Question

我有两个具有外键关系的表。我尝试过搜索方法，它始终会导致OneToMany和ManyToOne映射。我有这两个桌子。

user_role

用户

我试图显示所有用户并通过将位置列的值显示到user_role表中来显示其位置。

这是我的课程

User.java

@Entity
@Table(name="user")
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;

    ..........

    private UserRole userRole;

    @ManyToOne()
    @JoinColumn(name="role_id")
    public UserRole getUserRole() {
        return userRole;
    }

    public void setUserRole(UserRole userRole) {
        this.userRole = userRole;
    }

    .......
}

UserRole.java

@Entity
@Table(name="user_role")
public class UserRole {

    .......

    private List<User> user;

    @OneToMany(targetEntity=User.class, mappedBy="userRole",cascade=CascadeType.ALL, fetch = FetchType.LAZY)
    public List<User> getUser() {
        return user;
    }

    public void setUser(List<User> user) {
        this.user = user;
    }

    public long getRole_id() {
        return role_id;
    }

    public void setRole_id(long role_id) {
        this.role_id = role_id;
    }

    .........

}

我不断收到此错误。

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaConfiguration.class]: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory; nested exception is org.hibernate.MappingException: Could not determine type for: java.util.List, at table: user_role, for columns: [org.hibernate.mapping.Column(user)]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1699) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:573) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:495) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.lambda$doGetBean$0(AbstractBeanFactory.java:317) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:315) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:199) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.context.support.AbstractApplicationContext.getBean(AbstractApplicationContext.java:1089) ~[spring-context-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:859) ~[spring-context-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:550) ~[spring-context-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.boot.web.servlet.context.ServletWebServerApplicationContext.refresh(ServletWebServerApplicationContext.java:140) ~[spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
    at org.springframework.boot.SpringApplication.refresh(SpringApplication.java:780) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
    at org.springframework.boot.SpringApplication.refreshContext(SpringApplication.java:412) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:333) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1277) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1265) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
    at com.rtc_insurance.AdminApplication.main(AdminApplication.java:10) [classes/:na]
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory; nested exception is org.hibernate.MappingException: Could not determine type for: java.util.List, at table: user_role, for columns: [org.hibernate.mapping.Column(user)]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:402) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:377) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.afterPropertiesSet(LocalContainerEntityManagerFactoryBean.java:341) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1758) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1695) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    ... 16 common frames omitted
Caused by: org.hibernate.MappingException: Could not determine type for: java.util.List, at table: user_role, for columns: [org.hibernate.mapping.Column(user)]
    at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:456) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:423) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.hibernate.mapping.Property.isValid(Property.java:226) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:597) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.hibernate.mapping.RootClass.validate(RootClass.java:265) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.hibernate.boot.internal.MetadataImpl.validate(MetadataImpl.java:329) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:461) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:892) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
    at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:57) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:365) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:390) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
    ... 20 common frames omitted

这是我第一次使用spring，所以我并不那么熟悉。任何帮助将不胜感激。

Answer 1

您可以像这样将关系移动到相关对象吗

@ManyToOne()
@JoinColumn(name="role_id", referencedColumnName = "role_id", insertable = false, updatable = false)    
private UserRole userRole;

和userRole一样

@OneToMany(targetEntity=User.class, mappedBy="userRole",cascade=CascadeType.ALL, fetch = FetchType.LAZY)    
private List<User> user = new ArrayList<>();

