Develop Reference

Check if dates consecutives in array with starting date today

I am trying to check if in this following array, the dates are consecutives with starting date today (07/01/2020).
var arrayDate = [
'06/30/2020', '06/29/2020', '06/28/2020', '06/26/2020'
]
It should return
var nbDatesConsecutives = 3
On the other hand, this following example should return 0 :
var arrayDate = [
'06/29/2020', '06/28/2020', '06/26/2020', '06/25/2020'
]
I have tried many times to resolve it but I still blocked. Here is one of my attempts :
let arrayDiff = []
arrayDate.map((element, i) => {
arrayDiff.push(today.diff(moment(element), 'days'));
});
let previousValue = null;
arrayDiff.map((element, i) => {
let currentValue = arrayDiff[i];
if (i > 0) {
if (currentValue > previousValue) {
strike++;
}
}
previousValue = currentValue;
})
Thanks !

Your idea of mapping to the day diff is good. Let me build on that:
You could...
Get "today" as the start of the current day
Map the dates to their difference to today, in days
Find the first array index where this difference is no longer equal to the index plus one (since you expect an array like [1, 2, 3, 4] in the perfect case, so e.g. array[2]=2 + 1=3)
This first mismatching index is already your result, except in the case where the whole array has the expected dates, so no index will mismatch - in that case you return the length of the array
Here you can see it working:
function getConsecutive (dates) {
// Note: I hardcoded the date so that the snippet always works.
// For real use, you need to remove the hardcoded date.
// const today = moment().startOf('day')
const today = moment('2020-07-01').startOf('day')
const diffs = dates.map(date => today.diff(moment(date, 'MM/DD/YYYY'), 'days'))
const firstIncorrectIndex = diffs.findIndex((diff, i) => diff !== i + 1)
return firstIncorrectIndex === -1 ? dates.length : firstIncorrectIndex
}
// Outputs 4:
console.log(getConsecutive(['06/30/2020', '06/29/2020', '06/28/2020', '06/27/2020']))
// Outputs 3:
console.log(getConsecutive(['06/30/2020', '06/29/2020', '06/28/2020', '06/26/2020']))
// Outputs 0:
console.log(getConsecutive(['06/29/2020', '06/28/2020', '06/26/2020', '06/25/2020']))
<script src="https://momentjs.com/downloads/moment.js"></script>

The mistake you are doing is 1) currentValue > previousValue instead you should have checked the difference, which must be 1 and when not 1 break the loop. So, here comes the mistake 2) you are using map function rather use simple for loop so that you can break.
`
function getConsecutiveDateCount(arrayDate) {
let arrayDiff = [];
let today = moment();
arrayDate.map((element, i) => {
arrayDiff.push(today.diff(moment(element), 'days'));
});
let strike = 0;
arrayDiff.unshift(0); /// insert 0 for today
let previousValue = arrayDiff[0];
for (let i = 1; i < arrayDiff.length; i++) {
currentValue = arrayDiff[i];
if (currentValue - previousValue === 1) {
strike++;
} else {
break;
}
previousValue = currentValue;
}
return strike;
}
`

How to aggregate arrays of days weekly in JS?

For example, take the time range from 05/10/2019 to 05/25/2019.
Dates on this interval need to be aggregated like this (2019 is omitted for brevity):
const result = [
[ '05-10', '05-11', '05-12'], // week 1
['05-13', '05-14', '05-15', '05-16', '05-17', '05-18', '05-19'], // week 2
['05-20', '05-21', '05-22', '05-23', '05-24', '05-25' ], // week 3
];
What is the best way to solve this problem with JS?
Is it possible to implement this by setting the beginning of the week on any day?
Will packages moment and moment-range help in this?

You can go through the dates, and if the day is 1 (Monday), create a new Array in your results:
const startDate = new Date('05-10-2019'),
endDate = new Date('05-25-2019'),
result = [];
function _twoDigits(x) {
return String(x).padStart(2, '0');
}
let tmp = startDate;
do {
if (tmp.getDay() === 1 || result.length === 0) {
// Create a week Array
result.push([]);
}
const str = `${_twoDigits(tmp.getMonth() + 1)}-${_twoDigits(tmp.getDate())}`;
// Add this date to the last week Array
result[result.length - 1].push(str);
// Add 24 hours
tmp = new Date(tmp.getTime() + 86400000);
} while (tmp.getTime() <= endDate.getTime());
console.log(result);
Note: MomentJS may help, but it's a big library. If you only need to do 2 or 3 basic things with dates, I would recommend not using it. If you need to do a lot of work with dates, then yes, it's a powerful library that will save you a lot of time.

