Google Gson - 反序列化列表<class>目的?(通用类型) [英] Google Gson - deserialize list&lt;class&gt; object? (generic type)

问题描述

我想通过 Google Gson 传输一个列表对象，但我不知道如何反序列化泛型类型.

I want to transfer a list object via Google Gson, but I don't know how to deserialize generic types.

我在查看 this 后尝试了什么(BalusC 的回答):

What I tried after looking at this (BalusC's answer):

MyClass mc = new Gson().fromJson(result, new List<MyClass>(){}.getClass());

但是后来我在 Eclipse 中收到一个错误，说类型 new List(){} 必须实现继承的抽象方法..."，如果我使用快速修复，我会得到超过 20 个方法存根的怪物.

but then I get an error in eclipse saying "The type new List(){} must implement the inherited abstract method..." and if I use a quick fix I get a monster of over 20 method stubs.

我很确定有一个更简单的解决方案，但我似乎找不到它！

I am pretty sure that there is an easier solution, but I seem unable to find it!

现在我有

Type listType = new TypeToken<List<MyClass>>()
                {
                }.getType();

MyClass mc = new Gson().fromJson(result, listType);

但是，我在fromJson"行得到以下异常:

However, I do get the following exception at the "fromJson" line:

java.lang.NullPointerException
at org.apache.harmony.luni.lang.reflect.ListOfTypes.length(ListOfTypes.java:47)
at org.apache.harmony.luni.lang.reflect.ImplForType.toString(ImplForType.java:83)
at java.lang.StringBuilder.append(StringBuilder.java:203)
at com.google.gson.JsonDeserializerExceptionWrapper.deserialize(JsonDeserializerExceptionWrapper.java:56)
at com.google.gson.JsonDeserializationVisitor.invokeCustomDeserializer(JsonDeserializationVisitor.java:88)
at com.google.gson.JsonDeserializationVisitor.visitUsingCustomHandler(JsonDeserializationVisitor.java:76)
at com.google.gson.ObjectNavigator.accept(ObjectNavigator.java:106)
at com.google.gson.JsonDeserializationContextDefault.fromJsonArray(JsonDeserializationContextDefault.java:64)
at com.google.gson.JsonDeserializationContextDefault.deserialize(JsonDeserializationContextDefault.java:49)
at com.google.gson.Gson.fromJson(Gson.java:568)
at com.google.gson.Gson.fromJson(Gson.java:515)
at com.google.gson.Gson.fromJson(Gson.java:484)
at com.google.gson.Gson.fromJson(Gson.java:434)

我确实捕获 JsonParseExceptions 并且结果"不为空.

I do catch JsonParseExceptions and "result" is not null.

我用调试器检查了 listType 并得到以下结果:

I checked listType with the debugger and got the following:

• 列表类型
  • args = ListOfTypes
    • list = null
    • resolvedTypes = Type[1]

    所以看起来getClass"调用没有正常工作.有什么建议...?

    so it seems the "getClass" invocation didn't work properly. Any suggestions...?

    我检查了 Gson 用户指南.它提到了在将泛型类型解析为 Json 期间应该发生的运行时异常.我做错了(上面没有显示)，就像在例子中一样，但根本没有得到那个异常.所以我按照用户指南的建议更改了序列化.但是没有帮助.

    I'v checked on the Gson User Guide. It mentions a Runtime Exception that should happen during parsing a generic type to Json. I did it "wrong" (not shown above), just as in the example, but didn't get that exception at all. So I changed the serialization as in the user guide suggested. Didn't help, though.

    已解决，请参阅下面的答案.

    Solved, see my answer below.

    推荐答案

    反序列化泛型集合的方法:

    Method to deserialize generic collection:

    import java.lang.reflect.Type;
    import com.google.gson.reflect.TypeToken;
    
    ...
    
    Type listType = new TypeToken<ArrayList<YourClass>>(){}.getType();
    List<YourClass> yourClassList = new Gson().fromJson(jsonArray, listType);
    

    由于评论中有几个人提到了它，这里解释如何使用 TypeToken 类.构造 new TypeToken<...>() {}.getType() 捕获编译时类型(在 < 和 >) 到运行时 java.lang.reflect.Type 对象.与 Class 对象只能表示原始(擦除)类型不同，Type 对象可以表示 Java 语言中的任何类型，包括泛型类型的参数化实例化.

    Since several people in the comments have mentioned it, here's an explanation of how the TypeToken class is being used. The construction new TypeToken<...>() {}.getType() captures a compile-time type (between the < and >) into a runtime java.lang.reflect.Type object. Unlike a Class object, which can only represent a raw (erased) type, the Type object can represent any type in the Java language, including a parameterized instantiation of a generic type.

    TypeToken 类本身没有公共构造函数，因为您不应该直接构造它.相反，您始终构建一个匿名子类(因此是 {}，这是此表达式的必要部分).

    The TypeToken class itself does not have a public constructor, because you're not supposed to construct it directly. Instead, you always construct an anonymous subclass (hence the {}, which is a necessary part of this expression).

    由于类型擦除，TypeToken 类只能捕获在编译时完全已知的类型.(也就是说，您不能对类型参数 T 执行 new TypeToken>() {}.getType().)

    Due to type erasure, the TypeToken class is only able to capture types that are fully known at compile time. (That is, you can't do new TypeToken<List<T>>() {}.getType() for a type parameter T.)

    有关详细信息，请参阅 文档code>TypeToken 类.

    For more information, see the documentation for the TypeToken class.

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