你可以使用TypeToken将json字符串加载到一个自定义对象中。

logs = gson.fromJson(br, new TypeToken<List<JsonLog>>(){}.getType());

文档:

Represents a generic type T. Java doesn't yet provide a way to represent generic types, so this class does. Forces clients to create a subclass of this class which enables retrieval the type information even at runtime. For example, to create a type literal for List<String>, you can create an empty anonymous inner class: TypeToken<List<String>> list = new TypeToken<List<String>>() {}; This syntax cannot be used to create type literals that have wildcard parameters, such as Class<?> or List<? extends CharSequence>.

科特林：

如果你需要在Kotlin中这样做，你可以这样做:

val myType = object : TypeToken<List<JsonLong>>() {}.type
val logs = gson.fromJson<List<JsonLong>>(br, myType)

或者你可以从其他选项中看到这个答案。

你的JSON样本是:

{
    "status": "ok",
    "comment": "",
    "result": {
    "id": 276,
    "firstName": "mohamed",
    "lastName": "hussien",
    "players": [
            "player 1",
            "player 2",
            "player 3",
            "player 4",
            "player 5"
    ]
}

所以如果你想在你的sharedpreferences中保存模块的数组列表，那么:

1-使用此方法将返回的数组列表转换为json格式

public static String toJson(Object jsonObject) {
    return new Gson().toJson(jsonObject);
}

2-保存在共享prefreneces中

PreferencesUtils.getInstance(context).setString("players", toJson((.....ArrayList you want to convert.....)));

3-检索它在任何时候得到JsonString从共享首选项

String playersString= PreferencesUtils.getInstance(this).getString("players");

4-再次转换为数组列表

public static Object fromJson(String jsonString, Type type) {
    return new Gson().fromJson(jsonString, type);
}

ArrayList<String> playersList= (ArrayList<String>) fromJson(playersString,
                    new TypeToken<ArrayList<String>>() {
                    }.getType());

如果你想解析对象的数组列表，这个解决方案也是可行的 希望对你使用Gson库有所帮助。

我不确定gson怎么样但这就是你和Jon相处的方式。样品希望一定有类似的方法使用gson

｛ “玩家”:( “玩家1”, “玩家2”, “玩家3”, “球员4”, “球员5” ］ ｝

===============================================

import java.io.FileReader;
import java.util.List;

import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;

public class JosnFileDemo {

    public static void main(String[] args) throws Exception
    {
        String jsonfile ="fileloaction/fileName.json";


                FileReader reader = null;
                JSONObject jsb = null;
                try {
                    reader = new FileReader(jsonfile);
                    JSONParser jsonParser = new JSONParser();
                     jsb = (JSONObject) jsonParser.parse(reader);
                } catch (Exception e) {
                    throw new Exception(e);
                } finally {
                    if (reader != null)
                        reader.close();
                }
                List<String> Players=(List<String>) jsb.get("Players");
                for (String player : Players) {
                    System.out.println(player);
                }           
    }
}

如果你想使用数组，这很简单。

logs = gson.fromJson(br, JsonLog[].class); // line 6

提供JsonLog作为数组JsonLog[].class

科特林

data class Player(val name : String, val surname: String)

val json = [
  {
    "name": "name 1",
    "surname": "surname 1"
  },
  {
    "name": "name 2",
    "surname": "surname 2"
  },
  {
    "name": "name 3",
    "surname": "surname 3"
  }
]

val typeToken = object : TypeToken<List<Player>>() {}.type
val playerArray = Gson().fromJson<List<Player>>(json, typeToken)

OR

val playerArray = Gson().fromJson(json, Array<Player>::class.java)

为什么没有人写这个简单的方式转换JSON字符串在列表?

List<Object> list = Arrays.asList(new GsonBuilder().create().fromJson(jsonString, Object[].class));

假设，你有一条这样的弦。

"[{"id":2550,"cityName":"Langkawi","hotelName":"favehotel Cenang Beach - Langkawi","hotelId":"H1266070"},
{"id":2551,"cityName":"Kuala Lumpur","hotelName":"Metro Hotel Bukit Bintang","hotelId":"H835758"}]"

然后你可以通过Gson把它转换成数组列表

var hotels = Gson().fromJson(historyItem.hotels, Array<HotelInfo>::class.java).toList()

您的HotelInfo类应该是这样的。

import com.squareup.moshi.Json

data class HotelInfo(

    @Json(name="cityName")
    val cityName: String? = null,

    @Json(name="id")
    val id: Int? = null,

    @Json(name="hotelId")
    val hotelId: String? = null,

    @Json(name="hotelName")
    val hotelName: String? = null
)

如果你想将Json转换为类型化的ArrayList，指定列表中包含的对象的类型是错误的。正确的语法如下:

 Gson gson = new Gson(); 
 List<MyClass> myList = gson.fromJson(inputString, ArrayList.class);

在芬兰湾的科特林 用于发送:如果发送Arraylist

Gson().toJson(arraylist)

接收:如果接收ArrayList

var arraylist = Gson().fromJson(argument, object : TypeToken<ArrayList<LatLng>>() {}.type)  

用于发送:如果您发送ModelClass(例如latlnmodel .class)

var latlngmodel = LatlngModel()
latlngmodel.lat = 32.0154
latlngmodel.lng = 70.1254

Gson().toJson(latlngModel)

用于接收:如果您接收ModelClass

var arraylist = Gson().fromJson(argument,LatLngModel::class.java )