Code: Select all
for i, v in ipairs(listOfPillars) do
if numberOfPillars >= 0 then
table.remove(listOfPillars, 1)
numberOfPillars = numberOfPillars - 1
end
end
Code: Select all
for i, v in ipairs(listOfPillars) do
if numberOfPillars >= 0 then
table.remove(listOfPillars, 1)
numberOfPillars = numberOfPillars - 1
end
end
Code: Select all
for i = 1, #listOfPillars do listOfPillars[i] = nil end
ayye that worked, thanksairstruck wrote:The most straightforward way would be something like this (assuming you only use numeric indices):
If you don't have too many references to that table floating around, you could just do listOfPillars = {} and update any references.Code: Select all
for i = 1, #listOfPillars do listOfPillars[i] = nil end
Haha I know I should but I just can't get around to it XDairstruck wrote:Time to read PIL
witchcraftzorg wrote:Edit your first post, and put [Solved] before the subject.
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