0 概述

本文主要讲述在实际工作中遇到的fastjson（版本1.2.34）相关问题。

1 常见问题

1.1缺少默认的构造函数

使用JSON形式的序列化技术时，所有通信传递的对象必须设置无参的默认构造函数。

public@H_403_12@ class@H_403_12@ User {

  private@H_403_12@ String name;
  private@H_403_12@ int@H_403_12@ age;

  public@H_403_12@ User@H_403_12@(String name,int@H_403_12@ age) {
    this@H_403_12@.name = name;
    this@H_403_12@.age = age;
  }

  public@H_403_12@ String getName@H_403_12@() {
    return@H_403_12@ name;
  }

  public@H_403_12@ void@H_403_12@ setName@H_403_12@(String name) {
    this@H_403_12@.name = name;
  }

  public@H_403_12@ int@H_403_12@ getAge@H_403_12@() {
    return@H_403_12@ age;
  }

  public@H_403_12@ void@H_403_12@ setAge@H_403_12@(int@H_403_12@ age) {
    this@H_403_12@.age = age;
  }

public@H_403_12@ class@H_403_12@ JsonTest {
  public@H_403_12@ static@H_403_12@ void@H_403_12@ main@H_403_12@(String[] args) {

    System.out@H_403_12@.println(JSON.parSEObject("{\"name\":\"test\",\"age\":12}\n"@H_403_12@,User.class));
  }
}

运行程序抛出如下异常，解决办法，添加默认构造函数即可。

Exception in@H_403_12@ thread "main"@H_403_12@ com.alibaba.fastjson.JSONException: default constructor not found
...@H_403_12@

1.2 序列化或反序列化结果不对

fastjson默认根据get和set方法来生成json体中的属性键值对，如果实体属性中没有对应的set或get方法将得不到结果。

public@H_403_12@ class@H_403_12@ User {

  private@H_403_12@ String name;
  private@H_403_12@ int@H_403_12@ age;

  public@H_403_12@ User@H_403_12@(String name,int@H_403_12@ age) {
    this@H_403_12@.name = name;
    this@H_403_12@.age = age;
  }

  public@H_403_12@ User@H_403_12@() {
  }

  public@H_403_12@ String getName@H_403_12@() {
    return@H_403_12@ name;
  }

  public@H_403_12@ void@H_403_12@ setName@H_403_12@(String name) {
    this@H_403_12@.name = name;
  }

  @Override
  public@H_403_12@ String toString@H_403_12@() {
    return@H_403_12@ "User{"@H_403_12@ +
        "name='"@H_403_12@ + name + '\''@H_403_12@ +
        ",age="@H_403_12@ + age +
        '}'@H_403_12@;
  }
}

public@H_403_12@ class@H_403_12@ JsonTest {
  public@H_403_12@ static@H_403_12@ void@H_403_12@ main@H_403_12@(String[] args) {
    System.out@H_403_12@.println(JSON.toJSON(new@H_403_12@ User("test"@H_403_12@,12@H_403_12@)));
    System.out@H_403_12@.println(JSON.parSEObject("{\"name\":\"test\",User.class));
  }
}

由于age属性没有get方法，序列化时候就没有age信息，反序列化也类似 输出： {“name”:”test”} User{name=’test’,age=0}