如何（完全）将json反序列化为通用列表？

问题描述

当使用 ObjectMapper 将json String 转换为实体时，我可以将其设为通用： / p>

When using an ObjectMapper to transform a json String into an entity, I can make it generic as:

public <E> E getConvertedAs(String body, Class<E> type) throws IOException {
    return mapper.readValue(body, type);
}

现在让我说我想读集合。我可以这样做：

Now let's say I want to read collections. I can do:

List<SomeEntity> someEntityList = asList(mapper.readValue(body, SomeEntity[].class));
List<SomeOtherEntity> someOtherEntityList = asList(mapper.readValue(body, SomeOtherEntity[].class));

我想写一个上述的等效方法，但是对于集合。由于你不能在java中使用泛型数组，所以这样的东西不起作用：

I would like to write an equivalent method of the above, but for collections. Since you can't have generic arrays in java, something like this won't work:

public <E> List<E> getConvertedListAs(String body, Class<E> type) {
    return mapper.readValue(body, type[].class);
}

有一个几乎可行的解决方案：

Here there is a solution that almost works:

mapper.readValue(jsonString, new TypeReference<List<EntryType>>() {});

问题是它没有反序列化为 E ，但 LinkedHashMap.Entry 。有什么方法可以更进一步，如下所示？

The problem is that it doesn't deserialize into a list of E, but of LinkedHashMap.Entry. Is there any way of going going a step further, something like the following?

public <E> List<E> getConvertedListAs(String body, Class<E> type) {
    mapper.readValue(body, new TypeReference<List<type>>() {}); // Doesn't compile
}

推荐答案

此方法可以帮助读取 json 到一个或多个对象：

This method can help to read json to an object or collections:

public class JsonUtil {
    private static final ObjectMapper mapper = new ObjectMapper();

    public static <T>T toObject(String json, TypeReference<T> typeRef){
        T t = null;
        try {
            t = mapper.readValue(json, typeRef);
        } catch (IOException e) {
            e.printStackTrace();
        }
        return t;
    }
}

读json到列表：

List<Device> devices= JsonUtil.toObject(jsonString,
                            new TypeReference<List<Device>>() {});

将json读取到对象：

Read json to object:

Device device= JsonUtil.toObject(jsonString,
                                new TypeReference<Device>() {});

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