我正在尝试在Java 9上运行我的Spring Boot应用程序,并且我遇到了JAXB问题,该问题在指南中有所描述,但对我来说并不适用.我添加了对JAXB api的依赖,应用程序开始工作.如果您收到以下异常,则由于缺少使用Java版本> = 9的JAXB缺失实现:
javax.xml.bind.JAXBException: Implementation of JAXB-API has not been found on module path or classpath.
at javax.xml.bind.ContextFinder.newInstance(ContextFinder.java:177) ~[jaxb-api-2.3.0.jar:2.3.0]
at javax.xml.bind.ContextFinder.find(ContextFinder.java:364) ~[jaxb-api-2.3.0.jar:2.3.0]
at javax.xml.bind.JAXBContext.newInstance(JAXBContext.java:508) ~[jaxb-api-2.3.0.jar:2.3.0]
at javax.xml.bind.JAXBContext.newInstance(JAXBContext.java:465) ~[jaxb-api-2.3.0.jar:2.3.0]
at javax.xml.bind.JAXBContext.newInstance(JAXBContext.java:366) ~[jaxb-api-2.3.0.jar:2.3.0]
at com.sun.jersey.server.impl.wadl.WadlApplicationContextImpl.<init>(WadlApplicationContextImpl.java:107) ~[jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.wadl.WadlFactory.init(WadlFactory.java:100) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.RootResourceUriRules.initWadl(RootResourceUriRules.java:169) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.RootResourceUriRules.<init>(RootResourceUriRules.java:106) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.WebApplicationImpl._initiate(WebApplicationImpl.java:1359) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.WebApplicationImpl.access$700(WebApplicationImpl.java:180) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.WebApplicationImpl$13.f(WebApplicationImpl.java:799) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.WebApplicationImpl$13.f(WebApplicationImpl.java:795) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.spi.inject.Errors.processWithErrors(Errors.java:193) [jersey-core-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.WebApplicationImpl.initiate(WebApplicationImpl.java:795) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.server.impl.application.WebApplicationImpl.initiate(WebApplicationImpl.java:790) [jersey-server-1.19.1.jar:1.19.1]
at com.sun.jersey.spi.container.servlet.ServletContainer.initiate(ServletContainer.java:509) [jersey-servlet-1.19.1.jar:1.19.1]
at com.sun.jersey.spi.container.servlet.ServletContainer$InternalWebComponent.initiate(ServletContainer.java:339) [jersey-servlet-1.19.1.jar:1.19.1]
at com.sun.jersey.spi.container.servlet.WebComponent.load(WebComponent.java:605) [jersey-servlet-1.19.1.jar:1.19.1]
at com.sun.jersey.spi.container.servlet.WebComponent.init(WebComponent.java:207) [jersey-servlet-1.19.1.jar:1.19.1]
at …我无法弄清楚我做错了什么.我正在学习JPA映射到关系数据库,通过在网上学习一些教程,但找不到一个直截了当的教程.当我运行我的项目时,它给了我一个错误.我想这是坚持不懈的em.persist();.如果我评论该行,所有看起来都很好,没有错误,但显然没有数据写入表.这是我的代码:
persistence.xml(生成,未触及)
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="RESTappPU" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>restapp.entities.ContactList</class>
<properties>
<property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost:5432/postgres"/>
<property name="javax.persistence.jdbc.user" value="postgres"/>
<property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver"/>
<property name="javax.persistence.jdbc.password" value="postgres"/>
</properties>
</persistence-unit>
</persistence>
实体类(生成,未触及) - 我是否需要在此处添加一些额外的"关系"方法？
package restapp.entities;
imports [...]
@Entity
@Table(name = "ContactList")
@NamedQueries({
@NamedQuery(name = "ContactList.findAll", query = "SELECT c FROM ContactList c"),
@NamedQuery(name = "ContactList.findByFirstname", query = "SELECT c FROM ContactList c WHERE c.firstname = :firstname"),
@NamedQuery(name = "ContactList.findByLastname", query = "SELECT c FROM ContactList c …