Javascript获取阵列中的连续日期交叉周和月

Question

I have an array with the following values (example):

[
  1491408000000,
  1491494400000,
  1491753600000,
  1493222400000,
  1493308800000,
  1493568000000
]

Where the index is a date time. The date time will always be at 12:00:00 on a date.

In this example, the first 3 dates are consecutive cross weekend (weekend is holiday so count as leave), then another group of 3 dates cross weekend and month.

Now, what I am trying to do is find sequential dates (cross week and month) and put them into an array as follows:

[
  1491408000000,
  1491494400000,
  1491753600000
],
[
  1493222400000,
  1493308800000,
  1493568000000
]

I have tried the following code to get the sequential dates but this cannot cross week and month, how to modify the code to get above result? Any help would be much appreciated!

var timeValue = new Date(dateReview).getTime(); 
valueCon.push(timeValue);                                           

var k = 0;   
sortedValue[k] = [];

valueCon.sort( function ( a, b ){
    return +a > +b ? 1 : +a == +b ? 0: -1;
})
.forEach( function( v , i ){
    var a = v,b = valueCon[i+1]||0;

    sortedValue[k].push( +a );

    if ( (+b - +a) > 86400000) {                                
        sortedValue[++k] = []
    }
    return 1;
});                     

sortedValue.sort( function ( a,b ){
    return a.length > b.length ? -1: 1;
});

Answer 1

Manipulation of timestamps is a pain. JavaScript has a built-in Date type, as you know, and I would suggest you use it. Date#getUTCDay returns the day of the week as an integer (for reference, 4 is Friday, or the day before a weekend), while Date#setUTCDate and Date#getUTCDate together allow you to adjust the date in day increments (and have it overflow/underflow to the next/previous month). Thus, to determine whether a timestamp b follows "sequentially" (excluding weekends) after a , you can use:

function sequential (a, b) {
  a = new Date(a)
  return a.setUTCDate(a.getUTCDate() + (a.getUTCDay() === 4 ? 3 : 1)) === b
}

Grouping is just an exercise after that; the code above contains all of the real logic behind this solution.

---

Example Snippet

 var dates = [ 1491408000000, 1491494400000, 1491753600000, 1493222400000, 1493308800000, 1493568000000 ] function sequential (a, b) { a = new Date(a) return a.setUTCDate(a.getUTCDate() + (a.getUTCDay() === 4 ? 3 : 1)) === b } function groupSequential(dates) { if (dates.length < 2) return [dates.slice()] dates.sort(function(a, b) { return a - b }) var result = [], group for (var i = 0; i < dates.length; i++) { sequential(dates[i - 1], dates[i]) || result.push(group = []) group.push(dates[i]) } return result } console.log(groupSequential(dates)) 

Answer 2

This requires help from a function to test if two dates are in the same week. The following goes over the set of time values provided in an array and puts the first value into an array within the array. For each subsequent value, it tests if it's in the same week as the first value in each array within the outer array.

If it's in the same week as the first value in any existing array, it's pushed into that array. Otherwise, it's put in a new array and pushed into the outer array.

There may be a neater way to implement the algorithm, but I'll leave that for others.

Due to time zone differences, they are adjusted to the host time zone based on the original time values representing noon in the source time zone.

 // Given 2 dates, return true if they are in the same week (Mon to Sun). // Otherwise, return false function sameWeek(a, b){ var e = new Date(+a); // Week starts at 00:00:00.000 on Monday on or before date var s = new Date(e.setDate(e.getDate() - ((e.getDay()||7) -1))); s.setHours(0,0,0,0); // Week ends at 23:59:59.999 the following Sunday e.setDate(e.getDate() + 6); e.setHours(23,59,59,999); // Test b and return value return b >= s && b <= e; } // Given time value for UTC-0400, adjust to same date and time // in local time zone and return a date function adjust(n) { var d = new Date(n); d.setMinutes(d.getMinutes() - 240 + d.getTimezoneOffset()); return d; } var result = [1491408000000,1491494400000,1491753600000,1493222400000,1493308800000,1493568000000 ].reduce(function(acc, n) { var d = adjust(n); var used; if (acc.length != 0) { used = acc.some(function(arr) { if (sameWeek(adjust(arr[0]), d)) { arr.push(n); return true; } }); } if (!used || acc.length == 0) { acc.push([n]); } return acc; },[]); // Result array console.log(result); // Printed as date strings adjusted to same host local time result.forEach(arr => { arr.forEach(n => console.log(adjust(n).toString())) console.log('\\n'); });