Question

我有两个描述所有数据的类和一个描述数据类型的枚举

class Category(val name: String,
               val id: String,
               val type: CategoryType,
               val items: List<Item>)

class Item(val name: String,
           val id: String,
           val type: CategoryType,
           val data: Any?,
           val categoryIds: Array<String>)

enum class CategoryType {
    TYPE_A,
    TYPE_B,
    TYPE_C
}

并遵循JSON输入

{
  "categories": [
    {
      "id": "category_id_1",
      "name": "Category 1",
      "type1": "TYPE_A",
      "type2": "TYPE_C"
    },
    {
      "id": "category_id_2",
      "name": "Category 2",
      "type1": "TYPE_B"
    },
    {
      "id": "category_id_3",
      "name": "Category 3",
      "type1": "TYPE_A"
    },
    {
      "id": "category_id_4",
      "name": "Category 4",
      "type1": "TYPE_C"
    }
  ],
  "items": [
    {
      "id": "item_id_1",
      "name": "Item 1",
      "type2": "TYPE_A",
      "categoryIds": [
        "category_id_4",
        "category_id_3"
      ]
    },
    {
      "id": "item_id_2",
      "name": "Item 2",
      "type1": "TYPE_C",
      "categoryIds": [
        "category_id_1"
      ]
    }
  ]
}

有两个主要要求：

我需要创建两个Category的实例，如果有的话id 在那里列出了type1和type2。
如果项目具有相同的Item和/或Category和type1的{{}，我需要在每个type2个实例中放置Category的相同实例1}}列在id的{{1}}。

对于每个Item，我可以使用任何类型的容器，例如categoryIds。

所以问题是：如果不使用庞大的CategoryType构造，我怎么能这样做呢？是否可以使用val typedAsTYPE_A: List<Category>完成，或者我应该使用类似com.google.gson.JsonDeserializer的内容？

Answer 1

这适用于您提供的Json。

Gson gson = new Gson();
String json = YOURJSON
ContainerObject myObj = gson.fromJson(json, ContainerObject.class);
//above code is placed where you want to deserialize the object

import java.util.ArrayList;

public class ContainerObject {
    private ArrayList<Categories> categories;
    private ArrayList<Item> items;

    public ContainerObject(){

    }

    public ArrayList<Categories> getCategories() {
        return categories;
    }

    public void setCategories(ArrayList<Categories> categories) {
        this.categories = categories;
    }

    public ArrayList<Item> getItems() {
        return items;
    }

    public void setItems(ArrayList<Item> items) {
        this.items = items;
    }

}

import java.util.List;

public class Categories {
    private String id;
    private String name;
    private CategoryType type1;
    private CategoryType type2;

    public Categories(){

    }

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public CategoryType getType1() {
        return type1;
    }

    public void setType1(CategoryType type1) {
        this.type1 = type1;
    }

    public CategoryType getType2() {
        return type2;
    }

    public void setType2(CategoryType type2) {
        this.type2 = type2;
    }

}


import java.util.ArrayList;

public class Item {
    private String id;
    private String name;
    private CategoryType type1;
    private CategoryType type2;
    private ArrayList<String> categoryIds;

    public Item(){

    }

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public CategoryType getType1() {
        return type1;
    }

    public void setType1(CategoryType type1) {
        this.type1 = type1;
    }

    public CategoryType getType2() {
        return type2;
    }

    public void setType2(CategoryType type2) {
        this.type2 = type2;
    }

    public ArrayList<String> getCategoryIds() {
        return categoryIds;
    }

    public void setCategoryIds(ArrayList<String> categoryIds) {
        this.categoryIds = categoryIds;
    }
}


public enum CategoryType {
    TYPE_A,
    TYPE_B,
    TYPE_C
}

Answer 2

免责声明：我对Gson并不熟悉，但既然你想使用它，我就试着把它放在一起。最有可能有更好的方法来实现相同的结果。代码可能不健壮，可能包含错误。

我希望我的代码足够可理解。基本上就是这些步骤：

使用Gson解析com.google.gson.JsonObject
使用JsonObject迭代每个项目并将其添加到适当的列表中（如果不存在则创建列表）
遍历每个类别并使用已生成的列表添加项目

以下是代码：

// Helper extension property for easily mapping Strings
// to their enum CategoryType
val String.catType: CategoryType?
    get() = CategoryType.values().find { it.toString() == this }

fun deserialize(json: String): Map<CategoryType, List<Category>> {
    // initialize the return value
    val categories = CategoryType.values().map { it to (mutableListOf<Category>()) }.toMap()
    // parse json
    val input = Gson().fromJson(json, com.google.gson.JsonObject::class.java)


    // accumulate itemLists for constructing Categories later
    // those lists will be looked up via CategoryType and CategoryId
    val itemLists = CategoryType.values().map { it to (mutableMapOf<String, MutableList<Item>>()) }.toMap()
    input["items"].asJsonArray.map { it.asJsonObject }.forEach { item ->
        // read properties for item
        val types = item.entrySet().filter { it.key.startsWith("type") }.map { it.value.asString.catType }
        val ids = item.getAsJsonArray("categoryIds").map { it.asString }
        val name = item["name"].asString
        val item_id = item["id"].asString
        val data = item["data"]
        //
        types.forEach { type ->
            type?.let {
                val itemObj = Item(name, item_id, type, data, ids.toTypedArray())
                ids.forEach { id ->
                    with(itemLists[type]!!) {
                        if (!this?.containsKey(id)) this[id] = mutableListOf(itemObj)
                        else this[id]?.add(itemObj)
                    }
                }
            }
        }
    }

    // go through all categories and create appropriate objects
    input["categories"].asJsonArray.map { it.asJsonObject }.forEach { cat ->
        val id = cat["id"].asString
        val name = cat["name"].asString
        val types = cat.entrySet().filter { it.key.startsWith("type") }.map { it.value.asString.catType }
        types.forEach { type ->
            type?.let {
                categories[type]
                        ?.add(Category(
                                name = name,
                                id = id,
                                type = type,
                                items = itemLists[type]?.get(id) ?: listOf()))
            }
        }
    }

    return categories
}

Gson：反序列化链接对象的列表

2 个答案: