JsonMappingException：找到了类型没有合适的构造函数[简单类型，等级]：无法从JSON对象实例化

问题描述

我正试图获得一个JSON请求并处理它>如果出现以下错误

I am getting the following error when trying to get a JSON request and process it>

组织codehaus.jackson.map.JsonMappingException：找到了类型没有合适的构造函数[简单类型，类com.myweb.ApplesDO]：无法从JSON对象实例化（需要添加/启用类型的信息？）

下面是JSON我想送：

Here is the JSON I am trying to send:

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

在控制器，我有以下方法签名

In Controller , I have the following method signature

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

AllApplesDO是ApplesDO的包装：

AllApplesDO is a wrapper of ApplesDO :

public class AllApplesDO {

    private List<ApplesDO> applesDO;

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

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

我想，杰克逊无法JSON转换成Java对象为sublclasses。请与配置参数杰克逊JSON转换成Java对象的帮助！我使用Spring框架

I am thinking that JACKSON is unable to convert JSON into JAVA objects for sublclasses. Please help with the configuration parameters for JACKSON to convert JSON into JAVA Objects! I am using Spring Framework

编辑：包括是导致在上面的示例类这一问题的主要错误 - 敬请期待解决方案接受的答案

Included the major bug that is causing this problem in the above sample class - Please look accepted answer for solution.

推荐答案

所以，最后我意识到的问题是什么。这不是杰克逊的配置问题，因为我怀疑。

So, finally I realized what the problem is. It is not a Jackson configuration issue as I doubted.

其实这个问题在 ApplesDO 类：

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
    }
}

有是使得它的默认构造函数的类定义的自定义构造函数。引入一个虚拟的构造做出了错误走开：

There was a custom constructor defined for the class making it the default constructor. Introducing a dummy constructor has made the error to go away:

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() {
    }

}

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