Trying to open all visible backpacks

[pedegie]

Im trying to open all visible backpacks. I have persistent which counts items inside backpacks so its fine if I open each one just for a moment only. I have many backpacks inside other backpacks with diffrent colors. For now i have something like this :

for i = 0, 15 do

	local cont = getcontainer(i)

	for j = 1, cont.itemcount do

		local info = iteminfo(cont.items[j].id)

		if info.iscontainer then

			local findIndex = 0

			

			local contIN = getcontainer(info.name)

				--if not contIN.isopen then

					for _, v in pairs(backpacks) do

						if v.name == info.name then

							findIndex = v.index

								v.index = v.index + 1

								break					

						end

					end

					--print(findIndex)

					openitem(info.name,cont,true,findIndex)

				--end

		end

	end

end

And here is other file file with temporary table (only for tests)

backpacks = {

{name = 'grey backpack',index = 1},

{name = 'red backpack',index = 1},

{name = 'brown backpack',index = 1},

{name = 'camouflage backpack', index = 1}

}

Im still so far from solution, if someone have any idea could be nice for sharing with me
@Raphael could you help here? I just want to sell all items from depot, I see 2 ways. First is try to sell all avaiable items from items.xml file, the second one is what Im trying to do. I need to open all containers from depot, then my persistent can count that

[Raphael]

You have to open the container before calling getcontainer() on it.

[pedegie]

@Raphael Ye you were right. Anyway when i do that i see i must change my mind. I cant open backpacks as new windows because of screen :


So I have to open one by one in same window. Do you have any suggestions? I thinking about counting same colors backpacks, then I will know which one i need to open after higherwindows() if i get this on most inside. For example i have:



1 brown backpack - > (2 blue bp, 3 red bp)
................................. |................ |
..........................4 blue bp..... (1 brown, 1 yelow)
.............................. |
....................... (3 green, 2 red)
............................ |
........................... etc..


My idea is when I open brown backpack, then i save amount of blue(2) and red(3) bp, next im going to the first blue bp(1) with 4 blue inside, then savying amount of these backpacks etc etc.. When there isnt any backpack inside, Im using higherwindows() and check which backpack was last opened, then opening the next one.

[Raphael]

Your best bet is to use recursion.