Java JPA中的CriteriaBuilder-where子句
我是JPA的新手。使用where子句生成select查询。我需要从表Contacts中选择所有ContactName,这些ContactName等于字符串名称的值 使用以下代码创建数据库表:Java JPA中的CriteriaBuilder-where子句,java,hibernate,jpa,hibernate-criteria,Java,Hibernate,Jpa,Hibernate Criteria,我是JPA的新手。使用where子句生成select查询。我需要从表Contacts中选择所有ContactName,这些ContactName等于字符串名称的值 使用以下代码创建数据库表: CREATE TABLE Contacts ( ContactId BIGINT UNSIGNED NOT NULL PRIMARY KEY AUTO_INCREMENT, ContactName VARCHAR(100) NOT NULL, ContactEmailID VARCHAR(100) NOT
CREATE TABLE Contacts (
ContactId BIGINT UNSIGNED NOT NULL PRIMARY KEY AUTO_INCREMENT,
ContactName VARCHAR(100) NOT NULL,
ContactEmailID VARCHAR(100) NOT NULL,
UserName VARCHAR(100) NOT NULL,
INDEX Contact_Names (ContactName)
) ENGINE = InnoDB;
下面是我的实体类
@Entity
private String UserName;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "ContactId")
public long getContactId() {
return ContactId;
}
public void setContactId(long contactId) {
ContactId = contactId;
}
@Basic
@Column(name = "UserName")
public String getUserName() {
return UserName;
}
public void setUserName(String UserName) {
this.UserName = UserName;
}
下面是我试图编写CriteriaBuilder代码的ContactServlet类代码
EntityManager manager = null;
EntityTransaction transaction = null;
try{
manager = this.factory.createEntityManager();
transaction = manager.getTransaction();
transaction.begin();
CriteriaBuilder cb = manager.getCriteriaBuilder();
CriteriaQuery<Contact> q1 = cb.createQuery(Contact.class);
Root<Contact> postRoot = q1.from(Contact.class);
q1.select(postRoot).where(cb.equal(postRoot.get("UserName"), name));
TypedQuery<Contact> qry = manager.createQuery(q1);
List<Contact> result = qry.getResultList();
for (Contact contactInstance : result)
{
DBContactName = contactInstance.getContactName().trim();
DBContactEmail = contactInstance.getContactEmailID().trim();
EntityManager=null;
EntityTransaction=null;
试一试{
manager=this.factory.createEntityManager();
事务=manager.getTransaction();
transaction.begin();
CriteriaBuilder cb=manager.getCriteriaBuilder();
CriteriaQuery q1=cb.createQuery(Contact.class);
根postRoot=q1.from(Contact.class);
q1.选择(postRoot).其中(cb.equal(postRoot.get(“用户名”),name));
TypedQuery qry=manager.createQuery(q1);
列表结果=qry.getResultList();
对于(联系人实例:结果)
{
DBContactName=contactInstance.getContactName().trim();
DBContactEmail=contactInstance.getContactEmailID().trim();
不知道我哪里出错了
以下是执行我的项目时出现的错误:
java.lang.IllegalArgumentException:无法解析org.hibernate.jpa.criteria.path.AbstractPathImpl.unknownAttribute(AbstractPathImpl.java:117)org.hibernate.jpa.criteria.path.AbstractPathImpl.locateAttribute(AbstractPathImpl.java:214)org.hibernate.jpa.criteria.path.AbstractPathImpl.get上的路径属性[用户名](AbstractPathImpl.java:185)在com.ContactServlet.doPost(ContactServlet.java:110)在javax.servlet.http.HttpServlet.service(HttpServlet.java:644)在javax.servlet.http.HttpServlet.service(HttpServlet.java:725)在org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:291)上org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239)org.apache.catalina.core.core.ApplicationFilterChainorg.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:219)org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:106)org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:505)org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:142)org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:79)org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:610)org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:88)位于org.apache.CoyoteAdapter.java:534的org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1081)的org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:658)位于org.apache.coyote.http11.Http11NioProtocol$Http11ConnectionHandler.process(Http11NioProtocol.java:222),位于org.apache.tomcat.util.net.niodempoint$SocketProcessor.doRun(niodempoint.java:1566),位于org.apache.tomcat.util.net.niodempoint$SocketProcessor.run(niodempoint.java:1523),位于java.util.concurrent.thpoolexecutor.runWorker(ThreadPoolExecutor.java:1142)在java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)在org.apache.tomcat.util.threads.TaskThread$wrappingranable.run(TaskThread.java:61)在java.lang.Thread.run(Thread.java:745)上运行(Thread.java:617)尝试:
q1.select(postRoot).where(cb.equal(postRoot.get("userName"), name));
用户名属性的u小写。我的第一个问题是,您需要将q1.select语句的结果传递给createQuery调用,例如manager.createQuery(q1.select(…);我尝试按照u的建议传递整个查询。但是没有结果…仍然是相同的错误。实际上,它无法识别该属性这个stmt.q1.select(postRoot)中的“UserName”。where(cb.equal(postRoot.get(“UserName”),name));你能解释一下为什么它在这种情况下有效吗?再次感谢。首先你应该看看。对于你面临的问题,JPA使用getter/setter约定,getUserName()/setUserName()与UserName相关,而不是UserName。