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Code: Select all

    for _, refinery in pairs(refineries) do 
      if refinery.caninsert(itemstack) then -- if this refinery can hold the items
        refinery.insert(itemstack)
        inventory.remove(itemstack) -- remove from inventory (the harvester)
        break
        -- do not continue looking at the next refinery...this might be optimized by
        -- removing the full refinery from the refineries table...
        -- but it has a few caveats, if the refinery can accept say, 
        -- copper but not iron, then if you had both in the harvester the refinery 
        -- might not be filled properly, you'd also need a check for if
        -- the one you removed was the last possible refinery
      end
    end

Code: Select all

        refinery.insert(itemstack)
        inventory.remove(itemstack) -- remove from inventory (the harvester)

Code: Select all

          for _, refinery in pairs(refineries) do 
            if refinery.caninsert(itemstack) then -- if this refinery can hold at least 1 item
              local prior = refinery.getitemcount(itemstack.name)
              refinery.insert(itemstack)
              local inserted = refinery.getitemcount(itemstack.name) - prior
              inventory.remove(inserted) -- remove from inventory (the harvester)
              if inserted ~= itemstack.count then
                itemstack.count = itemstack.count - inserted -- decrement count for next refinery
              else
                break -- move to next item stack
              end
            end
          end