【发布时间】：2019-03-06 15:24:27
【问题描述】：
我在部署应用程序时遇到错误。 我收到此错误： 原因：org.hibernate.hql.internal.ast.QuerySyntaxException: Newsletter is not mapped [SELECT n FROM Newsletter n where n.user.username = :username]
这是com.app.domain.Newsletter
@Entity(name="NEWSLETTER")
public class Newsletter {
@Id
@GeneratedValue(generator = "user_id_gen", strategy = GenerationType.SEQUENCE)
@SequenceGenerator(name = "user_id_gen", sequenceName = "user_id_seq", allocationSize = 1, initialValue = 1)
protected Long id;
@OneToOne(mappedBy = "newsletter")
@JoinColumn(name = "id_user", referencedColumnName = "id")
private User user;
@Column(name = "D_SUBSCRIBE_DATE")
@Temporal(TemporalType.TIMESTAMP)
private Date subscribeDate;
@Column(name = "B_ISSUB")
private Boolean isSub;
/**
*
* @return
*/
public User getUser() {
return user;
}
/**
*
* @param user
*/
public void setUser(User user) {
this.user = user;
}
/**
*
* @return
*/
public Date getSubscribeDate() {
return subscribeDate;
}
/**
*
* @param subscribeDate
*/
public void setSubscribeDate(Date subscribeDate) {
this.subscribeDate = subscribeDate;
}
/**
*
* @return
*/
public Boolean getSub() {
return isSub;
}
/**
*
* @param sub
*/
public void setSub(Boolean sub) {
isSub = sub;
}
}
这是com.app.domain.User
@Entity(name="USER")
public class User {
@Id
@GeneratedValue(generator = "user_id_gen", strategy = GenerationType.SEQUENCE)
@SequenceGenerator(name = "user_id_gen", sequenceName = "user_id_seq", allocationSize = 1, initialValue = 1)
private Long id;
@Column(name="USERNAME", nullable=false, length=20)
private String username;
@Column(name="PASSWORD_HASH", nullable=false, length=128)
private String passWordhash;
@Column(name="FIRST_NAME", nullable=false, length=50)
private String firstName;
@Column(name="LAST_NAME", nullable=false, length=50)
private String lastName;
private LocalDate birthDay;
private String address;
private String city;
private Integer postCode;
private Integer rn;
@Column(nullable=false, length=100)
private String email;
private String phoneNumber;
@OneToOne
private Newsletter newsletter;
public Long getId() {
return id;
}
public String getFirstName() {
return firstName;
}
public String getLastName() {
return lastName;
}
public LocalDate getBirthDay() {
return birthDay;
}
public String getAddress() {
return address;
}
public String getCity() {
return city;
}
public Integer getPostCode() {
return postCode;
}
public Integer getRn() {
return rn;
}
public String getEmail() {
return email;
}
public String getPhoneNumber() {
return phoneNumber;
}
public void setId(Long id) {
this.id = id;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
public void setBirthDay(LocalDate birthDay) {
this.birthDay = birthDay;
}
public void setAddress(String address) {
this.address = address;
}
public void setCity(String city) {
this.city = city;
}
public void setPostCode(Integer postCode) {
this.postCode = postCode;
}
public void setRn(Integer rn) {
this.rn = rn;
}
public void setEmail(String email) {
this.email = email;
}
public void setPhoneNumber(String phoneNumber) {
this.phoneNumber = phoneNumber;
}
public String getPassWordhash() {
return passWordhash;
}
public User setPassWordhash(String passWordhash) {
this.passWordhash = passWordhash;
return this;
}
public String getUsername() {
return username;
}
public User setUsername(String username) {
this.username = username;
return this;
}
public Newsletter getNewsletter() {
return newsletter;
}
public void setNewsletter(Newsletter newsletter) {
this.newsletter = newsletter;
}
}
这是我的 NewsletterDao ：
@Transactional
public interface NewsletterDao extends JpaRepository<Newsletter, Long> {
@Query("SELECT n FROM Newsletter n where n.user.username = :username")
Optional<Newsletter> findByUsername(@Param("username") String username);
}
这是我的 persistence.xml 文件配置。
<?xml version="1.0" encoding="UTF-8" ?>
<persistence 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_2.xsd"
version="2.2">
<persistence-unit name="test">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<mapping-file>META-INF/orm.xml</mapping-file>
<class>com.app.domain.User</class>
<class>com.app.domain.Newsletter</class>
</persistence-unit>
</persistence>
我收到此错误：
Caused by: java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: Newsletter is not mapped [SELECT n FROM Newsletter n where n.user.username = :username]
at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:138)
at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:181)
at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:188)
at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:713)
at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:103)
at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at java.base/jdk.internal.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.base/java.lang.reflect.Method.invoke(Method.java:566)
at org.springframework.orm.jpa.ExtendedEntityManagerCreator$ExtendedEntityManagerInvocationHandler.invoke(ExtendedEntityManagerCreator.java:350)
at com.sun.proxy.$Proxy45.createQuery(Unknown Source)
at org.springframework.data.jpa.repository.query.SimpleJpaQuery.validateQuery(SimpleJpaQuery.java:87)
... 82 more
谁能帮助我？我有点迷茫。
谢谢！
【问题讨论】：
-
如果从
@Entity中删除name参数会怎样？