排序并查找与当前时间最接近的时间

Question

I have an array of times

    ["10:00 PM", "08:00 AM", "12:00 AM", "01:00 AM", "12:00 PM", 
     "03:00 AM", "07:00 AM", "06:00 PM"]

I want to sort them and find the nearest time from the current time, For example, Assume the time is now 05:00 PM, the above array should return the 06:00 PM as response.

i can able to sort them with below code

    let sortedArray = arrayOfData.sort(function (a, b) {
    return parseInt(a.substring(0, 2)) - 
    parseInt(b.substring(0, 2));
    }) 

Can some one suggest a way to sort it properly and also to find the closest time from using the current time? Thanks in advance

Answer 1

Just add the difference between current hour and array hours in a separate array and sort it ascending and get the first element from array this will be the most appropriate hour.

Check the following snippet:

times = ["10:00 PM","7:00 PM", "08:00 AM", "12:00 AM", "01:00 AM", "12:00 PM", 
     "03:00 AM", "07:00 AM", "06:00 PM"];

const currentTime = new Date();
const timeDiff = [];

times.sort((a, b) => {
  return a.indexOf('PM');
})

times.filter(time => {
  const _meridianPosition = time.indexOf('AM') > -1 ? 'AM' : 'PM';

  let _time = parseInt(time);

    if(_meridianPosition === 'PM' && _time !== 12) {
      _time += 12;
    } else if(_meridianPosition === 'AM' && _time === 12) {
      _time = 0;
    }

    const k = Math.abs(currentTime.getHours() - _time);
     timeDiff.push({hour: time, diff: k});
});

timeDiff.sort((a,b) => {
  return a.diff - b.diff;
});

console.log(timeDiff[0].hour);

Working fiddle: https://jsbin.com/zojawagiyi/6/edit?js,console

Answer 2

You can use the solution below, that sort the array and after this find the nearest date from the current time.

First, the code adds the current time together with the array, and after this it gets the nearest date.

 let dates = ["10:00 PM", "08:00 AM", "12:00 AM", "01:00 AM", "12:00 PM", "03:00 AM", "07:00 AM", "06:00 PM"]; let currentDate = new Date(); let currentTime = currentDate.getHours() + ':' + currentDate.getMinutes() + (currentDate.getHours() > 12 ? ' PM' : ' AM'); dates.push(currentTime); dates = dates.sort(function(d1, d2) { return compareDates(d1, d2); }); console.log(dates); console.log(nearestDate(dates, currentTime)); function nearestDate(dates, current) { let currentIndex = dates.indexOf(current); if(currentIndex == 0) { return dates[currentIndex + 1]; } else if (currentIndex == dates.length - 1) { return dates[currentIndex - 1]; } let previousDate = dates[currentIndex - 1]; let nextDate = dates[currentIndex + 1]; let previousDiff = diffDates(previousDate, currentTime); let nextDiff = diffDates(nextDate, currentTime); if(previousDiff < nextDiff) { return previousDate; } else { return nextDate; } } function diffDates(d1, d2) { let diffHour = Math.abs(getHour(d2) - getHour(d1)); let diffMin = Math.abs(getMin(d2) - getMin(d1)); return diffHour + diffMin; } function compareDates(d1, d2) { let t1 = getHour(d1) + ':' + getMin(d1); let t2 = getHour(d2) + ':' + getMin(d2); if (getHour(d1) == getHour(d2) && getMin(d1) < getMin(d2)) { return -1; } else if(getHour(d1) == getHour(d2) && getMin(d1) > getMin(d2)) { return 1; } if (getHour(d1) < getHour(d2)) { return -1; } if (getHour(d1) > getHour(d2)) { return 1; } return 0; } function getHour(d) { let hour = parseInt(d.split(' ')[0].split(':')[0], 10); if (d.split(' ')[1] === 'PM' && !(hour == 12)) { hour += 12; } return hour; } function getMin(d) { return parseInt(d.split(' ')[0].split(':')[1], 10); } 

Answer 3

I think this code will do the work. You can try this.

 let currentTime = new Date(); let currentHour = parseInt(currentTime.getHours()); let availableDates = ["10:00 PM", "08:00 AM", "12:00 AM", "01:00 AM", "12:00 PM", "03:00 AM", "07:00 AM", "06:00 PM"]; let convertedHours = availableDates.map((date) => { let time = parseInt(date.split(' ')[0]); let period = date.split(' ')[1]; if(time === 12 && period === 'PM' ) return time; if(time < 12 && period === 'AM') return time; return time + 12; }); let getNearestTime = (convertedHours, currentHour) => { let nearestTime; let minValue = convertedHours[0] > currentHour ? (convertedHours[0] - currentHour) : (currentHour - convertedHours[0]); convertedHours.reduce((minVal, hour) => { let hourDiff = (currentHour > hour) ? currentHour - hour : hour - currentHour; if(hourDiff <= minVal) { nearestTime = hour; return hourDiff; } else { return minVal; } }, minValue) return availableDates[convertedHours.indexOf(nearestTime)]; }; console.log(getNearestTime(convertedHours, currentHour)); 

