Simple question!

Given a string that contains a date in ISO8601 format, print the first of the next month in ISO8601 format.

Example inputs:



Their respective outputs:



The rules:

  • Imports are free and don't count towards your score.
  • The date will be stored in a string called s.
  • The solution with the least characters, wins
  • 9
    \$\begingroup\$ Please avoid creating new tags unless they're really missing. We already have date and parsing. The beginner tag is an interesting idea, but I think that should be discussed in meta. On the other hand, you must include a tag telling what's the winning criterion. Is this code-golf? \$\endgroup\$ – Arnauld Dec 13 '19 at 10:27
  • 10
    \$\begingroup\$ I suggest removing The date will be stored in a string called s. and leave it up to the answerers to use any of the default I/O methods. Btw, reading from a pre-defined variable is not among those defaults. Also, not all languages have strings, and not all languages have variables. \$\endgroup\$ – Adám Dec 13 '19 at 10:31
  • 1
    \$\begingroup\$ Do you mean the output should be stored in s, or that the input would be stored in s? \$\endgroup\$ – Kobe Dec 13 '19 at 11:00
  • 1
    \$\begingroup\$ @Arnauld There was discussion around that topic (a "beginner" or "easy" tag) quite a while ago with no clear consensus. IMO that's still the case because it's very subjective. \$\endgroup\$ – AdmBorkBork Dec 13 '19 at 14:40
  • 1
    \$\begingroup\$ I'd say, the newly-added The solution with the least characters, wins could count as the winning criteria, but we prefer not to specify a variable that the input/output have to be stored inside \$\endgroup\$ – Shieru Asakoto Dec 20 '19 at 1:19

JavaScript (V8), 67, 57, 56 bytes

-10 thanks to Arnauld,
-1 joining template string and ternary


Test cases:



  • \$\begingroup\$ 57 bytes (without assuming that the input is stored in s, as this unusual requirement will hopefully be removed) \$\endgroup\$ – Arnauld Dec 13 '19 at 11:12
  • \$\begingroup\$ I never knew you could do that with split, thanks for the input. Might I also ask what's happening with the -~b part? @Arnauld \$\endgroup\$ – Kobe Dec 13 '19 at 11:25
  • \$\begingroup\$ -~b is -(-b-1) and works on non-numeric values. But since b is already coerced to a number by the modulo, you can actually just use b+1 here. \$\endgroup\$ – Arnauld Dec 13 '19 at 11:28
  • \$\begingroup\$ @Kobe -~b is basically b+1 (or actually -(-b-1)). It can be useful to skip parenthesis sometimes, although it's not really necessary in this case, so you could change it to b+1 being more readable if you'd prefer. Relevant tip. You might also find Tips for golfing in JavaScript and Tips for golfing in <all languages> interesting to read through. Welcome to CGCC! \$\endgroup\$ – Kevin Cruijssen Dec 13 '19 at 11:29
  • \$\begingroup\$ @KevinCruijssen Thanks for the comment and links, I'll definitely take a look at those soon :) \$\endgroup\$ – Kobe Dec 13 '19 at 11:31

Bash, 31, 24 bytes

7 bytes saved thanks to @manatwork and @nwellnhof

date -d${1%??}1month +%F

Try it online!

  • \$\begingroup\$ Wow! date's parsing ability keeps amazing me. BTW, I think you can skip the “0”. \$\endgroup\$ – manatwork Dec 13 '19 at 10:57

Ruby 49 bytes

d=Date.parse(s);print Date.new(d.year,d.mon,1)>>1

Online repl


PHP, 57 bytes

[$y,$m]=explode('-', $s);$m=++$m%13?:!!++$y;$d="$y-$m-1";

Test cases:

function next_month($s) {
    [$y,$m]=explode('-', $s);
    return "$y-$m-1";

$s = "2018-05-06";
echo next_month($s); // Gives 2018-6-1

$s = "2019-12-20";
echo next_month($s); // Gives 2020-1-1
  • \$\begingroup\$ Consider using Try it online! to demonstrate your code. \$\endgroup\$ – Neil Dec 21 '19 at 20:42

Red, 39 bytes

t: load s
t/month: t/month + t/day: 1

Retina 0.8.2, 33 bytes


Try it online! Link includes test cases. Explanation:


Increment with wrap-around...


... the last non-9 digit, plus trailing 9s, in either the year or the month, depending on whether the month is 12. (Digits in the following date parts also get incremented.)


Replace the day with 01. Also, if the month was 12 before, it will be 23 now, and also needs to be reset to 01.


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