Here is one possible implementation if you are interested in moment.js code.
But as blex said it's a large lib.
const start = moment('2019-05-10');
const end = moment('2019-05-25');
const array = [[]];
const from_date = moment(start).startOf('isoWeek');
const to_date = moment(end).endOf('isoWeek');
let j = 0;
let added = 0;
for (let currentDate = moment(from_date); currentDate <= to_date; currentDate.add(1, 'day')) {
if (added === 7) {
array.push([]);
j++;
added = 0;
}
if (currentDate.isBetween(start, end, null, '[]')) {
array[j].push(currentDate.format('MM-DD'));
}
else {
array[j].push('');
}
added++;
}
document.getElementById('output').innerText = JSON.stringify(array);
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.24.0/moment.min.js"></script>
<p id="output"></p>

How do I get the current school hour and time remaining?

I have this function:
function getInfoSchoolTime() {
var date = new Date();
var schoolBellTime = ["8:10","9:02","9:54","9:59","10:51","11:43","11:58","12:48","13:35","13:40","14:10","15:02","15:54"];
var remainingTime, currentHour;
for (var i = 0; i < schoolBellTime.length-1; i++) {
var startTime = schoolBellTime[i].split(":");
var endTime = schoolBellTime[i+1].split(":");
if (parseInt(startTime[0]) >= date.getHours() && parseInt(startTime[1]) >= date.getMinutes())
if (parseInt(endTime[0]) <= date.getHours() && parseInt(endTime[1]) <= date.getMinutes()) {
currentHour = i;
remainingTime=(parseInt(endTime[1])-date.getMinutes()+60)%60;
break;
}
}
if (currentHour == undefined)
return {current: -1, remaining: "not available"};
return {current: currentHour, remaining: remainingTime};
}
var info = getInfoSchoolTime();
console.log(info.current, info.remaining);
I have the schoolBellTime array that contains the timestamps of my school bell (I know, my school has strange bell times, these timestamps includes playtimes and lunchtime), this function is meant to return the 1st hour/2nd hour/3rd hour ... and the minutes that remains to the next hour/breaktime.
I checked all the code and can't find the error, it keeps returning {current: -1, remaining: "not available"}

The function at the top: setDateTime() takes a date and a time, and constructs a date object for that time.
Then I updated your function, I convert start and end to times on the current day, and then check if date.getTime() occurs between them. Then I simply subtract date.getTime() from end, and convert the result to minutes from milliseconds.
var setDateTime = function(date, str) {
var sp = str.split(':');
date.setHours(parseInt(sp[0], 10));
date.setMinutes(parseInt(sp[1], 10));
return date;
}
function getInfoSchoolTime() {
var date = new Date();
var schoolBellTime = ["8:10", "9:02", "9:54", "9:59", "10:51", "11:43", "11:58", "12:48", "13:35", "13:40", "14:10", "14:10", "15:02", "15:54"];
var remainingTime, currentHour, currentPeriod;
for (var i = 0; i < schoolBellTime.length - 1; i++) {
start = setDateTime(new Date(), schoolBellTime[i])
end = setDateTime(new Date(), schoolBellTime[i + 1])
if (date.getTime() > start.getTime() && date.getTime() < end.getTime()) {
currentHour = i
remainingTime = end.getTime() - date.getTime()
currentPeriod = ([schoolBellTime[i], schoolBellTime[i+1]]).join('-')
}
}
return {current: currentHour, currentPeriod: currentPeriod, remaining: Math.round(remainingTime * 0.0000166667)}
}
console.log(getInfoSchoolTime())

