Simple way to use v in for i,v in pairs to access child of object

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#1

Hi all,
I was just curious if there was a way for v to access the child of an object. The code I provided below obviously doesn’t work, but I’m sure you can see what I’m going for here. So my main question is: Is there any way for me to get this code to work other than having to manually create if statements for all values of v? Thanks

for i,v in pairs(copy.Connectors:GetChildren()) do
	copy.Connectors.v.Depth.Value = copy.Connectors.v.Depth.Value + 1
end
1 Like
#2

I don’t understand. Are you trying to get a child within copy.Connectors that has the same name as v?
Wouldn’t that be a little redundant? Just use v???

for i,v in pairs(copy.Connectors:GetChildren()) do
	local Depth = copy.Connectors:FindFirstChild(v.Name)
	Depth.Value += 1
end
#3

You don’t need to use pairs() anymore so I think what you want is the following:

for _, child in copy.Connectors:GetChildren() do
	child.Depth.Value += 1
end
4 Likes
#4

Are you sure? It doesn’t seem likely that you don’t have to use pairs()

#5

all that pairs() does it list out the objects in order. If you just want to loop through them in any order it actually runs faster to not use pairs()

#6

pairs does not list out objects in order, because a dictionary has no ‘correct’ order. The only way to parse a ‘table’ “in order” is by passing an array using ipairs. Using pairs otherwise does not guarantee that the table is parsed in the order that the developer writes it.

#7

I didn’t mention dictionaries. I was more leaning towards regular tables that are preset.

#8

It’s a fairly recent change iirc but you don’t.

#9

You are mistaking pairs and ipairs. pairs has no guarantee for order, ipairs does.
Omitting either behaves the same as pairs.

#10

with hashmaps, yes. It works just fine with regular tables

#11

Yes I’m positive. The Luau announcements for May of 2022 added generalized iterators.

#12

your code will not work, you should simply make:

for i, v in copy.Connectors:GetChildren() do
v.Depth.Value += 1
end