milo/apis/storage.lua

@@ -291,6 +291,89 @@ function Storage:listItems(throttle)
 	return self.cache
 end
 
+-- provide a raw list of all the items in all storage chests
+-- it might be beneficial to move this to the adapter class at some point and cache the raw item list in this class at some point
+function Storage:listItemsRaw()
+	local res = {}
+
+  for _, v in pairs(self.nodes) do
+    if v.category == "storage" then
+      local chest = device[v.name]
+			local items = chest.list()
+
+			for slot, item in pairs(items) do
+				items[slot] = itemDB:get(item, function() return chest.getItemMeta(slot) end)
+			end
+
+      res[v.name] = items
+    end
+  end
+
+  return res
+end
+
+-- provide a list of items and which chests provide them
+function Storage:listProviders()
+	local res = {}
+
+	local rawItems = self:listItemsRaw()
+
+	for chest, items in pairs(rawItems) do
+		for slot, item in pairs(items) do
+			local key = table.concat({item.name, item.damage, item.nbtHash}, ":")
+			if not res[key] then
+				res[key] = {}
+			end
+			table.insert(res[key], {item = item, device = device[chest], slot = slot})
+		end
+	end
+	return res
+end
+
+-- defrags the storage system
+function Storage:defrag()
+	local items = self:listProviders()
+
+	for _, providers in pairs(items) do
+		table.sort(providers, function(a, b)
+			return a.item.count < b.item.count
+		end)
+
+		-- We're done when we either compressed the stacks so far, that there's only one left (#providers == 1)
+		-- Or when we've compressed so far, that the there's only one stack which has a lower count than the maxCount
+		-- Because of the sorting, we know that this will be the stack in providers[1], so we check if providers[2] is at the maxCount
+		while #providers > 1 and providers[2].item.count ~= providers[2].item.maxCount do
+			local from = providers[1]
+			local to
+
+			-- We're pushing to the highest stack which is still below the maxCount, this way as many slots as possible will be filled
+			-- This loop is guarenteed to assign a value to "to", as the only cases where it wouldn't (#providers == 1 or no provider with less than maxCount)
+			-- are ruled out by the condition of the outer while loop
+			for i = 2, #providers do
+				if providers[i].item.count < providers[i].item.maxCount then
+					to = providers[i]
+				else
+					-- As this slot is already at maxCount, all the remaining ones will also be due to sorting
+					break
+				end
+			end
+
+			local toMove = math.min(to.item.maxCount - to.item.count, from.item.count)
+			from.device.pushItems(to.device.name, from.slot, toMove, to.slot)
+			to.item.count = to.item.count + toMove
+			from.item.count = from.item.count - toMove
+
+			if from.item.count <= 0 then
+				table.remove(providers, 1)
+			end
+
+			table.sort(providers, function(a, b)
+				return a.item.count < b.item.count
+			end)
+		end
+	end
+end
+
 function Storage:updateCache(adapter, item, count)
 	if not adapter.cache then
 		adapter.dirty = true