Here's a somewhat different approach, both to the code and the API. It uses two helper functions. Each should be obvious with a single example: pad(7) //=> "07" and pairs(['foo', 'bar', 'baz', 'qux']) //=> [['foo', 'bar'], ['bar', 'baz'], ['baz', 'qux']].
The main function takes a list of bell times and returns a function which itself accepts a date object and returns the sort of output you're looking for (period, remaining time in period.) This API makes it much easier to test.
const pad = nbr => ('00' + nbr).slice(-2)
const pairs = vals => vals.reduce((res, val, idx) => idx < 1 ? res : res.concat([[vals[idx - 1], val]]), [])
const schoolPeriods = (schoolBellTime) => {
const subtractTimes = (t1, t2) => 60 * t1.hour + t1.minute - (60 * t2.hour + t2.minute)
const periods = pairs(schoolBellTime.map(time => ({hour: time.split(':')[0], minute: +time.split(':')[1]})))
return date => {
const current = {hour: date.getHours(), minute: date.getMinutes()}
if (subtractTimes(current, periods[0][0]) < 0) {
return {message: 'before school day'}
}
if (subtractTimes(current, periods[periods.length - 1][1]) > 0) {
return {message: 'after school day'}
}
const idx = periods.findIndex(period => subtractTimes(current, period[0]) >= 0 && subtractTimes(period[1], current) > 0)
const period = periods[idx]
return {
current: idx + 1,
currentPeriod: `${period[0].hour}:${pad(period[0].minute)} - ${period[1].hour}:${pad(period[1].minute)}`,
remaining: subtractTimes(period[1], current)
}
}
}
const getPeriod = schoolPeriods(["8:10","9:02","9:54","9:59","10:51","11:43","11:58","12:48","13:35","13:40","14:10","14:10","15:02","15:54"])
console.log("Using current time")
console.log(getPeriod(new Date()))
console.log("Using a fixed time")
console.log(getPeriod(new Date(2017, 11, 22, 14, 27))) // will Christmas break ever come?!
I made a random guess at the behavior you would want if the date is outside the period range.
Internally, it creates a list of period objects that look like
[{hour:9, minute: 59}, {hour: 10, minute: 51}]
Perhaps it would be cleaner if instead of a two-element array it was an object with start and end properties. That would be an easy change.
Do note that for this to make sense, the bells need to be listed in order. We could fix this with a sort call, but I don't see a good reason to do so.

Here is an ES6 example using deconstruct (const [a,b]=[1,2]), array map, array reduce, partial application (closure) and fat arrow function syntax.
This may not work in older browsers.
//pass date and bellTimes to function so you can test it more easily
// you can partially apply bellTimes
const getInfoSchoolTime = bellTimes => {
//convert hour and minute to a number
const convertedBellTimes = bellTimes
.map(bellTime=>bellTime.split(":"))//split hour and minute
.map(([hour,minute])=>[new Number(hour),new Number(minute)])//convert to number
.map(([hour,minute])=>(hour*60)+minute)//create single number (hour*60)+minutes
.reduce(//zip with next
(ret,item,index,all)=>
(index!==all.length-1)//do not do last one, create [1,2][2,3][3,4]...
? ret.concat([[item,all[index+1]]])
: ret,
[]
);
return date =>{
//convert passed in date to a number (hour*60)+minutes
const passedInTime = (date.getHours()*60)+date.getMinutes();
return convertedBellTimes.reduce(
([ret,goOn],[low,high],index,all)=>
//if goOn is true and passedInTime between current and next bell item
(goOn && passedInTime<high && passedInTime>=low)
? [//found the item, return object and set goOn to false
{
current: index+1,
currentPeriod: bellTimes[index]+"-"+bellTimes[index+1],
remaining: high-passedInTime
},
false//set goOn to false, do not continue checking
]
: [ret,goOn],//continue looking or continue skipping (if goOn is false)
[
{current: 0, currentPeriod: "School is out", remaining: 0},//default value
true//initial value for goOn
]
)[0];//reduced to multiple values (value, go on) only need value
}
};
//some tests
const date = new Date();
//partially apply with some bell times
const schoolTime = getInfoSchoolTime(
[
"8:10", "9:02", "9:54", "9:59", "10:51",
"11:43", "11:58", "12:48", "13:35", "13:40",
"14:10", "14:10", "15:02", "15:54"
]
);
//helper to log time from a date
const formatTime = date =>
("0"+date.getHours()).slice(-2)+":"+("0"+date.getMinutes()).slice(-2);
date.setHours(11);
date.setMinutes(1);
console.log(formatTime(date),schoolTime(date));//11:01
date.setHours(15);
date.setMinutes(53);
console.log(formatTime(date),schoolTime(date));//15:53
date.setHours(23);
date.setMinutes(1);
console.log(formatTime(date),schoolTime(date));//23:01

