if the day is a Friday 13th, then display text?

[the_toolman]

Hi there,

I am trying to write a simple piece of code that will display a little text phrase on every Friday 13th in any year.

How would I best go about doing this?

Thank you.

[Barand]

if (date('j') == 13 && date('w')== 5) {
    echo "You should have stayed in bed today" ;
}

See date() function

[requinix]

I'd go with 'l' and 'j'. And you could use them both at once to check for "Friday 13".

[the_toolman]

Hi,

Thank you for the replies  That worked perfectly!

[Barand]

19 minutes ago, requinix said:

I'd go with 'l' and 'j'.

I went for "w" as it's locale/language neutral.

[cyberRobot]

For what it's worth, you could also use "w" and "j" with the same date() call.

if (date('w|j') == '5|13') {  //if Friday the 13th
    echo "You should have stayed in bed today";
}

The straight bar ("|") isn't necessary. Just using that to separate the values. Of course, that's not as intuitive as "Friday 13".

[Barand]

And other variations

if ( date('l j') == 'Friday 13' )
if ( date('l d') == 'Friday 13' )
if ( date('D j') == 'Fri 13' )
if ( date('D d') == 'Fri 13' )

 

[Psycho]

I didn't realize this was a challenge question. You're all being lazy relying on the date function 😁

function isFridayThirteenth($year, $month, $day)
{
    $m = (($month+9)%12)+1;
    $C = floor($year/100);
    $Y = $year%100-(($m<11)?0:1);
    $W = ($day + floor(2.6*$m - 0.2) - (2*$C) + $Y + floor($Y/4) + floor($C/4)) % 7;
    
    return ($W==5 && $day==13);
}

 

Edited September 17, 2019 by Psycho

[requinix]

5 minutes ago, Psycho said:

I didn't realize this was a challenge question.

It wasn't 😛

5 minutes ago, Psycho said:

You're all being lazy relying on the date function 😁

Yours doesn't work properly for the year 2000.

[Psycho]

7 minutes ago, requinix said:

Yours doesn't work properly for the year 2000.

You're right. I was only considering future dates, but the same problem would occur in the year 2100. Apparently the problem stems from a flaw in the PHP logic for modulus which will return a negative number whereas most other languages only return a positive integer. I must fix this in case someone refers back to this post 80 years from now! I could define a custom modulus function to only return positive integers, but it's simpler for this process to just "hack" it by adding a hard-coded constant of 35 so the modulus of 7 will still be the same, but will never be negative.

For those to whom it is not obvious, this is all sarcasm. I did this as a mental exercise and am not proposing that this is a preferred solution in any way. But, it does work

function isFridayThirteenth($year, $month, $day)
{
    $k = $day;
    $m = (($month+9)%12)+1;
    $C = floor($year/100); 
    $Y = $year%100-(($m<11)?0:1);
    $W = ($k + floor(2.6*$m - 0.2) - (2*$C) + $Y + floor($Y/4) + floor($C/4) + 35) % 7;

    return ($W==5 && $day=13);
}

 

[Barand]

I fails for a few historic dates

$dt1 = new DateTime('2019-09-13');
$di = new DateInterval('P1M');
while ($dt1->format('Y') >= 1901) {
    if ($dt1->format('D j') == 'Fri 13') {
        
        if ( !isFridayThirteenth($dt1->format('Y'), $dt1->format('n'), $dt1->format('j') ) ) {
            echo $dt1->format('Y-m-d') . ' - Failed<br>';
        }
    }
    $dt1->sub($di);
}

/* RESULTS :

        2015-03-13 - Failed
        2012-04-13 - Failed
        2011-05-13 - Failed
        2009-03-13 - Failed
        2008-06-13 - Failed
        2007-07-13 - Failed
        2007-04-13 - Failed
        2005-05-13 - Failed
        2004-08-13 - Failed
        2003-06-13 - Failed
        2002-09-13 - Failed
        2001-07-13 - Failed
        2001-04-13 - Failed
        2000-10-13 - Failed
        1914-03-13 - Failed
        1910-05-13 - Failed
        1908-03-13 - Failed
        1906-07-13 - Failed
        1906-04-13 - Failed
        1904-05-13 - Failed
        1903-03-13 - Failed
        1902-06-13 - Failed
        1901-09-13 - Failed
*/

EDIT:

P.S.

Confirmed you're good until August 2100

2100-08-13 - Failed
2101-05-13 - Failed
2102-10-13 - Failed
2103-04-13 - Failed
2103-07-13 - Failed
2104-06-13 - Failed
2105-03-13 - Failed
2106-08-13 - Failed
2107-05-13 - Failed
2108-04-13 - Failed
2108-07-13 - Failed
2110-06-13 - Failed
2111-03-13 - Failed
2112-05-13 - Failed
2114-04-13 - Failed
2116-03-13 - Failed
2200-06-13 - Failed

 

[Psycho]

1 hour ago, Barand said:

I fails for a few historic dates

EDIT:

P.S.

Confirmed you're good until August 2100

I guess you missed the update I posted in response to @requinix It now works for those previous dates as well as the years that end in "00". Tested from year 1 to 3000. I think that should be sufficient. I was really just interested in implementing a calculation to find the day of the week. I think I'll stick with using date() from now on.

[requinix]

15 minutes ago, Psycho said:

I think I'll stick with using date() from now on.

Traitor.