Code Golf Stack Exchange is a question and answer site for programming puzzle enthusiasts and code golfers. It only takes a minute to sign up.
Sign up to join this community
Simple question!
Given a string that contains a date in ISO8601 format, print the first of the next month in ISO8601 format.
Example inputs:
2018-05-06
2019-12-20
Their respective outputs:
2018-06-01
2020-01-01
The rules:
s.-10 thanks to Arnauld,
-1 joining template string and ternary
s=>([a,b]=s.split`-`,+a+!(b%=12))+`-${b++<9?'0'+b:b}-01`
Test cases:
f=s=>([a,b]=s.split`-`,+a+!(b%=12))+`-${b++<9?'0'+b:b}-01`
console.log(f('2018-12-20'))
console.log(f('2019-05-06'))
-~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\$
-~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\$
Dec 13, 2019 at 11:29
date's parsing ability keeps amazing me. BTW, I think you can skip the “0”.
\$\endgroup\$
Dec 13, 2019 at 10:57
T`9d`d`.9*(-12)?-..$
-23|-..$
-01
Try it online! Link includes test cases. Explanation:
T`9d`d`
Increment with wrap-around...
.9*(-12)?-..$
... 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.)
-23|-..$
-01
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.
[$y,$m]=explode('-', $s);$m=++$m%13?:!!++$y;$d="$y-$m-1";
Test cases:
<?php
function next_month($s) {
[$y,$m]=explode('-', $s);
$m=++$m%13?:!!++$y;
return "$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
Each date and time value has a fixed number of digits that must be padded with leading zeros, meaning output of 2018-6-1 or 2020-1-1 is not valid.
\$\endgroup\$
t: load s
t/month: t/month + t/day: 1
t
from datetime import *; from dateutil.relativedelta import relativedelta as r;
(datetime(*map(int, s.split('-')), 1) + r(months=1)).strftime('%Y-%m-01')
Your rules say that imports don't count, so I didn't count them.
from datetime import datetime as d to save a few bytes. Also, you can remove
\$\endgroup\$
Jun 20, 2020 at 3:01
print statement. To catch this, I suggest using TIO, such as by adding a link: Try It Online!. (I incorporated a couple more suggestions to shorten it to 69 bytes).
\$\endgroup\$
Jun 20, 2020 at 3:07
dateandparsing. Thebeginnertag 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 thiscode-golf? \$\endgroup\$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\$s, or that the input would be stored ins? \$\endgroup\$The solution with the least characters, winscould count as the winning criteria, but we prefer not to specify a variable that the input/output have to be stored inside \$\endgroup\$