Question

我有一个巨大的静态JSON文件，其中包含某个游戏中的所有冠军以及有关他们的一些信息。当前，我需要弄清楚如何找到其“关键”属性为266的冠军对象。可悲的是，我在弄清楚如何获取该对象方面遇到了一些困难。我假设我必须遍历对象，但是，通常当我遍历某些东西时，它是一个数组，但事实并非如此。

我在下面放了一个json结构的简化示例。我有一个整数266，现在我需要以某种方式使用“ key”访问该对象：“ 266”，这将是Aatrox。关于我该怎么做的任何线索？

"data": {
    "Aatrox": {
        "version": "8.19.1",
        "id": "Aatrox",
        "key": "266",
        "name": "Aatrox",
        "title": "the Darkin Blade"
    },
    "Ahri": {
        "version": "8.19.1",
        "id": "Ahri",
        "key": "103",
        "name": "Ahri",
        "title": "the Nine-Tailed Fox"
    }
}

Answer 1

您可以使用find

使用Object.entries从数据密钥获取条目
使用find来查找键属性等于我们期望的键的值

let obj = {"data": {"Aatrox": {"version": "8.19.1","id": "Aatrox","key": "266","name": "Aatrox","title": "the Darkin Blade"},"Ahri": {"version": "8.19.1","id": "Ahri","key": "103","name": "Ahri","title": "the Nine-Tailed Fox"}}}

let findByKey = (matchKey) => Object.entries(obj.data).find(([key,value]) => value.key === matchKey)

console.log(findByKey('266'))

Answer 2

这是一个通用函数，可用于搜索任何键和值：

const obj = {"data": {"Aatrox": {"version": "8.19.1","id": "Aatrox","key": "266","name": "Aatrox","title": "the Darkin Blade"},"Ahri": {"version": "8.19.1","id": "Ahri","key": "103","name": "Ahri","title": "the Nine-Tailed Fox"}}}

const searchKeyValue = {key:"key", value:"266"}
const search = data => skv => Object.entries(data).filter(([key,value])=>value[skv.key]===skv.value)

console.log(search(obj.data)(searchKeyValue))

因此您可以通过例如searchKeyValue = {key:"id", value:"Ahri"}或任何您喜欢的东西。

Answer 3

您也可以使用JSONPath，即

return step.PromptAsync("oauth", new PromptOptions { });

假设：

public class MainActivity extends AppCompatActivity implements TextureView.SurfaceTextureListener {

    private MediaPlayer mp1, mp2, mp3;
    private TextureView tv1, tv2, tv3;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        tv1 = findViewById(R.id.textureView1);
        tv2 = findViewById(R.id.textureView2);
        tv3 = findViewById(R.id.textureView3);

        tv1.setSurfaceTextureListener(this);
        tv2.setSurfaceTextureListener(this);
        tv3.setSurfaceTextureListener(this);

    }


    @Override
    public void onSurfaceTextureAvailable(SurfaceTexture surfaceTexture, int width, int height) {
        Surface surface = new Surface(surfaceTexture);
        mp1 = MediaPlayer.create(this, R.raw.a7);
        mp1.setSurface(surface);
//        mp1.prepareAsync(); //
        mp1.setOnPreparedListener(new MediaPlayer.OnPreparedListener() {
            @Override
            public void onPrepared(MediaPlayer mp1) {
                mp1.start();
            }
        });

        Surface surface2 = new Surface(surfaceTexture);
        mp2 = MediaPlayer.create(this, R.raw.a9);
        mp2.setSurface(surface2);
//        mp1.prepareAsync(); //
        mp2.setOnPreparedListener(new MediaPlayer.OnPreparedListener() {
            @Override
            public void onPrepared(MediaPlayer mp2) {
                mp2.start();
            }
        });

        Surface surface3 = new Surface(surfaceTexture);
        mp3 = MediaPlayer.create(this, R.raw.a10);
        mp3.setSurface(surface3);
//        mp1.prepareAsync(); //
        mp3.setOnPreparedListener(new MediaPlayer.OnPreparedListener() {
            @Override
            public void onPrepared(MediaPlayer mp3) {
                mp3.start();
            }
        });
    }

    @Override
    public void onSurfaceTextureSizeChanged(SurfaceTexture surface, int width, int height) {

    }

    @Override
    public boolean onSurfaceTextureDestroyed(SurfaceTexture surface) {
        return false;
    }

    @Override
    public void onSurfaceTextureUpdated(SurfaceTexture surface) {

    }

    @Override
    protected void onPause() {
        if (mp1 != null && mp1.isPlaying()) {
            mp1.pause();
        }
        super.onPause();
    }

    @Override
    protected void onResume() {
        if (mp1 != null) {
            mp1.start();
        }
        super.onResume();
    }

    @Override
    protected void onDestroy() {
        if (mp1 != null) {
            mp1.stop();
            mp1.release();
            mp1 = null;
        }
        super.onDestroy();
    }
}

我有一个带有data属性的JSON文件，其中包含包含有关冠军的信息的对象。如何获得具有特定KEY属性的冠军？

3 个答案: