[英]Deserialize Map<String, List<?>> using Gson
I serialize data in server: 我在服务器中序列化数据:
Gson gson = new Gson();
Map<String, List<?>> resultMap = BackendUpdateManager.getInstance()
.getUpdates(timeHolder, shopIdInt, buyerIdInt);
gson.toJson(resultMap);
and deserialize: 和反序列化:
Gson gson = new Gson();
Map<String, List<?>> resultMap = gson.fromJson(json,
new TypeToken<Map<String, List<?>>>() {
}.getType());
However, when I try use items from the Map
: 但是,当我尝试使用
Map
项目时:
List<ProductCategory> productCategoryList = (List<ProductCategory>)updateMap.get(key);
for(ProductCategory productCategory : productCategoryList) {
}
I get error: 我收到错误:
Caused by:
java.lang.ClassCastException
:com.google.gson.internal.LinkedTreeMap
cannot be cast tocom.example.model.entity.ProductCategory
引起:
java.lang.ClassCastException
:com.google.gson.internal.LinkedTreeMap
无法com.example.model.entity.ProductCategory
为com.example.model.entity.ProductCategory
How can I fix this error or otherwise create a Map
with List<different classes>
? 如何修复此错误或以其他方式创建带有
List<different classes>
的Map
?
I've tried creating classes with getters and setters, that contains List
s of different classes instead of Map<String, List<?>>
and use it for serialization and deserialization. 我尝试使用getter和setter创建类,其中包含不同类的
List
而不是Map<String, List<?>>
并将其用于序列化和反序列化。 But I'm looking for a better way. 但我正在寻找一种更好的方法。
How is Gson supposed to know that a particular Json string is a ProductCategory
? Gson如何知道特定的Json字符串是
ProductCategory
? For instance, if the definition of ProductCategory
is this: 例如,如果
ProductCategory
的定义是这样的:
package com.stackoverflow.vitvetal;
public class ProductCategory {
String name;
String type;
}
And the Json is this: 而Json是这样的:
{
"name":"bananas",
"type":"go-go"
}
Where is the link that tells Gson to create an instance of a com.stackoverflow.vitvetal.ProductCategory
? 告诉Gson创建
com.stackoverflow.vitvetal.ProductCategory
实例的链接在哪里?
So what gson does instead is, it creates a Map<String, String>
that looks like 那么gson所做的是,它创建一个看起来像的
Map<String, String>
"name" -> "bananas"
"type" -> "go-go"
If you want to do something different, the easiest thing to do - but also the least powerful - is to fully specify the parameterized type when you create your TypeToken
; 如果你想做一些不同的事情, 最简单的事情 - 也是最不强大的 - 是在你创建
TypeToken
时完全指定参数化类型; no wildcard <?>
allowed. 没有通配符
<?>
允许。
If you need to do something more powerful, like creating maps with more variety in objects, you need to create a custom deserializer , using a TypeAdapter<T>
, that teaches Gson how to handle your particular sort of object. 如果你需要做一些更强大的事情,比如在对象中创建更多种类的地图,你需要使用
TypeAdapter<T>
创建一个自定义反序列化器 ,它教会Gson如何处理你特定类型的对象。
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