Checking if date matches recurring date

I have a date variable containing a date, and an interval variable containing an array of numbers. Each number on the interval array represents a date, which is acquired by adding said number to the day count of the previous date. For example:
If date is equal to 2016-09-01 and interval is equal to [15, 15, 20], the resulting dates would be 2016-09-16, 2016-10-01, 2016-10-21, 2016-11-06 and so on.
I want to check if a given date matches that pattern. To do this, I tried using moment-recur, which does exactly what I want with the .every() function, but intervals with repeating numbers don't seem to work ([15, 15, 20] would be parsed as [15, 20] for example). How can I accomplish this, either with moment-recur or a different library?
Here's the desired output using moment-recur:
const now = moment();
const date = moment("2016-09-10", "YYYY-MM-DD");
console.log(date.recur().every([18, 18, 57]).days().matches(now));

The accumulative nature of what you are trying to do is a little tricky.
There may be a nicer way, but I think this works pretty well so long as your interval list isn't too big.
The main insight is that if you are looking for an accumulative interval of say [2, 10, 11] then you will be looking for every 2, 12, 23, 25, 35, 46, etc. This amount to looking for three different dates at regular intervals of the sum of your accumulator -- in this case 23. So you could just use moment's every() with each of the three cases and a single interval.
For example if you have:
const now = moment();
const date = moment("2016-10-22", "YYYY-MM-DD");
const interval = [18, 18, 57]
// sum of the intervals -- i.e. the main interval (93)
const main_int = interval.reduce( (prev, curr) => prev + curr );
// an array of days for each small interval
const dates = interval.map(i => moment(date.add(i ,'days')))
// moment mutates the original so this results in
// [ moment("2016-11-09T00:00:00.000"),
// moment("2016-11-27T00:00:00.000"),
// moment("2017-01-23T00:00:00.000") ]
// Now see if any of these days matches the main interval
const match = dates.some(d => d.recur().every(main_int).days().matches(now))

In a very kludgy way, what you're trying to do is generate new dates that accumulate according to your sequence, then see if at some point it matches a test date.
The following uses moment.js, but really doesn't need it. The moment functionality could be replaced with about 10 lines of code in a couple of separate functions.
/* #param {Date} sequenceStart - start of sequence
** #param {Array} sequence - sequence of intervals
** #param {Date} date - date for comparison
*/
function inSequence(sequenceStart, sequence, date) {
// Copy start date so don't affect original
var s = moment(sequenceStart);
// Get test date in suitable format
var d = moment(date).format('YYYYMMDD');
var i = 0;
do {
// debug
console.log(s.format('YYYYMMDD') + ' ' + d)
// If dates match, return true
if (s.format('YYYYMMDD') == d) {
return true;
}
// If didn't match, add the next value in the sequence
s.add(sequence[i++ % sequence.length], 'day');
// Stop if go past test date
} while (s.format('YYYYMMDD') <= d)
// If didn't return true, return false
return false;
}
var sequence = [15,15,20];
var start = new Date(2017,8,1); // 1 Sep 2017
var test = new Date(2017,10,5) // 5 Nov 2017
console.log(inSequence(start, sequence, test));
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.19.1/moment.min.js"></script>

You can do this without having to use moment-recur if you want to, using a similar approach to what #Mark_M described
The first step is to determine the number of days between the given date and the start date: moment(endDate).diff(startDate, 'days');.
Then, create an array of cumulative totals for the days since start date for each of the entries in the interval array:
function sumSoFar(nums) {
return nums.reduce((sums, num) => {
const prevSum = last(sums) || 0;
return sums.concat(num + prevSum);
}, []);
}
// ...
const sums = sumSoFar(interval);
Finally, the sum of the whole interval array is just the last entry in that list, and so we can find out which entry in the interval list it matches by taking the days difference modulo interval sum. If that is 0, or an entry in the sums array, then the date matches the interval. If not, then it doesn't.
Here is the complete code I came up with:
const startDate = moment('2016-09-01');
const interval = [15, 15, 20];
const last = (arr) => arr[arr.length - 1];
const sum = (nums) => nums.reduce((acc, num) => acc + num, 0);
function sumSoFar(nums) {
return nums.reduce((sums, num) => {
const prevSum = last(sums) || 0;
return sums.concat(num + prevSum);
}, []);
}
const validDates = [moment('2016-09-16'), moment('2016-10-01'), moment('2016-10-21'), moment('2016-11-05')];
function isValid(startDate, interval, date) {
const days = moment(date).diff(startDate, 'days');
const sums = sumSoFar(interval);
const remainingDays = days % last(sums);
return remainingDays === 0 || sums.indexOf(remainingDays) >= 0;
}
validDates.forEach(d => console.log(isValid(startDate, interval, d)));
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.19.1/moment.min.js"></script>

