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Fellow Spring Data REST愛好者，我正在運行我的Spring Data REST 1.1應用程序，並試圖使用curl使用「text/uri-list」內容添加實體關係類型，如鏈接中所述：Example-API-usage-with-curl。Spring Data REST：添加實體關係時出現靜默失敗
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，儘管服務器返回「201創建」，體內包含一個空的JSON對象，與實體的關係還沒有生成：
{
"links" : [ ],
"content" : [ ]
}
我希望看到一個SQL UPDATE正在執行，但分析SQL顯示，只有SELECT語句出現：
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「手動」在我的數據執行SQL語句添加的關係基礎客戶端：
UPDATE ENEMY SET HERO_ID = 1 WHERE ENEMY_ID = 1;
，然後執行卷曲聲明：
curl -v -H "Accept: application/json" http://localhost:8080/simplemvc/rest/heroes/1/defeatedEnemies
我得到的演示JSON表示 - 通過超媒體鏈接 - 這春數據REST承認一個一對多的關係英雄和敵方實體之間：
{
"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"
} ]
} ]
}
這是被添加了現有Spring MVC的應用，其彈簧數據REST，使用下面的文章作爲指導：Adding-Spring-Data-REST-to-an-existing-Spring-MVC-Application
我已經嘗試使用H2和MySQL數據庫都有相同的結果。下面是我的JPA實體，春天JPA的數據存儲庫，應用程序上下文，和web.xml：
Hero.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;
...
}
Enemy.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> {
}
根的context.xml包括：
<jpa:repositories base-package="com.simple.simplemvc.repositories" />
<bean id="restConfig" class="org.springframework.data.rest.webmvc.config.RepositoryRestMvcConfiguration"/>
的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>
任何想法？謝謝！