GSON - ＆GT; ＆QUOT;类XYZ＆QUOT无参数的构造函数;使用数组

问题描述

我尝试反序列化一个JSON字符串GSON的帮助。虽然gson.fromJson我收到以下错误：

i try to deserialize a json string with the help of gson. While gson.fromJson I get the following error:

类XYZ无参数的构造函数;不存在。注册GSON的InstanceCreator这种类型来解决这个问题。

我试图用一个InstanceCreate工作，但我没有得到这个运行。
我希望你能帮助我。

I tried to work with an InstanceCreate but I didn't get this running.I hope you can help me.

JSON字符串

[
{
    "prog": "Name1",
    "name": "Name2",
    "computername": "Name3",
    "date": "2010-11-20 19:39:55"
},
{
    "prog": "Name1",
    "name": "Name2",
    "computername": "Name3",
    "date": "2010-11-20 12:38:12"
}

]

（[和]）
根据字符串与字符正确...？

according to gson I have to cut the first and last chars ("[" and "]")according to http://www.jsonlint.com/ the string is with the chars correct... :?:

在code看起来像：

    public class License {
   public String prog;
   public String name;
   public String computername;
   public String date;

   public License() {
      this.computername = "";
      this.date = "";
      this.name = "";
      this.prog = "";
       // no-args constructor
   }
}

---

               String JSONSerializedResources = "json_string_from_above"
           try
         {
              GsonBuilder gsonb = new GsonBuilder();
              Gson gson = gsonb.create();

              JSONObject j;

              License[] lic = null;
              j = new JSONObject(JSONSerializedResources);

              lic = gson.fromJson(j.toString(), License[].class);

               for (License license : lic) {
                  Toast.makeText(getApplicationContext(), license.name + " - " + license.date, Toast.LENGTH_SHORT).show();
            }
         }
         catch(Exception e)
         {
            Toast.makeText(getApplicationContext(), "Error: " + e.getMessage(), Toast.LENGTH_LONG).show();
             e.printStackTrace();
         }

至于克里斯

推荐答案

试着让你的构造公开，让GSON实际上可以访问它。

Try making your constructor public, so that gson can actually access it.

public License() {
  this.computername = "";
  this.date = "";
  this.name = "";
  this.prog = "";
  // no-args constructor
}

但自从Java创建一个默认的构造函数，你可以只使用：

but since Java creates a default constructor, you could just use:

public class License {
    public String prog = "";
    public String name = "";
    public String computername = "";
    public String date = "";
}

更新：

这真的很简单：的JSONObject 预计JSON对象{..}。您应该使用 JSONArray 其中预计[...]。

It's really quite trivial: JSONObject expects a Json object "{..}". You should use JSONArray which expects "[...]".

我想它和它的作品。你还是应该更改许可类如上所述。

I tried it and it works. You should still change License class as noted above.

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