How would I sort a table numerically while keeping a string value attached to it?
I am making a highscore board that will display your ranking. The current code creates a table of all the scores(which are stored in a group filled with int values, the value being the score, and the name of the value being the already defined player name) , and then sorts the scores from greatest to least. However, I also need the name of the player to go along with the scores.
Basically, how would I sort the table while keeping the names defined so I can put it next to their score? The actual player name is stored in v.Name (Which I put a comment in there for) I just need to figure out how to take that name and display it with the rest of them.
local values = {}
for _, v in pairs(PlayerNames:GetChildren()) do
--v.Name is where the string is stored!
table.insert(values, v.Value)
end
table.sort(values, function(a, b)--Sort the numbers
return (a > b)
end)
place=0
for index, value in ipairs(values) do--Creates a new frame that contains the player score
place=place+1
local FrameC=HighscoreScr.Frame1:clone()
FrameC.Parent=HSFrame
--FrameC.ImageLabel.PlayerText=PLAYERNAME I NEED TO FIND!
FrameC.Damage.Text=value
FrameC.Visible=true
FrameC.Place.Text=place
end
print("Number of Scores: "..(#values))
HSFrame.Visible=true
2 answers
table.sort can compare arbitrary things with the comparator function. You simply want to be comparing based on .Value:
local values = PlayerNames:GetChildren()
table.sort(values, function(a, b)
return a.Value > b.Value
end)
You now have values the list of NumberValues, in descending order of their .Value.
You can get their scores as simply value.Value and their names as value.Name.
I would approach this problem differently I would completely ignore the use of table.sort and do something like
local values = {}
for _, v in pairs(PlayerNames:GetChildren()) do
local index = #values+1
for i = 1, #values do
if v.Value < values[i].Value then
index = i
break;
end
end
table.insert(values, index, {Name=v.Name, Value=v.Value})
end
--[[Now to access the value, do
for i, v in ipairs(values)
v.Name is the name
v.Value is the value/score
]]
I'll just assume PlayerNames:GetChildren() is small (around 15 max), because this method of sorting is quite inefficient
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