各位 Spring Data REST 爱好者，我正在运行我的 Spring Data REST 1.1 应用程序，我正在尝试使用 curl 添加使用“text/uri-list”内容类型的实体关系，如链接中所述：Example-API-与卷曲一起使用。
curl -v -d "http://localhost:8080/simplemvc/rest/enemies/3" -H "Content-Type: text/uri-list" http://localhost:8080/simplemvc/rest/heroes/1/defeatedEnemies
不幸的是，虽然服务器返回“201 Created”，但主体包含一个空的 JSON 对象，并且实体关系没有被创建：
{
"links" : [ ],
"content" : [ ]
}
我希望看到正在执行 SQL UPDATE，但分析 SQL 显示只有 SELECT 语句发生：
Hibernate: select hero0_.HERO_ID as HERO1_1_0_, hero0_.name as name2_1_0_ from HERO hero0_ where hero0_.HERO_ID=?
Hibernate: select defeateden0_.HERO_ID as HERO4_1_1_, defeateden0_.ENEMY_ID as ENEMY1_0_1_, defeateden0_.ENEMY_ID as ENEMY1_0_0_, defeateden0_.description as descript2_0_0_, defeateden0_.HERO_ID as HERO4_0_0_, defeateden0_.name as name3_0_0_ from ENEMY defeateden0_ where defeateden0_.HERO_ID=?
Hibernate: select enemy0_.ENEMY_ID as ENEMY1_0_1_, enemy0_.description as descript2_0_1_, enemy0_.HERO_ID as HERO4_0_1_, enemy0_.name as name3_0_1_, hero1_.HERO_ID as HERO1_1_0_, hero1_.name as name2_1_0_ from ENEMY enemy0_ left outer join HERO hero1_ on enemy0_.HERO_ID=hero1_.HERO_ID where enemy0_.ENEMY_ID=?
有趣的是，如果我通过在数据库客户端中执行 SQL 语句“手动”添加关系：
UPDATE ENEMY SET HERO_ID = 1 WHERE ENEMY_ID = 1;
然后执行 curl 语句：
curl -v -H "Accept: application/json" http://localhost:8080/simplemvc/rest/heroes/1/defeatedEnemies
我得到了一个 JSON 表示，它通过超媒体链接演示 Spring Data REST 识别 Hero 和 Enemy 实体之间的一对多关系：
{
"links" : [ ],
"content" : [ {
"name" : "Red Ghost",
"description" : "Likes to chase",
"links" : [ {
"rel" : "self",
"href" : "http://localhost:8080/simplemvc/rest/enemies/1"
}, {
"rel" : "enemy.enemy.hero",
"href" : "http://localhost:8080/simplemvc/rest/enemies/1/hero"
} ]
} ]
}
这是一个现有的 Spring MVC 应用程序，正在向其中添加 Spring Data REST，使用以下文章作为指南：Adding-Spring-Data-REST-to-an-existing-Spring-MVC-Application
我尝试过使用 H2 和 MySQL 数据库，结果相同。下面是我的 JPA 实体、Spring Data JPA 存储库、应用程序上下文和 web.xml：
英雄.java
@Entity
@Table(name = "HERO")
public class Hero {
@TableGenerator(
name="heroGen",
table="ID_GEN",
pkColumnName="GEN_KEY",
valueColumnName="GEN_VALUE",
pkColumnValue="HERO_ID",
allocationSize=1)
@Id
@GeneratedValue(strategy=TABLE, generator="heroGen")
@Column(name = "HERO_ID")
private Integer id;
@Column
private String name;
@OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, mappedBy = "hero")
private Set<Enemy> defeatedEnemies;
...
}
敌人.java
@Entity
@Table(name = "ENEMY")
public class Enemy {
@TableGenerator(name="enemyGen",
table="ID_GEN",
pkColumnName="GEN_KEY",
valueColumnName="GEN_VALUE",
pkColumnValue="ENEMY_ID",
allocationSize=1)
@Id
@GeneratedValue(strategy=TABLE, generator="enemyGen")
@Column(name="ENEMY_ID")
private Integer id;
@Column
private String name;
@Column
private String description;
@ManyToOne
@JoinColumn(name = "HERO_ID")
private Hero hero;
...
}
HeroRepository.java
@RestResource(path = "heroes")
public interface HeroRepository extends CrudRepository<Hero, Integer> {
}
EnemyRepository.java
@RestResource(path = "enemies")
public interface EnemyRepository extends CrudRepository<Enemy, Integer> {
}
root-context.xml包括：
<jpa:repositories base-package="com.simple.simplemvc.repositories" />
<bean id="restConfig" class="org.springframework.data.rest.webmvc.config.RepositoryRestMvcConfiguration"/>
web.xml包括：
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<servlet>
<servlet-name>rest-dispatcher</servlet-name>
<servlet-class>org.springframework.data.rest.webmvc.RepositoryRestDispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>rest-dispatcher</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
有任何想法吗？谢谢！