Find the next closest date in MM/DD/YYYY format JavaScript

I have an array of dates formatted as MM/DD/YYYY. I need to find the next closest date in the future starting from today. Say today was 1/22/2016 then 2/19/2016 would return.
2/3/2015
7/5/2015
1/21/2016
2/19/2016
7/1/2016
I've tried doing substrings to get the month, day, year separate and attempting a sort based off those values but surely there has to be a better way.

There is no need for a sorting algorithm. You only need to iterate once and find the closest date that is greater or equals today.
Pseudocode
closest <- infinity
foreach date in dates:
if (date >= now and date < closest) then
closest <- d
return closest
JavaScript
const dates = [
'2/3/2035',
'7/5/2035',
'1/21/2036',
'2/19/2036',
'7/1/2036',
'10/22/2039',
'08/12/2039',
];
const now = new Date();
let closest = Infinity;
dates.forEach(function(d) {
const date = new Date(d);
if (date >= now && (date < new Date(closest) || date < closest)) {
closest = d;
}
});
console.log(closest);

Personally I would use a library such as the very good Moment.JS library, to handle all the horrible complexity of dates.
It has a difference method:
http://momentjs.com/docs/#/displaying/difference/
e.g.
var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b) // 86400000
It would then be trivial to Math.min() the differences of each date in your list.
There's also a moment.min, which might shortcut this entirely, if all your dates are in the future already:
http://momentjs.com/docs/#/get-set/min/

A naïve implementation would be to parse each date as a string and sort them in ascending order. Then, remove any dates that are in the past, and get the first child of the array of remaining dates. See this jsbin example:
var dates = [
'2/3/2015',
'7/5/2015',
'1/21/2016',
'2/19/2016',
'7/1/2016'
];
// parse each string as a Date object and sort them in ascending order
function sortDates(dates) {
return dates.map(function(date) {
return new Date(date).getTime();
}).sort(function(a, b) {
return a - b;
});
}
var orderedDates = sortDates(dates);
// remove any dates in the past, and get the first child of the array of remaining dates
var nextDate = orderedDates.filter(function(date) {
return (Date.now() - date) > 0;
})[0];
Keep in mind that this depends on the format of the date string that you pass to the Date object (in other words, is 1/12/2015 January 12th, or December 1st? JavaScript will parse it as January 12th.

You can use while loop, new Date()
var dates = ["2/3/2015","7/5/2015","1/21/2016","2/19/2016","7/1/2016"]
, d = "1/22/2016", n = -1, res = null;
while (++n < dates.length && new Date(dates[n]) < new Date(d));
res = dates[n] || d;
console.log(res)

Lots of answers, one more can't hurt.
Date strings should always be manually parsed. A library can help, but if you only have a single format, a simple function is all that's required.
The following uses reduce to loop over the array of dates and finds the closest future date. If no date is in the future, it returns null.
The returned value is the string from the array, not a Date.
function parseMDY(s) {
var b = (s || '').split(/\D/);
return new Date(b[2], b[0]-1, b[1])
}
function getClosestDateToToday(arr) {
var now = new Date();
now.setHours(23,59,59);
return arr.reduce(function (acc, s) {
var d = parseMDY(s);
return d < now? acc : (acc && d > parseMDY(acc)? acc : s);
}, null);
}
var dates = ['2/3/2015', '7/5/2015','1/21/2016',
'2/19/2016','7/1/2016'];
document.write(getClosestDateToToday(dates));

