Android 从innerJsonObject转换GSON/JSON
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
"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);