Question

首先，我通常会知道如何反序列化JSON对象。我遇到的特定问题是，我有一个Json对象，其中包含名为"1", "2", "3"的数组，等等。但是在Java中，我无法声明变量ArrayList<AnotherObject> 1;，还有比手工替换数字更好的方法吗？？

Json（大大减少了）：

{
    "object": {
        "1": [{...}],
        "2": [{...}],
        "3": [{...}],
        "4": [{...}],
        "5": [{...}],
        "6": [{...}],
        "7": [{...}]
    }
}

预先感谢

Answer 1

这是使用GSON来反序列化JSON的方式：

public static void main(String[] args) {
    final String json = "{\n" +
        "    \"object\": {\n" +
        "        \"1\": [{ \"id\" : 111 }],\n" +
        "        \"2\": [{ \"id\" : 222 }],\n" +
        "        \"3\": [{ \"id\" : 333 }]\n" +
        "    }\n" +
        "}\n";
    final Gson gson = new GsonBuilder()
        .create();
    final ObjectWrapper value = gson.fromJson(json, ObjectWrapper.class);

    System.out.println(value.object);
    System.out.println(value.object.keySet());
    System.out.println(value.object.get(1));
}

// This is top-most object we want to deserialize from JSON
static class ObjectWrapper {
    // Didn't bother about proper naming while it is better to give a meaningful name here
    private Map<Integer, List<Element>> object;
}

static class Element {
    // Added this attribute to demonstrate that objects within array are properly read
    private int id;
    @Override
    public String toString() {
        return "{id=" + id + "}";
    }
}

Answer 2

这些数字看起来很像数组的索引。如果是这样，您可以删除所有这些对象，并将整个对象设置为数组。这样您将获得一个数组数组。

Answer 3

我现在正在使用以下解决方法：

json.replace("\"1\"","\"one\"")
    .replace("\"2\"","\"two\"")
    .replace("\"3\"","\"three\"")
    .replace("\"4\"","\"four\"")
    .replace("\"5\"","\"five\"")
    .replace("\"6\"","\"six\"")
    .replace("\"7\"","\"seven\"");

编辑：此解决方案确实有效，但是非常难看。 @yegodm的解决方案效果更好！谢谢你！

Answer 4

解决此问题的方法是使代码通过以下方式转换JSON字符串：

{
    "object": {
        "a1": [{...}],
        "a2": [{...}],
        "a3": [{...}],
        "a4": [{...}],
        "a5": [{...}],
        "a6": [{...}],
        "a7": [{...}]
    }
}

当对象名称为Integer时，Java GSON反序列化JSON

4 个答案: