Currently we put in the number we want and it spits out whatever tokens will make out that much. Which is all fine and dandy, until we need a good amount of a specific number (obviously, not a problem when you need 20k tokens and all of them can be 1000 Point Wizard Token's). Given how yesterday we had the chance to turn 1 Point Wizard Token's into the Recycle Beast, we need the option to be more specific. Refreshing a solid 20,000 times is all kinds of fun, except for the added stress on an already breaking system. It could be a checklist where we input the number we want to take out and can select the tokens we want. Remove [400] 1 10 100 1000 [] [] [x] [] And we'll get four 100 tokens.
Yes, much needed.
Yes please. But I think it'd be easier to put in the quantity we want + the tokens (so 4 and check off the 100 box) instead of 400. It seems kind of confusing putting in 400, like you want to take 400x 100 point tokens out (but maybe that's me)
Ah, yes, that'd make sense. XD I was going off of what's on the page, ahaha
Yes please. Especially the way suggested. For the math-impaired, there might also be a confirmation box containing the total amount of tokens cashed out, like this:
Remove [4] 1 10 100 1000 [] [] [x] [] You're now cashing out 400 tokens.
(This would be nice especially if the list of tokens to choose from contains all token denominations. ...yes, I sometimes have problems with multiplying 15 :P)
I only did the "1/10/100/1000" because it saved time, of course they'd do all of the increments :p
Yeah, I thought so, but I was kind of thinking to myself whether all token variants were really needed here. Like that 15 point token :D Perhaps there's a need for those too.
I'll support this idea, because if more cases need wizard tokens they become also more meaningful.
I would really like to see this implemented. At Luminaire, I like to through small tokens in random users stockings, but it's a pain to get them out.
Formerly known as its_just_me
(I started in March)
Wine is the answer
What was the question?