Here is the jsbin link https://jsbin.com/piwuziqeje/edit?js,console

Answer 4

I tried a different/more intuitive approach than the ones I see here. It might be a little longer than some, but it's more clear in what it's doing in my opinion. The code works as you can check in the fiddle. Here is the code:

var times = ["10:00 PM", "08:00 AM", "12:00 AM", "01:00 AM", "12:00 PM", 
 "03:00 AM", "07:00 AM", "06:00 PM"];

//Sort the array
times.sort(function (a, b) {
return new Date('1970/01/01 ' + a) - new Date('1970/01/01 ' + b);
});

//Test Sorted Array
console.log(times);

var testTime = "05:00 PM";

function findNearestTime(times, currentTime) {

//Copy given array to new array
var allTimes = times.slice();

//Push current time to new arrray
allTimes.push(currentTime);

//Sort New array
allTimes.sort(function (a, b) {
return new Date('1970/01/01 ' + a) - new    Date('1970/01/01 ' + b);
});

//Nearest time will be either the item to the left or to the right of currentTime since array is sorted
//Now we just find which one is the closest
var indexOfCurrent = allTimes.indexOf(currentTime);

if (indexOfCurrent == 0) { //if current is first element, nearest will be item 
//after first element
return allTimes.slice(indexOfCurrent + 1, indexOfCurrent + 2 );
}else if (indexOfCurrent == allTimes.length - 1) { //current is last one, 
//nearest will be the item before current
return allTimes.slice(allTimes.length - 2, indexOfCurrent);
}else { //if neither case above, this is where magic happens
//Find the diff between left/right adjacent element and the current element in the new sorted array 
var currTime = new Date("01/01/2018 " + currentTime).getHours();

var currTimeLower = new Date("01/01/2018 " + allTimes.slice(indexOfCurrent - 1, 
indexOfCurrent)).getHours();

var currTimeUpper = new Date("01/01/2018 " + allTimes.slice(indexOfCurrent + 1, 
indexOfCurrent + 2)).getHours();

var leftDiff  = currTime - currTimeLower;
var rightDiff = currTimeUpper - currTime;

if(leftDiff < rightDiff) {
  return allTimes.slice(indexOfCurrent - 1, indexOfCurrent);
}
else {
  return allTimes.slice(indexOfCurrent + 1, indexOfCurrent + 2);
}

};
}

console.log(findNearestTime(times, testTime));

Here is the working fiddle. I tested with different times and it works. https://jsfiddle.net/b36fxpqr/13/

Answer 5

You can try this small code.

  var timeSrc = ["10:00 PM", "08:00 AM", "11:05 AM", "12:00 AM", "01:00 AM", "12:00 PM", "03:00 AM", "07:00 AM", "06:00 PM"]; var curDate = new Date(); curDate = curDate.toDateString(); var times = timeSrc.map((t) => { return new Date(curDate + " " + t); // Make the time as a datetime with current date. }); var now = new Date(); var min = Math.abs(now - times[0]); var result = ''; //Get the difference of each time with current time. The minimum difference is the closest. for(let i = 1; i < times.length; i++) { if (Math.abs(now - times[i]) <= min) { min = Math.abs(now - times[i]); result = timeSrc[i]; } } console.log(result); 

You can try it here

Answer 6

 var arrayofDate = ["10:00 PM", "08:00 AM", "12:00 AM", "01:00 AM", "12:00 PM", "03:00 AM", "07:00 AM", "06:00 PM"]; var railwayTime = arrayofDate.map((data, key) => { data = parseInt(data.substr(0,2)); if(arrayofDate[key].indexOf('PM') !== -1) { data = data + 12; } return data; }); var output = closestTime(new Date().getHours(), railwayTime); document.getElementById('result').innerHTML = arrayofDate[railwayTime.indexOf(output)]; function closestTime (num, arr) { var curr = arr[0]; var diff = Math.abs (num - curr); for (var val = 0; val < arr.length; val++) { var newdiff = Math.abs (num - arr[val]); if (newdiff < diff) { diff = newdiff; curr = arr[val]; } } return curr; } 

 <div id="result"></div> 

Answer 7

You can also try this, But havnt tested it with more test case, correct me if i'm wrong `

var a = ["10:00 PM", "08:00 AM", "12:00 AM", "01:00 AM", "12:00 PM","03:00 AM", "07:00 AM", "06:00 PM"]
var findhour = new Date().getHours()
var ans = ""
var arrayNum = ""
for(i=0;i<a.length;i++){    
   temp = a[i].split(':')
   if (a[i].includes('PM')){
      temp1 = (12 + +temp[0])%24
      document.write(temp1+"\n")
   }else{
      temp1 = temp[0]
      document.write(temp1+"\n")
   }
   if( Math.abs(ans-findhour) > Math.abs(temp1-findhour)){
      ans = temp1
      arrayNum = i;
      console.log(ans)
   }    
}
document.write("ans  " + a[arrayNum])

`