间的User_App_Bridge表（正如名字所暗示的那样，是一个桥梁），作为一对多基数。 ER图： 用户------ User_App_Bridge ------应用程序 p> 1用户与User_App_Bridge表中的许多应用程序相关联。 1 App与User_App_Bridge表中的许多用户相关联。 User_App_Bridge表的DDL： CREATE TABLE`User_App_Bridge`（`User_App_Bridge_Id` int（11）NOT NULL AUTO_INCREMENT，`User_Id` int（11）NOT NULL，`App_Id` int（11）NOT NULL， PRIMARY KEY（`User_App_Bridge_Id`）， KEY`App_Id`（`App_Id`）， KEY`User_Id`（`User_Id`）， CONSTRAINT`user_app_bridge_ibfk_1` FOREIGN KEY（`App_Id`）REFERENCES`App`（`App_Id`））ENGINE = InnoDB DEFAULT CHARSET = utf8; 假设我有以下映射到这些表的JPA注释类： @Entity 公共类应用程序{ @Id @GeneratedValue（strategy = GenerationType.AUTO） @Column（name =App_Id）私人长appId; @OneToMany @JoinTable （ name =UserAppBridge， joinColumns = {@JoinColumn（name =App_Id，referencedColumnName = {@JoinColumn（name =User_Id，referencedColumnName =User_Id）} ） private List< User>用户; @Column（name =App_Name） private String appName; // Getters& Setter方法 实体公共类用户 @Id @GeneratedValue（strategy = GenerationType.AUTO） @Column （name =User_Id） private long userId; @OneToMany @JoinTable （ name =UserAppBridge， joinColumns = {@JoinColumn（name =User_Id，referencedColumnName = User_Id）}， inverseJoinColumns = {@JoinColumn（name =App_Id，referencedColumnName =App_Id，unique = true）} ） private List< App>应用; @Column（name =User_Name） private String userName; // Getters& Setter方法 $ b $ @Entity public class UserAppBridge { @Id @GeneratedValue（strategy = GenerationType.AUTO） @Column（name =User_App_Bridge_Id）私人长userAppBridgeId; @Column（name =User_Id） private long userId; @Column（name =App_Id） private long appId; // Getters& Setter方法 code $ 问题： 这是（@JoinTable中的块）是用UserAppBridge为User和App实现一对多映射的正确方法吗？ 在@JoinTable中，列&参考列名被分配给SQL值（例如，name =User_Id，referencedColumnName =User_Id）或者它应该是Java引用名称（例如name =userId，referencedColumnName =userId）？ 里面的inverseJoinColumns代码为黑色，是unique = true必需的（它是什么）？ 我需要在UserAppBridge类中做任何其他的事情（用于连接到应用程序和用户）感谢您花时间阅读本文... 解决方案您的问题的答案： 1.不，它不是。您需要在User和UserAppBridge之间创建oneToMany映射，并在App和UserAppBridge之间创建oneToMany映射。这里是代码： 在用户实体中： @OneToMany （mappedBy =user） private设置< UserAppBridge> userAppBridgeSet; 在App实体中： @OneToMany（mappedBy =app） private设置< UserAppBridge> userAppBridgeSet; 在UserAppBridge实体中： @Entity @IdClass（UserAppBridgeId.class） public class UserAppBridge { @Id @Column（name =User_App_Bridge_Id ）私人长userAppBridgeId; @Id @Column（name =User_Id） private long userId; @Id @Column（name =App_Id） private long appId; @ManyToOne @PrimaryKeyJoinColumn（name =User_Id，referencedColumnName =User_Id）私人用户用户; @ManyToOne @PrimaryKeyJoinColumn（name =App_Id，referencedColumnName =App_Id）私人App应用; // Getters& SetApp方法} 在UserAppBridgeId类中： public class UserAppBridgeId { private long userAppBridgeId; private int userId; private int appId; // Getters& Setter方法} 有关如何创建高级manyToMany关系的更多信息，请参阅 wiki链接 它应该是一个SQL名称 使用unique = true ，没有unique = true，它会是manyToMany。 说明：让我们有以下一段代码： @OneToMany @JoinTable { //这里应该是表名和而不是实体名称 name =User_App_Bridge， joinColumns = {@JoinColumn（name =User_Id，referencedColumnName =User_Id）}， inverseJoinColumns = {@JoinColumn（name = App_Id，referencedColumnName =App_Id，unique = true）} } private List< App>应用; 如果存在unique = true属性，则映射表可以只有唯一的App_id相同的App_id在映射表中不会多次出现一次），而它可以具有任何User_id（相同的User_id可以在映射表中出现多次）。这意味着，一个User_id可能已经分配了更多的App_id，而一个App_id可能只分配了一个User_id。因此你强制一对多关系。 是的，您确实如此。如答案1所示，您必须为UserAppBridge实体创建一个Id类，并为用户和应用程序实体添加一个映射 Am new to JPA...I have the following preexisting tables inside a MySQL 5 database:AppUserUser_App_BridgeThe relationship(s) are mapped to the User_App_Bridge table (as the name implies, a bridge) amongst App and User as a One to Many cardinality.The E-R diagram:User ------ User_App_Bridge------ App1 User is associated with many apps inside the User_App_Bridge table.1 App is associated with many users inside the User_App_Bridge table.The DDL for the User_App_Bridge table:CREATE TABLE `User_App_Bridge` ( `User_App_Bridge_Id` int(11) NOT NULL AUTO_INCREMENT, `User_Id` int(11) NOT NULL, `App_Id` int(11) NOT NULL, PRIMARY KEY (`User_App_Bridge_Id`), KEY `App_Id` (`App_Id`), KEY `User_Id` (`User_Id`), CONSTRAINT `user_app_bridge_ibfk_1` FOREIGN KEY (`App_Id`) REFERENCES `App` (`App_Id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8;Lets