Question

我尝试了以下代码，我希望获得最新日期的记录。在以下示例中，我需要具有最新和相同日期的记录带键2和5的平均数组。我试过以下但没有成功。

var arr = [];

arr['0'] = { label : '2003', Date : '2003-11-17 00:00:00'};
arr['1'] = { label : '2007', Date : '2003-11-17 00:00:00'};
arr['2'] = { label : '2009', Date : '2010-11-17 00:00:00'};
arr['3'] = { label : '2008', Date : '2005-11-17 00:00:00'};
arr['4'] = { label : '2008', Date : '2005-11-17 00:00:00'};
arr['5'] = { label : '2010', Date : '2010-11-17 00:00:00'};

console.log(arr);

var max = arr.reduce(function(a, b) {
    return (moment(a.Date).isAfter(b.Date));
}, 0);

console.log(max);

Answer 1

首先根据日期索引数组

var dateMap = {};
arr.forEach( function(item){
  dateMap[ item.Date ] = dateMap[ item.Date ] || [];
  dateMap[ item.Date ].push( item );
});

现在对键进行排序，简单的字符串排序就可以了

var keys = Object.keys( dateMap );
keys.sort();

由于数组按升序排序，因此获取最大值为

的最后一个键

var maxDate = keys.pop();

现在获取与此最大日期相关联的所有数组项

return dateMap[ maxDate ];

Answer 2

您可以对数组进行排序和过滤：

＆＃13;

private void bindHandler() {
    new Handler().postDelayed(new Runnable() {
        @Override
        public void run() {
            callNext();
            Log.e("In HANDLER------>", "HANDLER");
        }
    }, DISPLAY_LENGTH * 1000);
}



  private void callNext() {
    if (counter < statusModel.getStories().get(position).getUserStories().size()) {
        if (statusModel.getStories().get(position).getUserStories().get(counter).getType().equalsIgnoreCase("photo")) {
            Log.e("PHOTOOOO--->", "PHOTOOOOO");

            String photo = statusModel.getStories().get(position).getUserStories().get(counter).getFileName();
            image.setVisibility(View.VISIBLE);
            video_view.setVisibility(View.GONE);
            Glide.with(getApplicationContext())
                    .load(photo)
                    .into(image);
            counter++;
            DISPLAY_LENGTH = Long.parseLong(statusModel.getStories().get(position).getUserStories().get(counter).getDisplayTime());
            Log.e("DIAPLAY TIME--->", String.valueOf(DISPLAY_LENGTH));
            bindHandler();
        } else {
            if (statusModel.getStories().get(position).getUserStories().get(counter).getType().equalsIgnoreCase("video")) {
                Log.e("VIDEOOOO--->", "VIDEOOOOO");
                image.setVisibility(View.GONE);
                video_view.setVisibility(View.VISIBLE);
                String VideoURL = statusModel.getStories().get(position).getUserStories().get(counter).getFileName();
                uri = Uri.parse(VideoURL);
                if (!TextUtils.isEmpty(VideoURL)) {
                    video_view.setVideoURI(uri);
                    video_view.requestFocus();
                    video_view.start();
                    video_view.setOnPreparedListener(new MediaPlayer.OnPreparedListener() {
                        @Override
                        public void onPrepared(MediaPlayer mediaPlayer) {

                        }
                    });

                    video_view.setOnCompletionListener(new MediaPlayer.OnCompletionListener() {
                        @Override
                        public void onCompletion(MediaPlayer mediaPlayer) {
                            Log.e("VIDEooooooo--->", "Completeeeee");

   Log.e("Thread Notifyyyy--->", "Hooo Gayaaaaaa");
                            counter++;
                            callNext();
                        }
                    });
                }
            }
        }
    }

}

＆＃13;

var arr = [];
arr['0'] = { label : '2003', Date : '2003-11-17 00:00:00'};
arr['1'] = { label : '2007', Date : '2003-11-17 00:00:00'};
arr['2'] = { label : '2009', Date : '2010-11-17 00:00:00'};
arr['3'] = { label : '2008', Date : '2005-11-17 00:00:00'};
arr['4'] = { label : '2008', Date : '2005-11-17 00:00:00'};
arr['5'] = { label : '2010', Date : '2010-11-17 00:00:00'};

var max = arr
  .sort((a, b) => moment(a.Date).isAfter(b.Date))
  .filter((elem, index, arr) => elem.Date === arr[arr.length - 1].Date);

console.log(max);

＆＃13;

Answer 3

这个简单的reduce函数将通过将当前日期与保留的最大值进行比较（使用累加器存储最大值）来完成工作。对于非常大的数组，这应该比排序和过滤更快，因为它只需要访问一次数组。

var max = arr.reduce((max, d) => max > d.Date ? max : d.Date, 0);

更详细代码中的等效版本：

var max = arr.reduce(function(max, d) {
    if (max > d.Date) {
    return max;
    } else { 
        return d.Date;
    }
}, 0);

如果需要完整值，可以使用对象作为累加器（注意对初始累加器值的更改）：

var max = arr.reduce((max, d) => max.Date > d.Date ? max : d, {Date: 0});

或

var max = arr.reduce(function(max, d) {
    if (max.Date > d.Date) {
    return max;
    } else { 
        return d;
    }
}, {Date: 0});

Answer 4

您可以使用array#reduce获取具有最大日期的对象，然后使用array#filter获取该日期的所有对象。

var arr = [];

arr['0'] = { label : '2003', Date : '2003-11-17 00:00:00'};
arr['1'] = { label : '2007', Date : '2003-11-17 00:00:00'};
arr['2'] = { label : '2009', Date : '2010-11-17 00:00:00'};
arr['3'] = { label : '2008', Date : '2005-11-17 00:00:00'};
arr['4'] = { label : '2008', Date : '2005-11-17 00:00:00'};
arr['5'] = { label : '2010', Date : '2010-11-17 00:00:00'};

var max = arr.reduce(function(prev,cur){
    return moment(prev.Date).isAfter(moment(cur.Date)) ? prev : cur;
});

var result = arr.filter(function(obj){
  return max.Date === obj.Date;
});

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.18.1/moment.min.js"></script>

Answer 5

使用Math.max，map和filter的其他解决方案。

var arr = [];
arr['0'] = { label: '2003', Date: '2003-11-17 00:00:00' };
arr['1'] = { label: '2007', Date: '2003-11-17 00:00:00' };
arr['2'] = { label: '2009', Date: '2010-11-17 00:00:00' };
arr['3'] = { label: '2008', Date: '2005-11-17 00:00:00' };
arr['4'] = { label: '2008', Date: '2005-11-17 00:00:00' };
arr['5'] = { label: '2010', Date: '2010-11-17 00:00:00' };

//Get max year
var year = Math.max.apply(null, arr.map(function(e) { return new Date(e.Date.replace(" ","T")).getFullYear() }));
//Filter array based on year
console.log(arr.filter(function (a,b) { return new Date(a.Date.replace(" ","T")).getFullYear() === year }))

基于多维数组Javascript中的键获取所有值，该键具有最高值

5 个答案: