Python 3, 50 bytes

from datetime import*
1/(date.today().weekday()-6)

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TI-Basic, 22 21 bytes

getDate
log(log(dayOfWk(Ans(1),Ans(2),Ans(3

-1 byte by using MarcMush's method but replacing ln( with log(.

Only works on TI-84+/SE. Assumes that the date is set correctly before the program is run. There is a newline at the end.

Vyxal O, 25 bytes

kðtD4/‟₀/N‟:400/kτṠ7%2<[←

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Vyxal doesn't have a weekday function, so I made one myself.

Zsh -G, 14 bytes

>`date`
dd Su*

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• date: get the date in the format Sun 28 Mar 07:55:54 BST 2021
• >: create a file named according to each of the words in the output
• dd: "copy and convert" - here it basically does nothing
  • this command expects no arguments
• Su*: search for a file starting with Su
  • if a file matching that exists (which happens when it is Sunday), then it is passed to dd
    • since dd expects no arguments, this produces an error
  • with the -G option: if no file matches (i.e. it is not Sunday), don't error as usual

This produces a bit of extra junk output to STDERR (not an error though) which may or may not be allowed; the question isn't really clear. Here's an alternative answer which doesn't:

Zsh -G, 18 bytes

>`date`
mv <-> Su*

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<-> matches any file whose name is a number, of which there are two: the day of the month, and the year.

When there is no Su* file, this is renames the day of the month to the year (overwriting the year), which works fine.

When Sun exists, it tries to move the day of the month and the year into the directory Sun, which fails because Sun is a file, not a directory.

Japt, 7 bytes

çKe ªUí

Test it

çKe ªUí
ç           :U=0 times repeat
 K          :  Current date
  e         :  0-based day of the week
    ª       :  Logical OR with
     Uí     :  The result of running the í method on U, which doesn't exist for numbers

Pyth, 7 bytes

/1-6.d9

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/1-6.d9
    .d9  // Current day of the week, 0 indexed on monday.
  -6     // 6 - day of week (0 if Sunday)
/1       // 1 ÷ ^

Excel, 24

Closing quote and paren not counted toward final score, as Excel will autocorrect both of those. Tested in Excel 2016.

=IF(MOD(TODAY()-1,7)^0,"")

I think this is different from the other submission enough to warrant another answer.

How it works:

• I've rolled the error into the conditional check. As it turns out, TODAY() mod 7 also gives the weekday. If it's a Sunday, this means that MOD(TODAY()-1,7) is 0, and 0^0 is an error in Excel.
  • Of course, I could also have divided 1 by the MOD() value or used +6 instead of -1.
• If it's not Sunday, the MOD(...,7) will be non-zero, which when raised to 0, returns 1, a truthy value. This makes the IF return an empty string (our "nothing").
• The if statement, therefore, cannot evaluate to FALSE, because it errors or returns nothing.

Alternative

Here's one that works just as well, but uses a Name error instead:

=IF(MOD(TODAY(),7)=1,A,"")

05AB1E, 54 52 bytes

žg¦¦D4÷že+•YFóåι•žf<è+žg4Öžf3‹&-ŽPjžg2£4%è++7%iõEëõ}

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Explanation

Formula from here.

žg¦¦                                                 # Take the last two digits of the year.
    D                                                # Save for later.
     4÷                                              # Divide by 4, discarding any fraction.
       že+                                           # Add the day of the month.
          •YFóåι•                                    # Push the month's key values.
                 žf                                  # Take the month.
                   <è                                # Find the month's key value.
                     +                               # Add the month's key value.
                      žg4Ö                           # Is this year a leap year?
                          žf3‹                       # Is it January or Feburary?
                              &                      # And the results of both questions.
                               -                     # Subtract 1 for January or February of a leap year.
                                ŽPj                  # Push 6420.
                                   žg2£              # Take the first two digits of the year.
                                       4%è           # Index the thing into the list.
                                          +          # Do step 6.
                                           +         # Add the last two digits of the year.
                                            7%       # Divide by 7 and take the remainder.
                                              i  ë } # Sunday is 1, so it goes in the if.
                                              iõ ë } # Push empty string.
                                              i Eë } # For loop.
                                              i  ëõ} # If not sunday, push empty string and implicit output.

Java 8, 34 bytes

Returns 1 on Mondays, 0 on other days, and throws an ArithmeticException on Sundays

v->1/new java.util.Date().getDay()

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SmileBASIC, 20 bytes

DTREAD OUT,,,W
W=W/W

DTREAD outputs the current year, month, day, and day of the week. Sunday is 0.

Zsh, 15 bytes

${(%):-%(w._.)}

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Prompt sequences can be really crazy sometimes...