say that I have the following JPA Annotated Classes which are mapped to these tables:@Entitypublic class App { @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name = "App_Id") private long appId; @OneToMany @JoinTable ( name = "UserAppBridge", joinColumns = { @JoinColumn(name="App_Id", referencedColumnName = "App_Id") }, inverseJoinColumns = { @JoinColumn(name="User_Id", referencedColumnName = "User_Id") } ) private List<User> users; @Column(name = "App_Name") private String appName; // Getters & Setter methods}@Entitypublic class User { @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name = "User_Id") private long userId; @OneToMany @JoinTable ( name = "UserAppBridge", joinColumns = { @JoinColumn(name="User_Id", referencedColumnName = "User_Id") }, inverseJoinColumns = { @JoinColumn(name="App_Id", referencedColumnName = "App_Id", unique = true) } ) private List<App> apps; @Column(name = "User_Name") private String userName; // Getters & Setter methods}@Entitypublic class UserAppBridge { @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name = "User_App_Bridge_Id") private long userAppBridgeId; @Column(name = "User_Id") private long userId; @Column(name = "App_Id") private long appId; // Getters & Setter methods}Question(s):Is this (the block inside the @JoinTable) the correct way to do the one to many mapping for User and App entites with the UserAppBridge?Inside the @JoinTable should the column & referencedColumnName be assigned to the SQL value (e.g. name="User_Id" , referencedColumnName = "User_Id")or should it be the Java reference name (e.g. name="userId" , referencedColumnName = "userId")?Inside the inverseJoinColumns code black, is the unique = true necessary (what is it for)?Do I need to do anything else inside the UserAppBridge class (for connecting to App and User)?Thank you for taking the time to read this... 解决方案 The answers for your questions:1.No, it is not. You need to create a oneToMany mapping between User and UserAppBridge, and oneToMany mapping between App and UserAppBridge. Here is the code:In User entity: @OneToMany(mappedBy = "user") private Set<UserAppBridge> userAppBridgeSet;In App entity: @OneToMany(mappedBy = "app") private Set<UserAppBridge> userAppBridgeSet;In UserAppBridge entity: @Entity @IdClass(UserAppBridgeId.class) public class UserAppBridge{ @Id @Column(name = "User_App_Bridge_Id") private long userAppBridgeId; @Id @Column(name = "User_Id") private long userId; @Id @Column(name = "App_Id") private long appId; @ManyToOne @PrimaryKeyJoinColumn(name="User_Id", referencedColumnName="User_Id") private User user; @ManyToOne @PrimaryKeyJoinColumn(name="App_Id", referencedColumnName="App_Id") private App app; // Getters & Setter methods }In UserAppBridgeId class: public class UserAppBridgeId{ private long userAppBridgeId; private int userId; private int appId; // Getters & Setter methods }For more information on how to create advanced manyToMany relationship please see a wiki linkIt should be an SQL namesWith "unique=true" you are forcing a oneToMany relationship, without a "unique=true" it would be manyToMany.Explanation: Lets have a following piece of code: @OneToMany @JoinTable{ //Here should be the table name and not the entity name name = "User_App_Bridge", joinColumns = { @JoinColumn(name="User_Id", referencedColumnName = "User_Id") }, inverseJoinColumns = { @JoinColumn(name="App_Id", referencedColumnName = "App_Id", unique = true) } } private List<App> apps;if there is a "unique=true" attribute, your mapping table can have only unique App_id (the same App_id can not occure more then once in mapping table), while it can have any User_id (same User_id can occur more then once in mapping table). What this means, is that one User_id may have assigned more App_id, while a single App_id may have assigned only one User_id. Hence you force a oneToMany relationship. Yes, you do. As shown in answer 1 , you will have to create an Id class for UserAppBridge entity and also add a mapping for User and App entities 这篇关于使用JPA设置一个对多个桌面的桌面的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！