排序并查找与当前时间最接近的时间

Question

我有很多次

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

我想对它们进行排序并查找与当前时间最近的时间，例如，假设时间现在是05:00 PM，则上面的数组应返回 06:00 PM 作为响应

我可以使用以下代码对其进行排序

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

有人可以提出一种对它进行正确排序并从当前时间中找到最接近时间的方法吗？预先感谢

Answer 1

只需在单独的array和sort中将当前小时与数组小时之间的差值加起来，然后从array中获得第一个元素，这将是最合适的小时。

检查以下代码段：

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);

正在工作的小提琴：https://jsbin.com/zojawagiyi/6/edit?js,console

Answer 2

您可以使用下面的解决方案，对数组进行排序，然后从当前时间查找最近的日期。

首先，代码将当前时间与数组相加，然后获得最近的日期。

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

我认为这段代码可以完成工作。你可以试试看。

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));

这是jsbin链接https://jsbin.com/piwuziqeje/edit?js,console

Answer 4

我尝试了一种与此处看到的方法不同/更直观的方法。它可能会比一些更长，但是我认为它的作用更加清楚。您可以在小提琴中检查代码的工作方式。这是代码：

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));

这是工作中的小提琴。我用不同的时间进行了测试，并且可以正常工作。 https://jsfiddle.net/b36fxpqr/13/

Answer 5

您可以尝试这个小代码。

   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);

您可以尝试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

您也可以尝试此方法，但是要用更多的测试用例进行测试，如果我错了请指正 `

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])

`