${(%):-%(w._.)}
${(%)         }   # expand as prompt sequence
     :-           # ${var:-fallback}, but without the var
       %( . .)    # Prompt ternary
         w        # If DoW matches given number (implied 0, which is Sunday)
          ._      # Then substitute _
            .     # Else substitute nothing

Since _ is not a command, it fails on Sunday.

Postgresql, 32

Not sure if you need to add ; at the end for valid answer

SELECT 1/EXTRACT(DOW FROM now())

C# (.NET Core), 39 43 41 bytes

_=>{if(1/(int)DateTime.Now.DayOfWeek<0);}

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Thanks to @caird coinheringaahing and @Jo King

Excel, 77 30 bytes

Yes, vastly more golfable.

=IF(WEEKDAY(TODAY())=1,1/0,"")

Simply checks if it's Sunday, and if so, finds the quickest way I know of to error. If not Sunday, returns "",the closest Excel has to not returning anything

VBA 18 bytes

This relies on the inbuilt function date() returning a day number that remainders 1 if divided by 7, so may be OS and/or CPU specific.

a=1/(date mod 7-1)

It runs in the VBA project Immediate window.

R, 31 bytes 30 bytes

if(format(Sys.Date(),'%u')>6)a

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No output on non-Sundays, Error: object 'a' not found on Sundays.

format(Sys.Date(),'%u') was the shortest way I could find to get weekday, it outputs a character-class number for day of week, with 7 for Sundays. We can compare to a numeric 7, and if true attempt to use an undefined object.

Saved a byte thanks to Giuseppe!

Clojure, 43 bytes

#(and(=(.getDay(java.util.Date.))7)(/ 1 0))

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Uses the fact that and doesn't evaluate the second argument unless necessary. I originally thought I could get away with using the Ratio literal 1/0 to save two bytes, but that unfortunately causes exceptions immediately. It must try to reduce the Ratio right away or something.

(defn sunday-fail []
  (and (= (.getDay (Date.)) 7)
       (/ 1 0)))

Octave, 23 22 bytes

(1:6)(weekday(now)-1);

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This will try to access an element in 1:6. When the day is Sunday it will try to access the element (1-1)=(0) which will result in an error as MATLAB is 1-indexed.

---

Original for 23.

assert(weekday(now)~=1)

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assert(...) will throw an error when the condition is false. weekday(now) returns the current day of the week where 1 = Sunday. Put the two together, the code will throw an error only on sundays when the condition becomes 1~=1.

Groovy, 16 bytes

1/new Date().day

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Swift 4, 78 bytes

import Foundation;if(Calendar.current.component(.day,from:Date()))==7{exit(1)}

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Pure bash BASH (interactive mode + no coreutils required), 17 20 19 bytes

PS1='`((1/D{%w}))&&:`'

Now only 19 bytes thanks to manatwork's comment below.

Bonus, if you put it in your bashrc it fails every sunday you log in :-) not just when you run it on sundays!

Perl 6,  29  21 bytes

die if now.Date.day-of-week>6

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die if now/86400%7+^3

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AWK, 27 25 23 21 bytes

END{1/strftime("%w")}

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Saved 4 bytes thanks to manatwork

Saved 2 bytes thanks to mik

Julia 0.5, 28 bytes

Dates.dayofweek(now())<7||~-

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This does not work with 0.6, but it does with 0.4.

Julia 0.6, 32 bytes

Thanks to @Dennis for pointing out < saves a byte over !=.

@assert Dates.dayofweek(now())<7

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Q, 20 Bytes

if[1=.z.d mod 7;'e]

.z.d returns the current date. mod does the modulo of the current date, which returns an int. If the date is a sunday, .z.d mod 7 returns 1. If 1=1, (on sunday), and error is raised using the ' operator For brevity the error is just the e character.

C, 35, 34 27 bytes

f(n){n/=time(0)/86400%7^3;}

-7 bytes with thanks to @MartinEnder and @Dennis

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Bash + coreutils, 15 14 bytes

`date|grep Su`

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MATL, 12 bytes

vZ'8XOs309>)

The error produced on Sundays is:

• Interpreter running on Octave:

  MATL run-time error: The following Octave error refers to statement number 9:  )
  ---
  array(1): out of bound 0
  

• Interpreter running on Matlab:

  MATL run-time error: The following MATLAB error refers to statement number 9:  )
  ---
  Index exceeds matrix dimensions
  

To invert behaviour (error on any day except on Sundays), add ~ after >.

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Explanation

This exploits the fact that

• indexing into an empty array with the logical index false is valid (and the result is an empty array, which produces no output); whereas

• indexing with true causes an error because the array lacks a first entry.