更新

JAR文件似乎已损坏。尝试从.m2 \ repository和mvn clean install

中删除内容

OR

然后右键单击您的项目，依次选择“ Maven”，“更新项目”，然后选中“强制更新快照/版本”。

Answer 2

我可以通过混合字段和属性之间的注释映射来重新创建您的异常。我看到您在上述问题中对字段ID进行了注释，但是对属性userRole进行了注释，如果执行相同的操作，则会得到相同的错误。我可以通过仅注释属性或字段来修复错误。这两种模型都可以工作：

using (SqlConnection con = new SqlConnection(conString))
using (SqlCommand cmd = new SqlCommand("GetLedger", con))
using (SqlDataAdapter sda = new SqlDataAdapter())
{
        cmd.Connection = con;
        cmd.CommandType = CommandType.StoredProcedure;
        cmd.Parameters.Add(new SQLParameter("@optb", Data.SqlDbType.VarChar);
        cmd.Parameters.Add(new SQLParameter("@startsession", Data.SqlDbType.DateTime));

        cmd.Parameters.Add(new SQLParameter("@endsession", Data.SqlDbType.DateTime));
        command.Parameters("@optb").Value = p1;
        command.Parameters("@startsession").Value = pp2;
        command.Parameters("@startsession").Value = pp3;

        sda.SelectCommand = cmd;

        using (ledgerdt ds = new ledgerdt())
        {
            sda.Fill(ds, "ledgerdt");
            return ds;
        }
}

这也有效：

@Entity
@Table(name = "USER")
public class User {


private Long id;
private UserRole userRole;
String userName;

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
public Long getId() {
    return id;
}


public void setId(Long id) {
    this.id = id;
}

@Column(name = "USER_NAME")
public String getUserName() {
    return userName;
}

public void setUserName(String userName) {
    this.userName = userName;
}

@ManyToOne()
@JoinColumn(name="role_id")
public UserRole getUserRole() {
    return userRole;
}

public void setUserRole(UserRole userRole) {
    this.userRole = userRole;
}
}

另一个模型对象注释字段的示例

@Entity
@Table(name = "USER")
public class User {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;


@ManyToOne()
@JoinColumn(name="role_id")
private UserRole userRole;


@Column(name = "USER_NAME")
String userName;

public Long getId() {
    return id;
}

public void setId(Long id) {
    this.id = id;
}

public String getUserName() {
    return userName;
}

public void setUserName(String userName) {
    this.userName = userName;
}

public UserRole getUserRole() {
    return userRole;
}

public void setUserRole(UserRole userRole) {
    this.userRole = userRole;
}
}

测试代码（我添加了这个代码是因为经常会混淆，即必须设置双向关系才能使用户和userRole都保存到UserRole并保持持久状态。）

@Entity
@Table(name = "USER")
public class User {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;


@ManyToOne()
@JoinColumn(name="role_id")
private UserRole userRole;


@Column(name = "USER_NAME")
String userName;

public Long getId() {
    return id;
}

public void setId(Long id) {
    this.id = id;
}

public String getUserName() {
    return userName;
}

public void setUserName(String userName) {
    this.userName = userName;
}

public UserRole getUserRole() {
    return userRole;
}

public void setUserRole(UserRole userRole) {
    this.userRole = userRole;
}
}

Answer 3

您必须注释Java字段，而不是像这样的getter注释：

@OneToMany(targetEntity=User.class, mappedBy="userRole",cascade=CascadeType.ALL, fetch = FetchType.LAZY)
private List<User> user;

和User.class

@ManyToOne
@JoinColumn(name="role_id")
private UserRole userRole;

此外，您是否在role_id表中定义了user列？我在屏幕截图中看不到

Answer 4

只是一个提示，请检查您的实体类，并确保在类声明语句之前包含@Entity注释。

    -- No of rows returned 510

    rs.absolute(500);
while(rs.next()){
//store data in the object
}

然后在@ManyToMany或@OneToMany关系下使用上述MyEntityClass的任何其他实体类上添加以下内容。

@Entity
public class MyEntityClass{
 //code
}

这种方法对我很有用，也许会有所帮助。

Answer 5

你可以写这样的一对多关系：位置 - 外键

public class UserRole {
.....
@OneToMany(targetEntity = User.class,cascade =CascadeType.ALL)
@JoinColumn(name="position",referencedColumnName = "role_id")
private List<User> users; 
......  
}

OneToMany和ManyToOne映射Spring Boot / Hibernate

5 个答案: