How to find position of an item in table?
Hey there.
Trying to scroll through table 'F' in my script
local E = A:WaitForChild('C'):WaitForChild('H')
local F = {"your","my","we","yeet",}
for x123,x124 in pairs(F) do
if E.Text == x124 then
E.Text = F[x123 + 1]
end
end
I plan to put this in a repeat to run it over and over again, where it fades in and shows the next message, then fades out and shows the next one
I thought where I put ``E.Text = F[x123 + 1] it would make E.Text equal to F[2], since x123 is one at the moment.
Where am I wrong
EDIT:
I updated my code to this, but I get 'nil' - you also get some bonus content
repeat
for nn = 0,1,0.05 do
E.TextTransparency = nn
wait(.05)
end
local abcd = #F
for a,b in pairs(F) do
if E.Text == b then
local function yeet1()
if b == #F then
return 1
else
return b + 1
end
end
print(tostring(F[tonumber(yeet1)]))
end
end
for mm = 1,0,-0.05 do
E.TextTransparency = mm
wait(.05)
end
until 1 == 2
while wait() do
for zz = 0,1,0.01 do
wait(.05)
E.TextColor3 = Color3.fromHSV(i,1,1)
end
end
2 answers
You're not wrong. It's printing as expected. Also, I'm using ipairs because your order when going through the table might be different than mine. With ipairs it'll iterate in order.
You're already finding the positions of items in the table with x123
local E = "your"
local F = {"your","my","we","yeet",}
for x123,x124 in ipairs(F) do
if E == x124 then
E = F[x123 + 1]
print("Matched at: " .. x123 .. " with " .. tostring(E) .. " and " .. tostring(x124))
else
print("No match: " .. x123)
end
end
My results:
Matched at 1 with my and your
Matched at 2 with we and my
Matched at 3 with yeet and we
Matched at 4 with nil and yeet
At index 1 it matched "your" to "your" so E becomes "my"
At index 2 it matched "my" to "my" so E becomes "we" and so on
To find the key of an arbitrary value in a table, we can use linear search:
local function LinearSearch(TargetTable, TargetValue)
for Key, Value in pairs(TargetTable) do
if Value == TargetValue then
return Key
end
end
end
local TargetTable = {"a", "b", "c"}
print(LinearSearch(TargetTable, "b")) --> 2
With linear search, the number of possible steps we have to take through the table in the worst case (value is at the last element iterated over in the table) increases linearly with the growth of the number of elements in a table.