Short Date into English Long Date

Question

Convert short date format into English long date in as few bytes as possible.

Input

Input will be in the form of a string with format, yyyy-mm-dd, with zero padding optional for all values. You can assume that this is syntactically correct, but not necessarily a valid date. Negative year values do not need to be supported.

Output

You must convert the date into the English long date format (e.g. 14th February 2017). Zero padding here is not allowed.

If the date is invalid (e.g. 2011-02-29), then this must be recognised in some way. Throwing an exception is allowed.

More examples can be seen below.

Test Cases

"1980-05-12" -> 12th May 1980
"2005-12-3"  -> 3rd December 2005
"150-4-21"   -> 21st April 150
"2011-2-29"  -> (error/invalid)
"1999-10-35" -> (error/invalid)

Answer 1

Python 3.6, 137 129 bytes

from datetime import*
def f(k):g=[*map(int,k.split('-'))];n=g[2];return f"{date(*g):%-d{'tsnrhtdd'[n%5*(n^15>4>n%10)::4]} %B %Y}"

Try it online!

Answer 2

PostgreSQL, 61 characters

prepare f(date)as select to_char($1,'fmDDth fmMonth fmYYYY');

Prepared statement, takes input as parameter.

Sample run:

Tuples only is on.
Output format is unaligned.
psql (9.6.3, server 9.4.8)
Type "help" for help.

psql=# prepare f(date)as select to_char($1,'fmDDth fmMonth fmYYYY');
PREPARE

psql=# execute f('1980-05-12');
12th May 1980

psql=# execute f('2005-12-3');
3rd December 2005

psql=# execute f('150-4-21');
21st April 150

psql=# execute f('2011-2-29');
ERROR:  date/time field value out of range: "2011-2-29"
LINE 1: execute f('2011-2-29');
                  ^
psql=# execute f('1999-10-35');
ERROR:  date/time field value out of range: "1999-10-35"
LINE 1: execute f('1999-10-35');
                  ^
HINT:  Perhaps you need a different "datestyle" setting.

Answer 3

JavaScript (ES6), 142 140 bytes

Outputs NaNth Invalid Date for invalid dates.

The code for ordinal numbers was adapted from this answer.

d=>`${s=(D=new Date(d)).getDate()+''}${[,'st','nd','rd'][s.match`1?.$`]||'th'} `+D.toLocaleDateString('en-GB',{month:'long',year:'numeric'})

f=
d=>`${s=(D=new Date(d)).getDate()+''}${[,'st','nd','rd'][s.match`1?.$`]||'th'} `+D.toLocaleDateString('en-GB',{month:'long',year:'numeric'})

console.log(
  f('2005-12-3'),
  f('1980-05-12'),
  f('2005-12-3'),
  f('150-4-21'),
  f('2011-2-29'),
  f('1999-10-35')
)

Answer 4

Python 3.6, 154 bytes

from datetime import*
s=[*map(int,input().split('-'))]
b=s[2]
print(date(*s).strftime(f"%-d{'th'if(3<b<21)+(23<b<31)else('st','nd','rd')[b%10-1]} %B %Y"))

Try it online! (Set input stream and then run.)

Thanks to good suggestions from commenters below.

Answer 5

PHP, 87 bytes

<?=checkdate(($a=explode("-",$argn))[1],$a[2],$a[0])?date("jS F Y",strtotime($argn)):E;

Run as pipe with -F or test it online. Always prints a 4 digit year; fails for years > 9999.

no validity check, 35 bytes:

<?=date("jS F Y",strtotime($argn));

Answer 6

Bash + coreutils, 115 78

• 2 bytes saved thanks to @manatwork.

d="date -d$1 +%-e"
t=`$d`
f=thstndrd
$d"${f:t/10-1?t%10<4?t%10*2:0:0:2} %B %Y"

Try it online.

Answer 7

C#, 147 143 bytes

s=>{var t=System.DateTime.Parse(s);int d=t.Day,o=d%10;return d+((d/10)%10==1?"th":o==1?"st":o==2?"nd":o==3?"rd":"th")+t.ToString(" MMMM yyy");}

Saved 4 bytes thanks to @The_Lone_Devil.

Answer 8

mIRC version 7.49 (197 bytes)

//tokenize 45 2-2-2 | say $iif($3 isnum 1- $iif($2 = 2,$iif(4 // $1 && 25 \\ $1||16//$1,29,28),$iif($or($2,6) isin 615,30,31))&&$2 isnum1-12&&1//$1,$asctime($ctime($+($1,-,$2,-,$3)date), doo mmmm yyyy))

Answer 9

Ruby, 104 103 102+8 = 112 111 110 bytes

Uses -rdate -p program flags.

-1 byte from manatwork.

sub(/.*-(\d*)/){Date.parse($&).strftime"%-d#{d=eval$1;(d<4||d>20)&&"..stndrd"[d%10*2,2]||:th} %B %-Y"}

Try it online!

Answer 10

C# (.NET Core), 167 197 bytes

s=>s.Equals(DateTime.MinValue)?"":s.Day+((s.Day%10==1&s.Day!=11)?"st":(s.Day%10==2&s.Day!=12)?"nd":(s.Day%10==3&s.Day!=13)?"rd":"th")+" "+s.ToString("MMMM")+" "+s.Year

Try it online!