This really depends upon your dates and data structures (the ones shown in original example are not so great for me).
From the other answers...
To take the example from Josh, you could also keep a pointer to which date you are using, or simply shift off of a sorted queue of dates to make it work, but it's really adding noise to your code, disrupting the purpose.
Frederik.L answer is really beautiful code, but it would still have to be executed multiple times, so I cannot recommend it.
Feedback warning
I've been given feedback in comments that Date.parse can behave inconsistently. I'll move to passing a date parsing callback function, and demonstrate Date.UTC usage in the callback for OP-specific date format. Please be careful when defining your own callbacks, and please do not copy-paste.
Suggestion
I'd suggest utilizing Date functions i.e. Date.parse; but also try where possible to get data sources sorted without needing application-level sorting. Then you can store-once and step through the array using array.shift() or similar;
Ideally also YYYY-MM-DD
Four-Digit Year
Two-Digit Month
Two-Digit Day
... (continue from least occurring to most occurring)
sample code
var dates = [
'2/3/2015',
'7/5/2015',
'7/1/2016',
'1/21/2016',
'2/19/2016'
]; // unsorted garbage dates
var DateList = function( dateList, getDate ) {
var sortedDates = dateList.sort( function(a, b) {
return getDate(a) - getDate(b);
});
this.next = function() {
var dt = sortedDates.shift();
sortedDates.push(dt); // comment to remove cyclical nature
return dt;
}
};
// specific implementation parser for this format
var getDisgustingDateFormat = function(dStr) {
var dParts = dStr.split('/');
return new Date(Date.UTC(dParts[2],dParts[0],dParts[1]));
};
var dl = new DateList( dates, getDisgustingDateFormat );
Usage
dl.next(); // "2/3/2015"
dl.next(); // "7/5/2015"
dl.next(); // "1/21/2016"
dl.next(); // "2/19/2016"
dl.next(); // "7/1/2016"
dl.next(); // "2/3/2015"
Hope this helps (Updated for clarity)

What about this version using for of and momentjs:
const getClosestFutureDate = (dates) => {
if (dates.length === 0) {
return null;
}
let minDiff = 0;
for (const date of dates) {
minDiff += minDiff + 30;
var currentDate = moment(date);
if (currentDate.isAfter(moment()) && currentDate.diff(moment(), "days") <= minDiff) {
break;
}
}
return currentDate;
};
Assuming now = 2019-08-21
console.log(getClosestFutureDate(["2019-05-07", "2019-06-01", "2019-07-13", "2019-11-09", "2019-11-10", "2019-11-11"]));
// 2019-11-09
I am fan of momentjs, but this can be easily refactored to use only vanilla Date.

const FindDate = (date, allDate) => {
// moment().diff only works on moment(). Make sure both date and elements in allDate list is in moment
let nearestDate = -1;
allDate.some(d => {
const currentDate = moment(d)
const difference = currentDate.diff(date); // Or date.diff(currentDate) depending on what you're trying to find
if(difference >= 0){
nearestDate = d
}
});
console.log(nearestDate)
}

In Livescript:
x =
* "2/3/2015"
* "7/5/2015"
* "1/21/2016"
* "2/19/2016"
* "7/1/2016"
sim-unix-ts = (date-str) ->
# Simulate unix timestamp like concatenating
# convert "MM/DD/YYYY" to YYYYMMDD (integer)
# so we can simply compare these integers
[MM, DD, YYYY] = date-str.split "/"
MM = "0#{MM}".slice -2 # apply zero padding
DD = "0#{DD}".slice -2 # apply zero padding
parse-int "#{YYYY}#{MM}#{DD}"
today = sim-unix-ts "2/18/2016"
date-list = [sim-unix-ts(..) for x]
# find next date
next-dates = [.. for date-list when .. > today]
next-date = next-dates.0
next-date-orig = x[date-list.index-of next-date]
alert [next-date, next-date-orig]
..in Javascript:
var x, simUnixTs, today, dateList, res$, i$, x$, len$, nextDates, y$, nextDate, nextDateOrig;
x = ["2/3/2015", "7/5/2015", "1/21/2016", "2/19/2016", "7/1/2016"];
simUnixTs = function(dateStr){
var ref$, MM, DD, YYYY;
ref$ = dateStr.toString().split("/"), MM = ref$[0], DD = ref$[1], YYYY = ref$[2];
MM = ("0" + MM).slice(-2);
DD = ("0" + DD).slice(-2);
return parseInt(YYYY + "" + MM + DD);
};
today = simUnixTs("2/18/2016");
res$ = [];
for (i$ = 0, len$ = x.length; i$ < len$; ++i$) {
x$ = x[i$];
res$.push(simUnixTs(x$));
}
dateList = res$;
res$ = [];
for (i$ = 0, len$ = dateList.length; i$ < len$; ++i$) {
y$ = dateList[i$];
if (y$ > today) {
res$.push(y$);
}
}
nextDates = res$;
nextDate = nextDates[0];
nextDateOrig = x[dateList.indexOf(nextDate)];
alert([nextDate, nextDateOrig]);