Regex Thread

\d+(00|11|22|33|44|55|66|77|88|99)

>>51533793

^dubs$

>>51533834
OP here, you fucked up it doesn't catch dubs

File: 7Vdv2vB.jpg (584KB, 1200x900px) Image search: [Google]

584KB, 1200x900px

/(\d)\1$/

>>51533793
(\d)\1$

i wrote a thing that would find all the dubs, trips, etc... in a number range. here's the function that checked if characters repeated:

def dubChecker(n):
    check=str(n)
    for i in range(len(check),1,-1):
        if len(set(check[-i:]))==1:
            return i
    return n

you could change

return n

to

return False

or something if you wanted to make it less ambiguous, but a single-digit integer should be sufficiently clear and allows you to use a dict that maps 3: trips, 2: dubs, etc...

if (postNumber == 00000000) ...
else if (postNumber = 00000011)...
else if (postNumber = 00000022)...
...
else if (postNumber = 99999999)...

>>51534200
Was going to call you stupid for str in range, but actually a clever use of len of str to quickly get digit size!

>>51534047
> not knowing how RegEx work
Stop being stupid anytime.

>>51534262
Its always nice to meet a fellow CS grad

>>51533793

/[[:checkem:]]/

>>51534262
I see you made a typo where you're assigning to postNumber instead of checking for equality. I'm going to need you to go ahead and manually correct all 10,000,000 typos.

File: 1448524484001.png (13KB, 664x381px) Image search: [Google]

13KB, 664x381px

>>51533793

<?php
function dubs($postNumber) {
  if(!is_int($postNumber) || $postNumber < 10) return false;
  list($n1,$n2) = sscanf(strrev($postNumber), "%01d%01d");
  return $n1 === $n2;
}
?>

>>51534315
i felt like the intuition of getting the length of the set was a little cleverer (since repeating digits will always resolve to a single-element set), but thanks

>>51534566
Nice fatal error, faggot. Might as well do what your process did and kill yourself.

>>51534674
don't hurt yourself on that edge, kid

\d+(\d)\1

>>51535575
Thanks for not fucking up

\d{2}

>>51533834
Dumbass