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Table removing value or nil this value in loop

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Table removing value or nil this value in loop

17 replies

Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 07:36:44 am

Hello FunkyFarts

, okno.

s.de.

I have problem with removing keys from a table.
I understand what's not working and why, but i can't find solution for that:
Let's think there is a table

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T={
[1]={},
[2]={},
[3]={}}

T[2]=nil

if i use loop and remove some [x] keys it would work perfectly, but leave missing ids, such as

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T={
[1]={},
[3]={}}

If i use

table.remove(t,pos)

it lead to loop working not correct, you know why.

Well, the first method would be great but how to make after that loop that table is fixed with this example:

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table={
	[1]="one",
	[2]="two",
	[3]="three",
	[4]="four",
	[5]="five"}
table[1]=nil
table[3]=nil
table[4]=nil

table will be

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table={
	[2]="one",
	[5]="five}

so i want 2 to be 1, 5 to be 2. And anytime i have this problem it would i fix it.
Any ideas?

Re: Table removing value or nil this value in loop

Yates
Reviewer Off

Offline

31.07.17 07:51:31 am

What kind of loop are you trying? You may need to loop over the table and reset each ID after the

table.remove

Re: Table removing value or nil this value in loop

Talented Doge
User Off

Offline

31.07.17 07:53:49 am

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t = {
[1] = "k",
[2] = "f",
[3] = "p"
}

table.remove(t, 2)
print(t[2])

I have tested it and it returned p as the result.

More specifically, from which table and what element are you trying to remove? Player list?

Re: Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 07:59:46 am

Yates: i'm trying

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for key, value in pairs(inv) do
	if v[1]==value.ID then
		key=nil
	end
end

how could i reset them?
@ user

Talented Doge: it's not a loop.
Inventory items.

Re: Table removing value or nil this value in loop

Yates
Reviewer Off

Offline

31.07.17 08:47:19 am

Mora has written

key=nil

If you're doing this, it won't work. Key and value are just temporary block scope variables for the for loop.

Replace it with this instead:

inv[key] = nil

But honestly I still don't get why

table.remove

isn't working for you.

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aVeryCoolTable = {
    [1] = "a",
    [2] = "b",
    [3] = "c",
    [4] = "d",
}

print(aVeryCoolTable[2]) -- print b
table.remove(aVeryCoolTable, 2) -- remove b
print(aVeryCoolTable[2]) -- print c

for key, value in ipairs(aVeryCoolTable) do
    if key == 2 then
        table.remove(aVeryCoolTable, 2) -- remove c
        print(aVeryCoolTable[2]) -- print d
    end
end

This all works like it should, test it at https://www.lua.org/cgi-bin/demo

Re: Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 08:54:49 am

Well, i'm already using for inv[key] and it's work perfectly, but you should know: while in LOOP you remove an item it lead to shit. Because the loop is not starting from 1-st key after removed item, but continue.
There is screen of inv[key]=nil
IMG:https://i.imgur.com/RLW2sUG.png

Re: Table removing value or nil this value in loop

Yates
Reviewer Off

Offline

31.07.17 08:59:23 am

Why are you using a loop to remove and display items?

Organize your data before you use it. That's a basic coding principle. Just use two loops.

Re: Table removing value or nil this value in loop

Talented Doge
User Off

Offline

31.07.17 09:00:44 am

This is simply because you removed the item, and the index go up, and when you want to remove the next item, the index is completely different. Thus the result is far from what you expected.

Re: Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 09:03:17 am

Yates How else i'd find certain value of key in mine inventory and delete this key when found what i needed?
Don't touch display items, they fcked bcs of inv[key]=nil.
user

Talented Doge. Well, i know.
IMG:https://i.imgur.com/RqVr4lR.png

That's why i asked here for help me:
IMG:https://i.imgur.com/heEEybl.png

edited 1×, last 31.07.17 09:05:51 am

Re: Table removing value or nil this value in loop

Yates
Reviewer Off

Offline

31.07.17 09:05:50 am

Mora: Use two loops and use table.remove or save the keys in a separate table and remove things afterwards (resulting in three loops eh).

Your approach is clearly wrong, I'm afraid if you cannot solve it we'll need to look at some code.

Re: Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 09:07:40 am

Nah, that's what im doing, i'm not using mine inventory, i'm using copy of it and work with it.
i'm using two loops:

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function MORA_MOD.cooking(id,slot)
	local inv=table.copy(PLAYERS[id].INVENTORY)
	local info={}
	for k, v in pairs(ITEMS[PLAYERS[id].RECIPES[slot]].REQ) do
	table.insert(info,{k=false})
	print(v[1],v[2])
		for key, value in pairs(inv) do
			if v[1]==value.ID then
				if value.STACK then
					if value.STACK>v[2] then
						value.STACK=value.STACK-v[2]
						info[k].k=true
						msg2(id,"-x")
					elseif value.STACK==v[2] then
						inv[key]=nil
						info[k].k=true
						msg2(id,"-1")
					else
						msg2(id,COLOUR[2].."[Cooking]: You dont have enough items of this component["..ITEMS[v[1]].NAME.."].")
						break
					end
				else
					inv[key]=nil
					info[k].k=true
				end
			end
		end
	end

then i do
PLAYERS[id].INVENTORY=inv
if i have all required items. And it's working but missing keys id.
IMG:https://i.imgur.com/heEEybl.png

Re: Table removing value or nil this value in loop

Yates
Reviewer Off

Offline

31.07.17 09:15:13 am

You can solve this by using a table and saving the keys you want to remove, then after the loop is complete you can loop over it to remove them then set your

PLAYERS[id].INVENTORY

table.

Basically: You should check if the player has the items first, then remove them.

Re: Table removing value or nil this value in loop

Talented Doge
User Off

Offline

31.07.17 09:27:00 am

This scenario can be done by this:

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x = {}

t = {
[1] = {"a", 1},
[2] = {"b", 3},
[3] = {"c", 4}
}

c = {2,3,5}

for k, v in ipairs(t) do
	if c[k] == v[2] then
		print("IM INSERTING "..v[1])
		table.insert(x,v[1])
	end
end

for i = 1, #x do
	for k = 1, #t do
		if t[k][1] == x[i] then table.remove(t, k) break end
	end
end

for i = 1, #t do
	print("THIS IS "..t[i][1])
end

Re: Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 09:28:45 am

[EDIT] nvm anymore, i fixed it.

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local BC=5
for i=1,5 do
	if inv[BC]==nil then
		table.remove(inv,BC)
	end
	BC=BC-1
end

edited 1×, last 31.07.17 09:33:57 am

Re: Table removing value or nil this value in loop

Talented Doge
User Off

Offline

31.07.17 09:32:39 am

To loop with for reversely:

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t = {"ME", "LOVE", "YOU"}

local x
for i = #t, 1, -1 do
 if not x then x = t[i] else x = x.." "..t[i] end
end
print(x)

This is included in the PIL book

edited 1×, last 31.07.17 09:36:24 am

Re: Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 09:35:06 am

Lel, i didn't know that, but it's almost identical like i said in previous post.
Anyway i prefer using yours if it would work.
Thank you for your help: user

Yates,

Talented Doge!
!!
Yates, bcs i'm not just removing 1 item.. if you read my post you will see that i'm using LOOP inside loop:
IMG:https://i.imgur.com/D6FbzqX.png

This is silly to make table if you have 1 item in it right?
If i'd use 1 item i would not have problems

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REQ={
{6,2},
{7,2}}

edited 1×, last 31.07.17 10:45:13 am

Re: Table removing value or nil this value in loop

Yates
Reviewer Off

Offline

31.07.17 11:00:16 am

Mora: I know. That's why I said to save the keys for later use.

I don't get why you had to edit your post just for that, user

Talented Doge basically wrote what I suggested.

Re: Table removing value or nil this value in loop

Mora OP
User Playing CS2D

31.07.17 11:14:02 am

Oh ye, i forget, i just clicked reply and prepared mine message, but site notice me i need to edit my reply, so i lost it and replied to your message.

I just did

PLAYERS[id].INVENTORY=inv

instead of

PLAYERS[id].INVENTORY=table.copy(inv)

Because i have stackable items and if i found this item in my inventory i'll add to .STACK value +1

Excuse me, my bad. Now anything is almost completely fixed.