使用@ElementCollection映射Set<enum>

Question

使用@ElementCollection映射Set<enum>

34

我有以下枚举：

package ir.raysis.tcs.rule.days;

public enum Days {
    SUNDAY, MONDAY, TUESDAY, WEDNESDAY,
    THURSDAY, FRIDAY, SATURDAY;
}

我尝试将它映射为一个 Set<Days>，代码如下：

@ElementCollection(targetClass = Days.class) 
@JoinTable(name = "days",joinColumns = @JoinColumn(name = "rule_id")) 
@Column(name ="daysOfWeek", nullable = false) @Enumerated(EnumType.STRING) 
private Set<Days> days = new HashSet<>();

然而，它会抛出以下异常：

Initial SessionFactory creation failed.org.hibernate.MappingException: Could not determine type for: java.util.Set, at table: rule, for columns: [org.hibernate.mapping.Column(days)]
Apr 14, 2013 4:15:17 PM org.apache.catalina.core.StandardWrapperValve invoke
SEVERE: Servlet.service() for servlet [jsp] in context with path [/ptcs] threw exception [javax.servlet.ServletException: java.lang.ExceptionInInitializerError] with root cause
org.hibernate.MappingException: Could not determine type for: java.util.Set, at table: rule, for columns: [org.hibernate.mapping.Column(days)]
    at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:306)
    at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:290)
    at org.hibernate.mapping.Property.isValid(Property.java:217)
    at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:464)
    at org.hibernate.mapping.RootClass.validate(RootClass.java:235)
    at org.hibernate.cfg.Configuration.validate(Configuration.java:1362)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1865)
    at ir.raysis.tcs.db.HibernateUtil.<clinit>(HibernateUtil.java:18)
    at ir.raysis.tcs.db.SaveUpadteDelete.save(SaveUpadteDelete.java:33)
    at ir.raysis.tcs.db.dBAllFunc.save(dBAllFunc.java:35)
    at ir.raysis.tcs.db.AbsDBObject.save(AbsDBObject.java:45)
    at ir.raysis.tcs.action.CreateMemberAction.execute(CreateMemberAction.java:31)
    at org.apache.jsp.action.createMember_jsp._jspService(createMember_jsp.java:79)
    at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:722)
    at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:432)
    at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:390)
    at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:334)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:722)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:305)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:225)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:123)
    at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:472)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:168)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:98)
    at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:927)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:118)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:407)
    at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1001)
    at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:579)
    at org.apache.tomcat.util.net.AprEndpoint$SocketProcessor.run(AprEndpoint.java:1770)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at java.lang.Thread.run(Unknown Source)

我该如何正确地映射它？

- MoienGK

3个回答

6

尝试使用@CollectionTable而不是@JoinTable

- Pace

谢谢回复，但那并没有解决问题 - 有什么想法吗？ - MoienGK

嗯...尝试去掉@ElementCollection，并将targetClass放在@JoinTable上。 - Pace

5

虽然另一个问题是正确的，但最简单的形式可能是：

@ElementCollection
@Enumerated
private Set<EnumName> enumValues;

其他所有的设置都将按照约定配置（连接表名、列）。

我强烈建议使用 @Enumerated(EnumType.STRING) - 了解一下为什么。而且，根据你正在做的事情（以及在哪里进行），你可能需要 @ElementCollection(fetch = FetchType.EAGER)。

- icyerasor

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- MoienGK · Accepted Answer

给未来的谷歌搜索者！最终我成功解决了问题，只需将注释放在我的代码的其他位置即可。

@ElementCollection(targetClass = Days.class)
@CollectionTable(name = "days", joinColumns = @JoinColumn(name = "rule_id"))
@Column(name = "daysOfWeek", nullable = false)
@Enumerated(EnumType.STRING)
public Set<Days> getDays() {
    return days;
}

如您所见，我在getter方法之前编写了注解代码(而不是将其放置在属性声明代码之前)，问题得到解决。如果有人能够解释是什么原因导致了这种情况，我将不胜感激。谢谢。