I have the following string which I ultimately need to have in the format of mm/yy
var expDate = 2016-03;
var formatExp = expDate.replace(/-/g , "/");
This gets me to 2016/03, but how can i get to 03/16?
Add a comment
|
3 Answers
one solution without regex:
var expDate = '2016-03';
var formatExp = expDate.split('-').reverse().join('/');
//result is 03/2016
alert('result: ' + formatExp);
var formatExpShort = expDate.substring(2).split('-').reverse().join('/');
//result is 03/16
alert('result short: ' + formatExpShort);2 Comments
JordanBarber
Thanks and that's super close but it does not drop the first two digits of the year?
JordanBarber
You the man! Thanks a ton. I'll mark as resolved in just a min
With a RegExp :
'2016-03'.replace(/^\d{2}(\d{2})-(\d{2})$/, '$1/$2')
1 Comment
Meiko Rachimow
this got a vote from me, i love regex but im too slow for that kind of solution ;)
Do you really need to use RegExp?
Why not creating a simple function that splits the exp Date and returns it the way you want it?
function parseDate(expDate){
var dateArray = expDate.split('-')
return dateArray[1] + '/' + dateArray[0].substring(2,4)
}
The split functions creates an array, the element in position 1 is the month, the element in position 2 is the year, on the latter you apply the substring function which extrapolates the last two digits.