Commented code:

v       % Concatenate stack. Gives empty array
Z'      % Push current date and time as a number
8XO     % Convert to date string with format 8: gives 'Mon', 'Tue' etc
s       % Sum of ASCII codes. Gives 310 for 'Sun', and less for others
309>    % Greater than 309? Gives true for 'Sun', false for others
)       % Index into the empty array
        % Implicit display. Empty arrays are not displayed (not even newline)

PowerShell, 24 20 Bytes

1/(date).DayOfWeek>1

Sunday triggers divide by zero same as most of the other solution here. Problem is that on other days that a double gets returned so redirect to nowhere useful trumps that output.

TI-Basic 84+, 23 bytes

getDate
0/(1-dayOfWk(Ans(1),Ans(2),Ans(3

Needs date & time commands, which are 84+ and higher only.

Funky, 21 bytes

if!os.date"%w"error()

os.date"%w" returns the current day of the week in 0-6 format, where 0 is sunday. Getting the logical not of that is only true when the weekday is 0, so Sunday. Then just a basic if(a){error()} will assure that this program only errors on sunday

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SAS, 36 bytes

%put %eval(1/(1-%index(&sysday,Su)))

Ocaml, 46 bytes

open Unix
let()=1/(gmtime(time())).tm_wday;()

and in the ocaml REPL, we can achieve better by removing the let and the final :():

$ open Unix;;1/(gmtime(time())).tm_wday;;<CR>

which is 41 bytes (incuding 1 byte for the carriage return).

Perl 5, 13 bytes

1/(gmtime)[6]

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Ported @biketire's answerj

removed 3 bytes with @mik's reminder

05AB1E, 45 44 bytes

As 05AB1E doesn't have a built in for getting the day of the week, I've used Zeller's Rule to calculate it.

Prints a newline to stderr in case of a Sunday (observable in the debug view on TIO)

žežf11+14%Ì13*5÷žgžf3‹-т%D4÷žgт÷©4÷®·(O7%i.ǝ

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Explanation

The general formula used is
DoW = d + [(13*(m+1))/5] + y + [y/4] + [c/4] - 2*c
Where DoW=day of week, d=day, m=month, y=last 2 digits of year, c=century and and expression in brackets ([]) is rounded down.

Each month used in the formula correspond to a number, where Jan=13,Feb=14,Mar=3,...,Dec=12
As we have the current month in the more common format Jan=1,...,Dec=12 we convert the month using the formula
m = (m0 + 11) % 14 + 1

As a biproduct of March being the first month, January and February belong to the previous year, so the calculation for determining y becomes
y = (year - (m0 < 3)) % 100

The final value for DoW we get is an int where 0=Sat,1=Sun,...,6=Fri.
Now we can explicitly throw an error if the result is true.

VBA / VBScript, 22 20 bytes

^{_{Saved 2 bytes thanks to Taylor Scott.}}

a=1/(Weekday(Now)-1)

This should be run in the Immediate Window. Weekday() returns 1 (Sunday) through 7 (Saturday) so this creates a divide by zero error on Sunday. Otherwise, no output.

Ruby, 15 bytes

1/Time.now.wday

wday will return 0 on Sunday causing a ZeroDivisionError: divided by 0 error. For example: 1/Time.new(2018,1,7).wday.

jq, 42 characters

(39 characters code + 3 characters command line option)

now|strftime("%w")|strptime("%d")|empty

Just trying a different approach here: parse week day number (0..6) as month day number (1..31).

Sample run:

bash-4.4$ TZ=UTC faketime 2018-01-06 jq -n 'now|strftime("%w")|strptime("%d")|empty'

bash-4.4$ TZ=UTC faketime 2018-01-07 jq -n 'now|strftime("%w")|strptime("%d")|empty'
jq: error (at <unknown>): date "0" does not match format "%d"

Note that jq only handles UTC dates.

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PHP, 15 bytes

<?@date(w)?:n;

Assumes default settings.

Output on Sundays

Fatal error: Undefined constant 'n' on line 1

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Java 8, 69 43 34 bytes

v->1/new java.util.Date().getDay()

-26 bytes thanks to @OlivierGrégoire.
-9 bytes thanks to @Neil.

Explanation:

Try it here.

• v->{...} (unused Void null parameter) is one byte shorter than ()->{...} (no parameter).
• new java.util.Date().getDay() will return 0-6 for Sunday-Saturday, so 1/... will give an java.lang.ArithmeticException: / by zero error if the value is 0, which only happens on Sundays.

C,  68  55 bytes

Thanks to @Ken Y-N for saving 13 bytes!

