我正在处理SpringMVC，Hibernate＆，JSON但出现此错误。
HTTP Status 500 - Could not write JSON: No serializer found for class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationConfig.SerializationFeature.FAIL_ON_EMPTY_BEANS) ) 请在下面检查我的实体
@Entity
@Table(name="USERS")
public class User {
@Id
@GeneratedValue
@Column(name="USER_ID")
private Integer userId;
@Column(name="USER_FIRST_NAME")
private String firstName;
@Column(name="USER_LAST_NAME")
private String lastName;
@Column(name="USER_MIDDLE_NAME")
private String middleName;
@Column(name="USER_EMAIL_ID")
private String emailId;
@Column(name="USER_PHONE_NO")
private Integer phoneNo;
@Column(name="USER_PASSWORD")
private String password;
@Column(name="USER_CONF_PASSWORD")
private String confPassword;
@Transient
private String token;
@Column(name="USER_CREATED_ON")
private Date createdOn;
@OneToMany(fetch=FetchType.EAGER,cascade=CascadeType.ALL)
@Fetch(value = FetchMode.SUBSELECT)
@JoinTable(name = "USER_ROLES", joinColumns = { @JoinColumn(name = "USER_ID") }, inverseJoinColumns = { @JoinColumn(name = "ROLE_ID") })
private List<ActifioRoles> userRole = new ArrayList<ActifioRoles>();
@OneToMany(fetch=FetchType.EAGER,cascade=CascadeType.ALL,mappedBy="userDetails")
@Fetch(value = FetchMode.SUBSELECT)
private List<com.actifio.domain.Address> userAddress = new ArrayList<com.actifio.domain.Address>();
@OneToOne(cascade=CascadeType.ALL)
private Tenant tenantDetails;
public Integer getUserId() {
return userId;
}
public void setUserId(Integer userId) {
this.userId = userId;
}
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
public String getEmailId() {
return emailId;
}
public void setEmailId(String emailId) {
this.emailId = emailId;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getConfPassword() {
return confPassword;
}
public void setConfPassword(String confPassword) {
this.confPassword = confPassword;
}
public Date getCreatedOn() {
return createdOn;
}
public void setCreatedOn(Date createdOn) {
this.createdOn = createdOn;
}
public List<ActifioRoles> getUserRole() {
return userRole;
}
public void setUserRole(List<ActifioRoles> userRole) {
this.userRole = userRole;
}
public String getMiddleName() {
return middleName;
}
public void setMiddleName(String middleName) {
this.middleName = middleName;
}
public Integer getPhoneNo() {
return phoneNo;
}
public void setPhoneNo(Integer phoneNo) {
this.phoneNo = phoneNo;
}
public List<com.actifio.domain.Address> getUserAddress() {
return userAddress;
}
public void setUserAddress(List<com.actifio.domain.Address> userAddress) {
this.userAddress = userAddress;
}
public Tenant getTenantDetails() {
return tenantDetails;
}
public void setTenantDetails(Tenant tenantDetails) {
this.tenantDetails = tenantDetails;
}
public String getToken() {
return token;
}
public void setToken(String token) {
this.token = token;
}
}我该如何解决？
请显示
—
堆栈跟踪
在不了解代码要做什么的情况下，调试起来有些困难，但是由于您使用的是Jackson和Hibernate ，因此您可能想查看github.com/FasterXML/jackson-datatype-hibernate
—
geoand
您是否正在尝试通过此类制作JSON？在这种情况下，JSON序列化程序将尝试写入所有属性，以及多对多关系的HashSet；这会导致一个懒惰的初始化程序异常
—
Angelo Immediata 2014年
@ user2963481 ...很好，非常有帮助的问题，兄弟。
—
Brain