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Handle locked chests

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Luca_S
2020-05-04 23:23:40 +02:00
parent db3c43f8c3
commit f6d67a61f3
1 changed files with 29 additions and 8 deletions
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37
milo/apis/storage.lua
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@@ -327,7 +327,7 @@ function Storage:listProviders(throttle)
if not res[key] then
res[key] = {}
end
table.insert(res[key], {item = item, device = device[chest], slot = slot})
table.insert(res[key], {item = item, device = device[chest], lockedToThis = Util.contains(Util.keys(self.nodes[chest].lock or {}), key) and true or false, slot = slot})
end
end
return res
@@ -338,10 +338,29 @@ function Storage:defrag(throttle)
local items = self:listProviders(throttle)
local slotsSaved = 0
for _, providers in pairs(items) do
table.sort(providers, function(a, b)
-- This will make sure the table is sorted in the following order:
-- Unlocked stacks with less than maxCount items | Locked stacks with less than maxCount items | stacks with more than maxCount items
-- This way the locked stacks will be filled first
local function sortFunction(a, b)
local preferenceA, preferenceB
preferenceA = (a.item.count == a.item.maxCount and 3)
or (a.lockedToThis and 2)
or 1
preferenceB = (b.item.count == b.item.maxCount and 3)
or (b.lockedToThis and 2)
or 1
if preferenceA < preferenceB then
return true
elseif preferenceB > preferenceA then
return false
else
return a.item.count < b.item.count
end)
end
end
for _, providers in pairs(items) do
table.sort(providers, sortFunction)
-- 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
@@ -354,10 +373,14 @@ function Storage:defrag(throttle)
-- 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
-- Give preference to locked chests
if not to.lockedToThis and providers[i].lockedToThis then
to = providers[i]
elseif (to.lockedToThis == providers[i].lockedToThis) and 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
-- If any of the remaining providers is locked that doesn't matter. We wouldn't have been able to push there anyways
break
end
end
@@ -389,9 +412,7 @@ function Storage:defrag(throttle)
slotsSaved = slotsSaved + 1
end
table.sort(providers, function(a, b)
return a.item.count < b.item.count
end)
table.sort(providers, sortFunction)
end
end