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Android 从innerJsonObject转换GSON/JSON_Android_Json_Gson - Fatal编程技术网

Android 从innerJsonObject转换GSON/JSON

Android 从innerJsonObject转换GSON/JSON,android,json,gson,Android,Json,Gson,我需要创建以下JSON对象,然后使用GSON库(toJson(Object))将其转换为字符串。但是,GSON会在每个JSON对象后面附加nameValuePair,我需要做什么 "Param_1": { "SubParam_1": { type: String, required: true } } 我将这些JSONObject放入ParameterMap(android HTTP客户端)并使用GSON将映射转换为JSON字符串。首先,您编写的JSON“对象”甚至不是有效的JSON 如果有

我需要创建以下JSON对象,然后使用GSON库(toJson(Object))将其转换为字符串。但是,GSON会在每个JSON对象后面附加nameValuePair,我需要做什么

"Param_1": 
{
"SubParam_1": { type: String, required: true }
}
我将这些JSONObject放入ParameterMap(android HTTP客户端)并使用GSON将映射转换为JSON字符串。

首先,您编写的JSON“对象”甚至不是有效的JSON

如果有的话,您可以创建如下JSON:

{
    "Param_1": {
        "SubParam_1": { "type": "String", "required": true }
    }
}
public class YourClass {
  private Param Param_1;
}

public class Param {
  private SubParam SubParam_1;
}

public class SubParam {
  private String type;
  private boolean required;
}
Gson gson = new Gson();
String jsonString = gson.toJson(yourObject);
这很简单,你只需要有正确的类模型,类似这样:

{
    "Param_1": {
        "SubParam_1": { "type": "String", "required": true }
    }
}
public class YourClass {
  private Param Param_1;
}

public class Param {
  private SubParam SubParam_1;
}

public class SubParam {
  private String type;
  private boolean required;
}
Gson gson = new Gson();
String jsonString = gson.toJson(yourObject);
然后可以使用方法
.toJson()
,如下所示:

{
    "Param_1": {
        "SubParam_1": { "type": "String", "required": true }
    }
}
public class YourClass {
  private Param Param_1;
}

public class Param {
  private SubParam SubParam_1;
}

public class SubParam {
  private String type;
  private boolean required;
}
Gson gson = new Gson();
String jsonString = gson.toJson(yourObject);