在 JavaScript/ES6 中获取第二大日期

Question

I have a problem getting the second highest date in ES6. I'm using moment.js too. Its supposed to be getting the id of 3.

 const datas = [ { id: 1, date: moment(String('Apple & Banana - 20072021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() }, { id: 2, date: moment(String('Apple & Oranges - 30082021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() }, { id: 3, date: moment(String('Lemon & Oranges - 30102021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() }, { id: 4, date: moment(String('Honeydew - 30112021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() } ]; const secondLatestDate = new Date(datas.map(file => new Date(file.date)).sort().reverse()[1]); const finalResult = datas.find(file => file.date.getTime() === secondLatestDate.getTime()); console.log(finalResult)

 <script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.29.1/moment.min.js"></script>

Answer 1

You should use custom sort function as:

datas.sort((a, b) => a.date - b.date)

There is no need to use find when you are reverse ing the array and getting the index 1 from it.

Note: I deliberately change the order of the datas array

 const datas = [{ id: 1, date: moment(String('Apple & Banana - 20072021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() }, { id: 2, date: moment(String('Apple & Oranges - 30082021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() }, { id: 4, date: moment(String('Honeydew - 30112021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() }, { id: 3, date: moment(String('Lemon & Oranges - 30102021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate() } ]; const secondLatestDate = datas.sort((a, b) => a.date - b.date).reverse()[1]; console.log(secondLatestDate);

 <script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.29.1/moment.min.js"></script>

or you can directly find the second largest after sort. There is no need to reverse the array

datas.sort((a, b) => a.date - b.date)[datas.length - 2]

 const datas = [{ id: 1, date: moment( String('Apple & Banana - 20072021').match(/[0-9]/g).join(''), 'DDMMYYYY' ).toDate(), }, { id: 2, date: moment( String('Apple & Oranges - 30082021').match(/[0-9]/g).join(''), 'DDMMYYYY' ).toDate(), }, { id: 4, date: moment( String('Honeydew - 30112021').match(/[0-9]/g).join(''), 'DDMMYYYY' ).toDate(), }, { id: 3, date: moment( String('Lemon & Oranges - 30102021').match(/[0-9]/g).join(''), 'DDMMYYYY' ).toDate(), }, ]; const secondLatestDate = datas.sort((a, b) => a.date - b.date)[datas.length - 2]; console.log(secondLatestDate);

 <script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.29.1/moment.min.js"></script>

Answer 2

If you're at all concerned with time complexity you could do this in one 'n' by implementing a function that keeps track of both the highest and second highest values. Maybe something like this:

const datas = [{
    id: 1,
    date: moment(String('Apple & Banana - 20072021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate()
  },
  {
    id: 2,
    date: moment(String('Apple & Oranges - 30082021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate()
  },
  {
    id: 4,
    date: moment(String('Honeydew - 30112021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate()
  },
  {
    id: 3,
    date: moment(String('Lemon & Oranges - 30102021').match(/[0-9]/g).join(""), 'DDMMYYYY').toDate()
  }
]


function getSecondHighest(dates){
    let highest = 0;
    let secondHighest = 0;

    for(const val of dates){
        const currentDate = val.date
        if(currentDate > highest){
            secondHighest = highest;
            highest = currentDate;
        } else if(currentDate > secondHighest){
            secondHighest = currentDate
        } else {
            continue;
        }
    }

    return secondHighest;
}