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使用@ElementCollection映射Set<enum>
EN

Stack Overflow用户
提问于 2013-04-14 20:01:42
回答 3查看 20.1K关注 0票数 26

我有以下枚举:

package ir.raysis.tcs.rule.days;

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

我尝试将其映射为Set<Days> days,如下所示:

@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<>();

但是,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)

我如何正确地映射它?

EN

回答 3

Stack Overflow用户

回答已采纳

发布于 2013-04-15 19:44:17

给未来的谷歌人!最后,我设法解决了这个问题,我只需要将注释放在代码中的其他地方,

@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;
}

正如您所看到的,我在getter方法之前和上面编写了注释代码(而不是将其放在属性声明代码之前)并解决了问题,任何人如果能向我解释导致这种情况的原因,将不胜感激。谢谢

票数 66
EN

Stack Overflow用户

发布于 2013-04-14 20:31:32

尝试使用@CollectionTable而不是@JoinTable

票数 5
EN

Stack Overflow用户

发布于 2020-08-04 20:27:18

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;

其他一切都是按照约定设置的,而不是配置(join-table-name、column)。

我强烈推荐使用@Enumerated(EnumType.STRING) --查找原因。您可能需要@ElementCollection(fetch = FetchType.EAGER),这取决于您要做什么(以及在哪里)。

票数 2
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/15998824

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