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java - 无法将 int 字段 com.java.hibernate.practise.User.id 设置为 java.util.HashSet

转载 作者:行者123 更新时间:2023-11-29 08:46:09 25 4
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I have hibernate @OneToMany mapping I am getting the mentioned error. Does not understand the reason. As getters and setters are public

Below are the entities

@Entity
@Table(name="USER_DETAILS")
public class User implements Serializable {

/**
*
*/
private static final long serialVersionUID = 1L;

@Id
@Column(name="USER_ID")
@GeneratedValue(strategy=GenerationType.AUTO)
private int id;

@Column(name="USER_FIRSTNAME",nullable=false, length=50)
private String userFirstName;

@Column(name="USER_LASTNAME",nullable=false, length=50)
private String userLastName;

@Column(name="USER_MIDDLENAME",length = 30)
private String userMiddleName;

@Column(name="USER_AGE")
private int userAge;

@Column(name="USER_SEX")
private String userSex;


@OneToMany(cascade=CascadeType.ALL, mappedBy="userAddress", targetEntity=Address.class)
private Set<Address> address = new HashSet<Address>();

public String getUserFirstName() {
return userFirstName;
}

public void setUserFirstName(String userFirstName) {
this.userFirstName = userFirstName;
}

public String getUserLastName() {
return userLastName;
}

public void setUserLastName(String userLastName) {
this.userLastName = userLastName;
}

public String getUserMiddleName() {
return userMiddleName;
}

public void setUserMiddleName(String userMiddleName) {
this.userMiddleName = userMiddleName;
}

public int getId() {
return id;
}

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

public int getUserAge() {
return userAge;
}

public void setUserAge(int userAge) {
this.userAge = userAge;
}

public String getUserSex() {
return userSex;
}

public void setUserSex(String userSex) {
this.userSex = userSex;
}

public Set<Address> getAddress() {
return address;
}

public void setAddress(Set<Address> address) {
this.address = address;
}


}

@Entity
@Table(name = "ADDRESS")
public class Address implements Serializable {

/**
*
*/
private static final long serialVersionUID = 1L;

@Id
@Column(name = "ADDRESS_ID")
@GeneratedValue(strategy=GenerationType.AUTO)
private int id;

@Column(name = "ZIP_CODE")
private String zipCode;

@Column(name="ADDRESS_USER_ID", insertable=false, updatable=false)
private Long addressUserID;

@Column(name = "ADDRESS_SEC")
private String addressSec;

@Column(name = "STREET")
private String street;

@Column(name = "CITY")
private String city;

@Column(name = "COUNTRY")
private String country;

@ManyToOne(cascade=CascadeType.ALL, targetEntity=User.class)
@JoinColumn(name="ADDRESS_USER_ID")
private Set<User> userAddress = new HashSet<User>();

public Long getAddressUserID() {
return addressUserID;
}

public void setAddressUserID(Long addressUserID) {
this.addressUserID = addressUserID;
}

public int getId() {
return id;
}

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

public String getZipCode() {
return zipCode;
}

public void setZipCode(String zipCode) {
this.zipCode = zipCode;
}

public Set<User> getUserAddress() {
return userAddress;
}

public void setUserAddress(Set<User> userAddress) {
this.userAddress = userAddress;
}


public String getStreet() {
return street;
}

public void setStreet(String street) {
this.street = street;
}

public String getCity() {
return city;
}

public void setCity(String city) {
this.city = city;
}

public String getAddressSec() {
return addressSec;
}

public void setAddressSec(String addressSec) {
this.addressSec = addressSec;
}

public String getCountry() {
return country;
}

public void setCountry(String country) {
this.country = country;
}

}

Part of Stack Trace are:

线程“main”中的异常 org.hibernate.PropertyAccessException:无法通过 com.java.hibernate.practise.User.id 的反射 getter 获取字段值 在... org.hibernate .property.DirectPropertyAccessor$DirectGetter.get(DirectPropertyAccessor.java:62)引起:java.lang.IllegalArgumentException:无法将 int 字段 com.java.hibernate.practise.User.id 设置为 java.util.HashSet...

我正在使用 hibernate.hbm2ddl.auto= cerate-drop 生成模式
请指导。

最佳答案

通常当我们使用 1..n 双向实体映射时,拥有方通常是多方,应该只有一个对一方对象的实例引用(不是集合 - 多对多),并且要使用的连接列是 on side 类中的主键。我们不需要像您一样在多边明确使用 FK 列。

所以如果这是你的关系 User [1]..[N] Address,那么你应该有更像的东西

public class User {
...
@OneToMany(mappedBy = "user", cascade=CascadeType.ALL)
private Set<Address> addresses;
}

public class Address {
// private Long addressUserID; // Don't need this property. We get it below
...
@ManyToOne
@JoinColumn(name = "USER_ID")
private User user;
}

关于java - 无法将 int 字段 com.java.hibernate.practise.User.id 设置为 java.util.HashSet,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25251014/

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