✓ ANSWER ACCEPTED

How to check if the child does not have a part?

Asked Nov 30, 2020247 views2 answers

Hello everyone, In my code, I would like to check if the parent doesn't an attachment. Since there is a :IsA, I tried to do the opposite. Here is what I can do in javascript:

Javascript:

if !(part.Name == "lkjhgd;") {
	//random code....
}

Lua:

				for _, v in ipairs(part:GetChildren()) do
					if v:IsA("Attachment") then
						
					end
				end

So, is there a way I can do that in lua? Thanks for the help!

AnonymousNov 30, 2020

2 answers

✓ Accepted answer

Okay so from my understanding you're trying to check if the part doesn't have a child thats an attachment. In this case its pretty simple to make a check and the rest is up to you.

local function hasAttachment(part)
	for _, v in pairs(part:GetChildren()) do
		if v:IsA("Attachment") then
			return true
		end
		return false
	end
end

if hasAttachment(part) then
	print("The part has an attachment!")
else
	print("The part doesn't have an attachment!")
end

Sorry if I misinterpreted what you said, reply to my answer if so, otherwise tag me as answered if this solved your issue.

AnonymousNov 30, 2020Permalink

This answer is similar to Bandit's one but there is simpler way to achieve what he has done: FindFirstChildWhichIsA or FindFirstChildOfClass (use the first).

local Attachment = Parent:FindFirstChildWhichIsA('Attachment');

if(Attachment)then;
	
else
	
end;

But if you wan't to revert is as you said, !== in JavaScript if i remember correctly, there is keyword not which reverts all given values:

not true -- false
not false -- true

--/In your case:
local Attachment = Parent:FindFirstChildWhichIsA('Attachment');

if(not Attachment)then
	print('no attachment');
end

I tried to use code format like in java script so maybe it will look better for you

AnonymousNov 30, 2020Permalink

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