+30 bytes for

using System;

DateTime.Parse()

Answer 11

Excel, 212 bytes

=ABS(RIGHT(A1,2))&IF(ABS(ABS(RIGHT(A1,2))-12)<2,"th",SWITCH(RIGHT(A1,1),"1","st","2","nd","3","rd","th"))&TEXT(MID(A1,FIND("-",A1)+1,FIND("-",REPLACE(A1,1,FIND("-",A1),""))-1)*30," mmmm ")&LEFT(A1,FIND("-",A1)-1)

If you break it into chunks at every ampersand, you get these pieces:

• ABS() pulls the day number from the last two characters in the string. Since that may include a hyphen, ABS converts it to positive.
• IF((ABS-12)<2,"th",SWITCH()) adds the ordinal. The -12 bit is because 11, 12, and 13 don't follow the normal rule and they all get th instead of st, nd, and rd. This corrects for that.
  • Note: The SWITCH function is only available in Excel 2016 and later. (Source) It's shorter than CHOOSE in this case because it can return a value if no match is found whereas CHOOSE requires numeric input and must have a corresponding return for each possible value.
• TEXT(MID()*30," mmmm ") extracts the month name. MID() pulls out the month number as a string and multiplying by 30 returns a number. Excel sees that number as a date _{^{(1900-01-30, 1900-02-29, 1900-03-30, etc.)}} and TEXT() formats it as a month name with a space on both ends. 28 and 29 would have also works but 30 looks "nicer".
• LEFT() extracts the year number.

---

Now, given all that, it would have been way easier if the test cases were all in a date range that Excel can handle as an actual date: 1900-01-01 to 9999-12-31. The big advantage is that the entire date is formatted at once. That solution is 133 bytes:

=TEXT(DATEVALUE(A1),"d""" & IF(ABS(ABS(RIGHT(A1,2))-12)<2,"th",SWITCH(RIGHT(A1,1),"1","st","2","nd","3","rd","th")) & """ mmmm yyyy")

The other big hurdle was having to include the ordinal. Without that, the solution is just 34 bytes:

=TEXT(DATEVALUE(A1),"d mmmm yyyy")

Answer 12

Swift 3 : 298 bytes

let d=DateFormatter()
d.dateFormat="yyyy-MM-dd"
if let m=d.date(from:"1999-10-3"){let n=NumberFormatter()
n.numberStyle = .ordinal
let s=n.string(from:NSNumber(value:Calendar.current.component(.day, from:m)))
d.dateFormat="MMMM YYY"
print("\(s!) \(d.string(from:m))")}else{print("(error/invalid)")}

Try it online!

Answer 13

T-SQL, 194 bytes

DECLARE @ DATE;SELECT @=PARSE('00'+i AS DATE)FROM t;PRINT DATENAME(d,@)+CASE WHEN DAY(@)IN(1,21,31)THEN'st'WHEN DAY(@)IN(2,22)THEN'nd'WHEN DAY(@)IN(3,23)THEN'rd'ELSE'th'END+FORMAT(@,' MMMM yyy')

Input is via text column i in pre-existing table t, per our IO standards.

Works for dates from Jan 1, 0001 to Dec 31, 9999. The year is output with at least 3 digits (per 150AD example).

Invalid dates will result in the following ugly error:

Error converting string value 'foo' into data type date using culture ''.

Different default language/culture settings might change this behavior. If you want a slightly more graceful error output (NULL), add 4 bytes by changing PARSE() to TRY_PARSE().

Format and explanation:

DECLARE @ DATE;
SELECT @=PARSE('00'+i AS DATE)FROM t;
PRINT DATENAME(d,@) + 
    CASE WHEN DAY(@) IN (1,21,31) THEN 'st'
         WHEN DAY(@) IN (2,22)    THEN 'nd'
         WHEN DAY(@) IN (3,23)    THEN 'rd'
         ELSE 'th' END
    + FORMAT(@, ' MMMM yyy')

The DATE data type introduced in SQL 2008 allows much wider range than DATETIME, from Jan 1, 0001 to Dec 31, 9999.

Some very early dates can be parsed wrong with my US locality settings ("01-02-03" becomes "Jan 2 2003"), so I pre-pended a couple extra zeros so it knows that first value is the year.

After that, its just a messy CASE statement to add the ordinal suffix to the day. Annoyingly, the SQL FORMAT command has no way to do that automatically.

Answer 14

q/kdb+ 210 bytes, non-competing

Solution:

f:{a:"I"$"-"vs x;if[(12<a 1)|31<d:a 2;:0];" "sv(raze($)d,$[d in 1 21 31;`st;d in 2 22;`nd;d in 3 23;`rd;`th];$:[``January`February`March`April`May`June`July`August`September`October`November`December]a 1;($)a 0)};

Examples:

q)f "2017-08-03"
"3rd August 2017"
q)f "1980-05-12"
"12th May 1980"
q)f "2005-12-3"
"3rd December 2005"
q)f "150-4-21" 
"21st April 150"
q)f "2011-2-29"       / yes it's wrong :(
"29th February 2011"
q)f "1999-10-35"
0

Explanation:

This is a horrible challenge as there is no date formatting, so I have to create months from scratch (95 bytes) as well as generating the suffix.

Ungolfed solution is below, basically split the input string and then join back together after we've added the suffix and switched out the month.

f:{
   // split input on "-", cast to integers, save as variable a
   a:"I"$ "-" vs x;
   // if a[1] (month) > 12 or a[2] (day) > 31 return 0; note: save day in variable d for later
   if[(12<a 1) | 31<d:a 2;
     :0];
   // joins the list on " " (like " ".join(...) in python)
   " " sv (
           // the day with suffix
           raze string d,$[d in 1 21 31;`st;d in 2 22;`nd;d in 3 23;`rd;`th];
           // index into a of months, start with 0 as null, to mimic 1-indexing
           string[``January`February`March`April`May`June`July`August`September`October`November`December]a 1;
           // the year cast back to a string (removes any leading zeroes)
           string a 0)
  };

Notes:

Dates in q only go back to ~1709 so I don't have a trivial way of validating the date, hence this is a non-competing entry... The best I can do is check whether the day is > 31 or month is > 12 and return 0.