关于数据序列化(4),FastJson的两个bug

bug1 fastJson循环引用引发一个bug

fastJson可以将一个对象序列化为json，也可以通过反序列化出一个完整的对象。且支持循环引用。

package com.sincetimes.website.core.common.support;

import com.sincetimes.website.core.common.vo.ToStringAbstract;

/** ::new */
public class DataSimpleVO {
    public String name;
    public Object value;
    public Object value1;

    public DataSimpleVO() {
    }

    public DataSimpleVO(String name,Object value) {
    this.name = name;
    this.value = value;
    }

    @Override
    public String toString() {
    return "DataSimpleVO [name=" + name + ",value=" + value + ",value1=" + value1 + "]";
    }

}

DataSimpleVO a = new DataSimpleVO("a",1);
    DataSimpleVO b = new DataSimpleVO("b",2);
    b.value = a;
    Map<String,Object> map = new HashMap<>();
    map.put(a.name,a);
    b.value1 = map;
    String jsonStr = JSON.toJSONString(b);
    System.out.println(jsonStr);
    DataSimpleVO obj = JSON.parSEObject(jsonStr,DataSimpleVO.class);
    System.out.println(obj.toString());

执行结果

{"name":"b","value":{"name":"a","value":1},"value1":{"a":{"$ref":"$.value"}}}
DataSimpleVO [name=b,value={"name":"a",value1={"a":{"name":"a","value":1}}]

反序列化成功

改一下代码

DataSimpleVO a = new DataSimpleVO("a",2);
    b.value1 = a;
    Map<String,a);
    b.value = map;
    String jsonStr = JSON.toJSONString(b);
    System.out.println(jsonStr);
    DataSimpleVO obj = JSON.parSEObject(jsonStr,"value":{"a":{"name":"a","value":1}},"value1":{"$ref":"$.value.a"}}
DataSimpleVO [name=b,value={"a":{"name":"a",value1=null]

b中的value1为空,反序列化失败

bug2 要序列化的类含有Class类型属性引起的循环递归无法结束最后内存溢出

要序列化、反序列化的类

public class Request implements Serializable {

    private static final long serialVersionUID = -3145939364922415428L;

    private Class<?> clazz;

    private String method;

    private Object param;

    public Class<?> getClazz() {
    return clazz;
    }

    public void setClazz(Class<?> clazz) {
    this.clazz = clazz;
    }

    public String getmethod() {
    return method;
    }

    public void setMethod(String method) {
    this.method = method;
    }

    public Object getParam() {
    return param;
    }

    public void setParam(Object param) {
    this.param = param;
    }

    public Object invoke(Object bean) throws Exception {
    return clazz.getmethod(method,param.getClass()).invoke(bean,param);
    }
}

触发bug

Request r = new Request();
    r.setClazz(Integer.class);
    String s = JSON.toJSONString(r,SerializerFeature.WriteClassName);
    System.out.println(s);
    JSON.parSEObject(s);//bug 触发

我们看看内部

public static JSONObject parSEObject(String text) {
        Object obj = parse(text);
        if (obj instanceof JSONObject) {
            return (JSONObject) obj;
        }

        return (JSONObject) JSON.toJSON(obj);
    }

如果改为JSON.parSEObject(s,Request.class);或者直接使用JSON.parse(str)就没问题。

最后的建议： 不要使用JSON.parSEObject(…)只使用JSON.parse(str);