对作为日期的字符串数组进行排序 JavaScript

Question

I've tried using underscorejs, min and max methods but they can't handle strings. From what i've read and learnt anyway, since I get infinite back from both.

My array looks like : dateData = ["26/06/2016", "04/06/2016", "13/05/2016", "20/07/2016"]

How can I grab the last and the first date in these?

I tried using sort also that looks like : _.chain(dateData).sort().first().value() but I get back the last item in the array rather then the last date in the array.

Answer 1

 var dateData = ["26/06/2016", "04/06/2016", "13/05/2016", "20/07/2016"]; function dateToNum(d) { // Convert date "26/06/2016" to 20160626 d = d.split("/"); return Number(d[2]+d[1]+d[0]); } dateData.sort(function(a,b){ return dateToNum(a) - dateToNum(b); }); console.log( dateData );

To retrieve the first, last date:

var firstDate = dateData[0];
var lastDate  = dateData[dateData.length -1];

---

Basically, if you first convert all your 26/06/2016 to a date Number like 20160626 you can .sort() those numbers instead.

so you're basically sorting:

20140626  
20140604  
20140513  
20140720  

resulting in:

[  
  "13/05/2016",  
  "04/06/2016",  
  "26/06/2016",  
  "20/07/2016"  
]

Answer 2

If we can format the dateStrings in a particular format, then sorting them as strings also sorts them as dates eg YYYY-MM-DD.

You can use localeCompare to compare strings.You can use following code to sort the dates:

 dateData = ["26/06/2016", "04/06/2016", "13/05/2016", "20/07/2016"] dateData.sort(function(a, b){ var A = a.split("/"); var B = b.split("/"); var strA = [ A[2], A[1], A[0] ].join("/"); var strB = [ B[2], B[1], B[0] ].join("/"); return strA.localeCompare( strB ); }); console.log( dateData );

Once sorted, you can get the min and max dates as:

var minDate = dateData[0];
var maxDate = dateData[ dateData.length - 1 ];

Answer 3

The getTime() method returns the numeric value corresponding to the time for the specified date according to universal time. Date.getTime()

 dateData = ["26/06/2016", "04/06/2016", "13/05/2016", "20/07/2016"] .map(a=>a.split('/').reverse().join('/')) .sort((a,b)=>new Date(a).getTime() - new Date(b).getTime()); console.log(dateData);

Answer 4

A number people have already touched on this, but you need to convert the date strings to something that can be compared in the sort function. The one thing I haven't seen shared is how to get the first and last dates. This should do the trick:

//original date array
var dateData = ["04/06/2016", "13/05/2016", "20/07/2016","26/06/2016"];

//map through the original array and convert the dates to js date objects
var formattedDates = dateData.map(function(date){
    var splitDate = date.split("/")
    return new Date(splitDate[2],splitDate[1]-1,splitDate[0])
})

//sort the dates
formattedDates.sort(function(a,b){
  // Turn your strings into dates, and then subtract them
  // to get a value that is either negative, positive, or zero.
  return new Date(a) - new Date(b);
});

//Now you can get the first and last dates:
var firstDate = formattedDates[0]
var lastDate = formattedDates[formattedDates.length-1];

//log to check:
console.log('first date: ', firstDate)
console.log('last date: ', lastDate)

Answer 5

To be easier for the subsequest operations, change the format to be like this: YYYY/MM/DD. This way regular sorting will get you the min and max properly, and you won't need further parsing. Helper function to sort would be like this:

for(var i=0;i<dateData.length;++i)
{
    var split = dateData[i].split("/");
    dateData[i] = split.reverse().join("/");
}

Answer 6

One way I know to do this is using the .sort() function for a string. https://msdn.microsoft.com/en-us/library/4b4fbfhk(v=vs.94).aspx

You would have to change your array into YYYY-MM-DD

then you could have the following code

var dateData = ["2016/06/26", "2016/06/04", "2016/05/13", "2016/07/20"]; dateData.sort(); var first = dateData[0]; var last = dateData[dateData.length-1];

where first is the earliest date and last is the latest date

Answer 7

You can use Array#reduce , because you need only two values and this can be achieved with a single loop.

 var dates = ["26/06/2016", "04/06/2016", "13/05/2016", "20/07/2016"], values = dates.reduce(function (r, a, i) { function getISO(s) { return s.replace(/(.+)\\/(.+)\\/(.+)/, '$3$2$1'); } if (!i) { return { min: a, max: a }; } if (getISO(a) < getISO(r.min)) { r.min = a; } if (getISO(a) > getISO(r.max)) { r.max = a; } return r; }, {}); console.log(values); 

Answer 8

Roko's answer worked for me, and +1. And Jose had a similar thought to me, +1...
...There's an easier and more robust way: .valueOf()

converting date string to number :

const dateAsNumber = new Date(dateAsString).valueOf()

JS has a built in method/function for calculating the number of milliseconds that have passed since a date; that's .valueOf(), which can be called on a Date object. So, turn your date string into a Date object (with "new Date()" with the date string as the argument), and then convert to milliseconds.

After that, the normal .sort() works fine. As shown below, for your convenience:

const arrayOfDateStrings = ["5/01/2012", "10/01/2020", "10/01/2019", "11/30/2016", "10/01/2021", "02/01/2020"];

const sortedArray = arrayOfDateStrings.sort((a,b)=>new Date(a).valueOf() - new Date(b).valueOf());

console.log(sortedArray);

Or Moment.js can be used instead of the built-in Date object/functions, that works in a very similar way.