Json (fasterxml) 计算器异常

Question

尝试序列化类别时出现计算器溢出问题。

异常

Warning: StandardWrapperValve[dispatcher]: Servlet.service() for servlet dispatcher threw exception java.lang.WhosebugError at java.lang.ClassLoader.defineClass1(Native Method) at java.lang.ClassLoader.defineClass(ClassLoader.java:760) at org.apache.felix.framework.BundleWiringImpl$BundleClassLoader.findClass(BundleWiringImpl.java:2279) at org.apache.felix.framework.BundleWiringImpl.findClassOrResourceByDelegation(BundleWiringImpl.java:1501) at org.apache.felix.framework.BundleWiringImpl.access0(BundleWiringImpl.java:75) at org.apache.felix.framework.BundleWiringImpl$BundleClassLoader.loadClass(BundleWiringImpl.java:1955) at java.lang.ClassLoader.loadClass(ClassLoader.java:357) at com.fasterxml.jackson.databind.ser.std.BeanSerializerBase.serializeFields(BeanSerializerBase.java:660) at com.fasterxml.jackson.databind.ser.BeanSerializer.serialize(BeanSerializer.java:152) at com.fasterxml.jackson.databind.ser.impl.IndexedListSerializer.serializeContents(IndexedListSerializer.java:100) at com.fasterxml.jackson.databind.ser.impl.IndexedListSerializer.serializeContents(IndexedListSerializer.java:21) at com.fasterxml.jackson.databind.ser.std.AsArraySerializerBase.serialize(AsArraySerializerBase.java:183) at com.fasterxml.jackson.databind.ser.BeanPropertyWriter.serializeAsField(BeanPropertyWriter.java:541) at com.fasterxml.jackson.databind.ser.std.BeanSerializerBase.serializeFields(BeanSerializerBase.java:644) at com.fasterxml.jackson.databind.ser.BeanSerializer.serialize(BeanSerializer.java:152)

Category.java

@Entity
public class Category implements DataObject, Serializable {

    @Id
    @GeneratedValue 
    private Long id;
    private String title;
    private String description;

    @ManyToOne @JsonIgnore 
    private Category parent;


@Override
public long getId() {
    return id;
}

@Override
public void setId(long id) {
    this.id = id;
}

public String getTitle() {
    return title;
}

public void setTitle(String title) {
    this.title = title;
}
public String getDescription() {
    return description;
}

public void setDescription(String description) {
    this.description = description;
}

public Category getParent() {
   return null;//return parent;
}

public void setParent(Category parent) {
   // this.parent = parent;
}

public boolean isMainCategory()
{
   return true;// return this.parent == null;
}

/**
 * Returns the chain of parent categories with the main category on index 0
 * @return Chain of categories 
 */
public List<Category> getParentChain()
{

   List<Category> cats = new ArrayList<>();
    Category current = this;
    while(!current.isMainCategory())
    {
        cats.add(current);
        current = current.getParent();
    }
    cats.add(current);
    Collections.reverse(cats);
    return cats;
}

@Override
public String toString()
{
    return this.title;
}

@Override
public boolean equals(Object o)
{
    if(!(o instanceof Category))return false;
    Category c = (Category)o;

    return c.title.equals(this.title);
}

@Override
public int hashCode()
{
    return super.hashCode();
} 
}

休息控制器功能

@RequestMapping(value="/{id}", method=RequestMethod.GET)
public ResponseEntity<Category> get(@PathVariable("id") long categoryId)
{
    Category c  =  service.getCategoryRepository().ReadValue(categoryId);
    if(c == null)
        return new ResponseEntity<>(HttpStatus.NOT_FOUND);
    return new ResponseEntity<>(c,HttpStatus.OK);
}

备注

即使我将 return new ResponseEntity<>(c,HttpStatus.OK); 替换为 return new ResponseEntity<>(new Category(),HttpStatus.OK);，我也会得到一个 Whosebug whilist none 字段包含一个值。

它与我的其他 classes 一起工作正常，只是这个 class 导致了 Whosebug。

Answer 1

如果您发表评论 private Category parent;，您可能不会拥有 Whosebug。我在一个有循环依赖的项目中遇到了同样的问题。

解决这个问题的最好方法是使用父代的 id 而不是 Class 像：

private Long parentId;

编辑：

问题出在试图序列化的 getParentChain() 上。通过在方法前添加@JsonIgnore 问题得到解决。

Answer 2

没问题，@JsonIgnore 就可以了。但是，如果我们需要在 JSON 输出中忽略字段怎么办？

解决方法很简单。

我们在关系的一侧用 @JsonManagedReference 注释来注释我们的 'guilty' 字段（这意味着我们的 @ManyToMany 注释）。

和 @JsonBackReference 在关系的另一边（@OneToMany 已被放置）。

就是这样。没有更多的递归循环。

Answer 3

一个注解解决了你的问题。

在 class 上添加以下注释。

@JsonIdentityInfo(
      generator = ObjectIdGenerators.PropertyGenerator.class, 
      property = "id")

其他方法是在集合上注释 @JsonManagedReference 用于前向，@JsonBackReference. 用于映射中的后向。

示例：

public class User{
    @JsonManagedReference
    @OneToMany(mappedBy = "user")
    Set<Address> s = new Hashset<>();
}

public class Address{
    @JsonBackReference
    @ManyToOne
    @JoinColumn
    User user;
}

Answer 4

这就是我为避免这种递归地狱所做的事情。

将 @JsonIgnore 添加到您的 JPA 实体中的每个 @OneToMany(mappedBy="xxxx")

JsonIgnore 来自 jackson-annotations

    <!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-annotations -->
    <dependency>
        <groupId>com.fasterxml.jackson.core</groupId>
        <artifactId>jackson-annotations</artifactId>
        <version>2.10.0</version>
    </dependency>

JPA 实体示例：

package model;

import java.io.Serializable;
import javax.persistence.*;
import javax.xml.bind.annotation.XmlRootElement;

import com.fasterxml.jackson.annotation.JsonIgnore;

import java.util.List;


/**
 * The persistent class for the categoria database table.
 * 
 */
@Entity
@NamedQuery(name="Categoria.findAll", query="SELECT c FROM Categoria c")
@XmlRootElement(name = "categoria")
public class Categoria implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @Column(name="categoria_id")
    private int categoriaId;

    private String descripcion;

    @JsonIgnore
    //bi-directional many-to-one association to Establecimiento
    @OneToMany(mappedBy="categoria")
    private List<Establecimiento> establecimientos;

    public Categoria() {
    }

    public int getCategoriaId() {
        return this.categoriaId;
    }

    public void setCategoriaId(int categoriaId) {
        this.categoriaId = categoriaId;
    }

    public String getDescripcion() {
        return this.descripcion;
    }

    public void setDescripcion(String descripcion) {
        this.descripcion = descripcion;
    }

    public List<Establecimiento> getEstablecimientos() {
        return this.establecimientos;
    }

    public void setEstablecimientos(List<Establecimiento> establecimientos) {
        this.establecimientos = establecimientos;
    }

    public Establecimiento addEstablecimiento(Establecimiento establecimiento) {
        getEstablecimientos().add(establecimiento);
        establecimiento.setCategoria(this);

        return establecimiento;
    }

    public Establecimiento removeEstablecimiento(Establecimiento establecimiento) {
        getEstablecimientos().remove(establecimiento);
        establecimiento.setCategoria(null);

        return establecimiento;
    }

}

Json (fasterxml) 计算器异常

Json (fasterxml) stackoverflow exception

java

data-binding

spring

json

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