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How to do that if a minute passed, remove 1 from the minute value?
01 | function doClock(undif) |
02 | if debounce2 == false then |
03 | local undif = 0 |
04 | local debounce2 = true |
05 | local DefaultTimeMinutes = mapStorage.DefaultTimeMinutes.Value |
06 | local MinutesLeft = DefaultTimeMinutes |
07 | local SecondsLeft = (MinutesLeft * 60) - 1 |
08 | local MinutesLeft = MinutesLeft - 1 |
09 | while SecondsLeft ~= 0 do |
10 | local SecondsLeft = SecondsLeft - 1 |
11 | -- i wanna check if a minute passed, remove 1 from the minute value |
12 | wait(1) |
13 | end |
14 | end |
15 | end |
Any idea why on how to do that if a minute passed, remove 1 from the minute value? (as said on the comment)
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So I used delay to not stop the entire script while waiting becuase you not only have to subtract seconds but minutes. bruno13bruno13 78 — 4y
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Just add a delay() timer. When a certain time passes, an function will execute.
01 | local minutesLeft |
02 | local secondsLeft |
03 |
04 | function removeMinutes() |
05 | minutesLeft = minutesLeft - 1 |
06 | print(minutesLeft.. " minutes and: " .. secondsLeft.. " seconds left.") -- Prints the time left |
07 | end |
08 | function removeSeconds() |
09 | secondsLeft = secondsLeft - 1 |
10 | print(minutesLeft.. " minutes and: " .. secondsLeft.. " seconds left.") -- Prints the time left. |
11 | end |
12 |
13 | function doClock() |
14 | local DefaultTimeMinutes = mapStorage.DefaultTimeMinutes.Value |
15 | local MinutesLeft = DefaultTimeMinutes |
16 | local secondsM = MinutesLeft * 60 |
17 | delay(1, removeSeconds()) |
18 | delay(60, removeMinutes()) |
19 | end |
So what I just made it's use a function called "delay()" that when certain times passes (in seconds, first argument), then a callback will be fired, in this case the functions.
Hope it helped.
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You don't need t print the time left, but it was just to prove the point to you. bruno13bruno13 78 — 4y