#import<time.h>
f(n){time(&n);n/=gmtime(&n)->tm_wday;;}

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Batch, 88 bytes

@for /f "skip=1" %%d in ('wmic path win32_localtime get dayofweek')do @set/a1/%%d&exit/b

Tries to divide by zero on Sunday.

Unfortunately neither date nor powershell date work on my PC to give me the day of the week.

PHP 7, 12 bytes

1%date("w");

On PHP 7 it throws an exception of type DivisionByZero on Sundays. The same happens if it is interpreted using HHVM.

On PHP 5 it displays a warning (on stderr) on Sundays:

PHP Warning:  Division by zero in Command line code on line 1

On any PHP version, it doesn't display anything on the other days of the week.

Run using the CLI:

php -r '1%date("w");'

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Two more bytes can be squeezed by stripping the quotes (1%date(w);) but this triggers a notice (that can be suppressed by properly set error_reporting = E_ALL & ~E_NOTICE in php.ini).

Gema, 40 characters

A=@subst{Su=@err{S};*=;@datime}@end

Had to specify an error message, so choose a short one: “S”.

Sample run:

bash-4.4$ faketime 2018-01-06 gema 'A=@subst{Su=@err{S};*=;@datime}@end'

bash-4.4$ faketime 2018-01-07 gema 'A=@subst{Su=@err{S};*=;@datime}@end'
S

APL (Dyalog), 23 bytes

⎕CY'dfns'
o←÷7|⌊days⎕TS

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R, 40 bytes

stopifnot(weekdays(Sys.Date(),T)!="Sun")

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weekdays returns the weekday of the date, with an optional argument abbreviate, which shortens Sunday to Sun, saving a single byte.

stopifnot throws an error if, for each argument, not all are TRUE, and throws an error with a message indicating the first element of which isn't TRUE, so the error is Error: "Sun" is not TRUE

Haskell + Data.Dates, 55 bytes

import Data.Dates
succ.dateWeekDay<$>getCurrentDateTime

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This uses the fact that Sunday is the last day of the week. dateWeekDay returns the day of the week as a WeekDay type, which is simply defined as

data WeekDay = Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday

WeekDay is an instance of Enum, thus we can use succ and pred to get the successor or predecessor of a weekday, e.g. succ Monday yields Tuesday.

However, Sunday is the last enum entry, so calling succ Sunday results in the following error:

fail_on_sunday.hs: succ{WeekDay}: tried to take `succ' of last tag in enumeration
CallStack (from HasCallStack):
  error, called at .DataDates.hs:56:34 in dates-0.2.2.1-6YwCvjmBci55IfacFLnAPe:Data.Dates

Edit 1: Thanks to nimi for -3 bytes!
Edit 2: -11 bytes now that functions are allowed.

---

Full program: 88 81 74 69 66 bytes

import Data.Dates
main=pure$!succ.dateWeekDay<$>getCurrentDateTime

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pure is needed to lift the resulting WeekDay back into the IO Monad. However, Haskell sees that the value is not output in any way by the program, so lazy as it is, the expression is not evaluated, so even on Sundays the program would not fail. This is why $! is needed, which forces the evaluation even if Haskell would normally not evaluate the expression.

---

Previous approach with Data.Time: 127 124 bytes

import Data.Time.Clock
import Data.Time.Calendar.WeekDate
c(_,_,d)|d<7=d
main=getCurrentTime>>=(pure$!).c.toWeekDate.utctDay

Try it online! These are some impressive imports. Change d<7 to e.g. d/=5 to test failure on a Friday. Fails with the following exception: Non-exhaustive patterns in function c.

Pyth, 8 7 bytes

 l-6.d9

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Explanation

    .d9 # Get the current day of week (0 = Monday, 6 = Sunday)
  -6    # Subtract 6 from the day
 l      # Try to calculate the log base 2 of the result of the previous operation raising a "ValueError: math domain error" on sundays
        # there is an extra space at the start, to supress the output on the other days

C# (.NET Core), 55 54 48 bytes

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Saved 1 byte thanks to Shaggy

Saved 5 byte thanks to Emigna

Saved 1 byte thanks to Kevin Cruijssen

_=>{var k=1/(int)System.DateTime.Now.DayOfWeek;}

Lucky that Sunday is indexed 0 in enum or else it would've needed to be (System.DayOfWeek)7

JavaScript, 23 Bytes

Date().slice(1)>'um'&&k

Full program.

The variable k must not be defined.

JavaScript, 20 bytes by Rick Hitchcock

/Su/.test(Date())&&k

JavaScript, 19 bytes by apsillers

Date().match`Su`&&k

Python 3, 33 bytes

import time
"Su"in time.ctime()>q

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Python 3, 50 bytes

from datetime import*
datetime.now().weekday()>5>q

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Saved ~3 bytes thanks to Rod.