Whenever I need my blood pressure raising, I go to the skin and accent problems forum thread and read the rejection reasons
this code is supposed to be 'if the first character is dan, change background of the blurb to green' but all my attempts have failed so far D:
.blurb:has(li.characters:first-of-type a[href*="Dan" i]) {
background: green !important;
}
do you think this isn't possible or is my syntax just wrong ?
I thought this was impossible for a good while, but ended up figuring it out! Use the following:
.blurb:has(.relationships + .characters a[href="Dan" i]),.blurb:has(.warnings + .characters a[href="Dan" i]) {background: green!important;}
And here is my understanding of what went wrong
actually you know what that's exactly it i would rather someone add 5 parantheticals after every sentence than use tone indicators it's 1. accomplishing SO much more in terms of clarity 2. extremely funny to look at depending on how they're used
notn advice from a veteran: if there's site lag, refresh more and click harder. this will 100% allow you to get through it more efficiently. remember, you must sell everything instantaneously once the holiday starts, lest you miss out on those sweet sweet first day price spikes. fuck anyone else trying to use the fr servers
My Twitter has been officially WIPED. It was good while it lasted, now for more time on Tumblr and Bluesky.
got a major pest problem this year actually
uk people, sign and share
non uk people, share but don't sign
terfs, get in the bin and stay there
Gang, I don't think anything will be topping a user having to explain the concept of a 3D shape to a mod
Whenever I need my blood pressure raising, I go to the skin and accent problems forum thread and read the rejection reasons
It only took 2 days to fix instead of the 1 and half months it took last time π₯³
Very lovely, Tumblr has shadowbanned this account. Sorry that I cannot reply to anyone sending me nice things right now.
He/Him β Axo β’ Gabe β’ Killie β Adult πππ β πΎπ-ππ β πΆππππ β π΄π3 β π·π΄ β π΅πππ¦ β πΊππ‘βπ’π β πΉπππβπ‘ π ππ ππππΉππππ’ππ 1 ππππππππ β πππ‘βππππππ ππππππππ
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