使用Gson库,如何将JSON字符串转换为自定义类JsonLog的数组列表?基本上,JsonLog是由我的Android应用程序制作的不同类型的日志实现的接口——短信日志、通话日志、数据日志——这个数组列表是所有这些日志的集合。第6行总是出错。

public static void log(File destination, JsonLog log) {
    Collection<JsonLog> logs = null;
    if (destination.exists()) {
        Gson gson = new Gson();
        BufferedReader br = new BufferedReader(new FileReader(destination));
        logs = gson.fromJson(br, ArrayList<JsonLog>.class); // line 6
        // logs.add(log);
        // serialize "logs" again
    }
}

编译器似乎不明白我引用的是一个类型化的数组列表。我该怎么办?


当前回答

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。

其他回答

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);

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。

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)