我对GSON还很陌生,并且得到了这种格式的JSON响应(只是一个更简单的例子,所以这些值没有意义):
{
"Thomas": {
"age": 32,
"surname": "Scott"
},
"Andy": {
"age": 25,
"surname": "Miller"
}
}
我想让GSON把它变成一个Map,PersonData显然是一个Object。名称字符串是PersonData的标识符。
正如我所说的,我对GSON非常陌生,只尝试过这样的东西:
Gson gson = new Gson();
Map<String, PersonData> decoded = gson.fromJson(jsonString, new TypeToken<Map<String, PersonData>>(){}.getType());
但这引发了错误:
Exception in thread "main" com.google.gson.JsonSyntaxException: java.lang.IllegalStateException: Expected BEGIN_ARRAY but was STRING at line 1 column 3141
感谢您的帮助:)
以下适用于我的
static class PersonData {
int age;
String surname;
public String toString() {
return "[age = " + age + ", surname = " + surname + "]";
}
}
public static void main(String[] args) {
String json = "{"Thomas": {"age": 32,"surname": "Scott"},"Andy": {"age": 25,"surname": "Miller"}}";
System.out.println(json);
Gson gson = new Gson();
Map<String, PersonData> decoded = gson.fromJson(json, new TypeToken<Map<String, PersonData>>(){}.getType());
System.out.println(decoded);
}
并打印
{"Thomas": {"age": 32,"surname": "Scott"},"Andy": {"age": 25,"surname": "Miller"}}
{Thomas=[age = 32, surname = Scott], Andy=[age = 25, surname = Miller]}
所以,也许你的PersonData
类非常不同。
您可以使用gson.toJsonTree(Object o)
将自定义对象转换为JSON格式。
以下对我有效:
private static class PersonData {
private int age;
private String surname;
public PersonData(int age, String surname) {
this.age = age;
this.surname = surname;
}
}
public static void main(String[] args) {
PersonData first = new PersonData(24, "Yovkov");
PersonData second = new PersonData(25, "Vitanov");
Gson gson = new Gson();
JsonObject jsonObject = new JsonObject();
jsonObject.add("kocko", gson.toJsonTree(first));
jsonObject.add("deyan", gson.toJsonTree(second));
System.out.println(gson.toJson(jsonObject));
}
并打印:
{"kocko":{"age":24,"surname":"Yovkov"},"deyan":{"age":25,"surname":"Vitanov"}}