一尘不染

JsonMappingException:找不到适合类型[简单类型,类]的构造函数:无法从JSON对象实例化

java

JsonMappingException:找不到适合类型[简单类型,类]的构造函数:无法从JSON对象实例化尝试获取JSON请求并处理它时出现以下错误:

org.codehaus.jackson.map.JsonMappingException:未找到类型[简单类型,类com.myweb.ApplesDO]的合适构造函数:无法从JSON对象实例化(需要添加/启用类型信息吗?)

这是我要发送的JSON:

{
  "applesDO" : [
    {
      "apple" : "Green Apple"
    },
    {
      "apple" : "Red Apple"
    }
  ]
}

在Controller中,我具有以下方法签名:

@RequestMapping("showApples.do")
public String getApples(@RequestBody final AllApplesDO applesRequest){
    // Method Code
}

AllApplesDO是ApplesDO的包装:

public class AllApplesDO {

    private List<ApplesDO> applesDO;

    public List<ApplesDO> getApplesDO() {
        return applesDO;
    }

    public void setApplesDO(List<ApplesDO> applesDO) {
        this.applesDO = applesDO;
    }
}

ApplesDO:

public class ApplesDO {

    private String apple;

    public String getApple() {
        return apple;
    }

    public void setApple(String appl) {
        this.apple = apple;
    }

    public ApplesDO(CustomType custom){
        //constructor Code
    }
}

我认为Jackson无法将JSON转换为子类的Java对象。请帮助Jackson的配置参数将JSON转换为Java对象。我正在使用Spring Framework。

编辑:在上面的示例类中包括导致此问题的主要错误-请寻找已接受的答案作为解决方案。


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2020-03-21

共1个答案

一尘不染

所以,最后我意识到了问题所在。我怀疑这不是杰克逊的配置问题。

实际上问题出在ApplesDO类中:

public class ApplesDO {

    private String apple;

    public String getApple() {
        return apple;
    }

    public void setApple(String apple) {
        this.apple = apple;
    }

    public ApplesDO(CustomType custom) {
        //constructor Code
    }
}

为该类定义了一个自定义构造函数,使其成为默认构造函数。引入虚拟构造函数使错误消失了:

public class ApplesDO {

    private String apple;

    public String getApple() {
        return apple;
    }

    public void setApple(String apple) {
        this.apple = apple;
    }

    public ApplesDO(CustomType custom) {
        //constructor Code
    }

    //Introducing the dummy constructor
    public ApplesDO() {
    }

}
2020-03-21