Polling a database with items of different priority - lua

Question

Given a list of items that have priorities ranging between 1 and 4 where one is the most important and 4 is the least important, how can I create a system that polls a central database for updates on the list of items, but polling the higher priority items more frequently than the lower priority items?
for example:
I have items a, b, and c that have priority values 1, 2, and three respectively.
In a perfectly functioning program, item a should be polling for updates more often than b and c, and b should be polling more frequently than c.
My main issue is designing a system in which all items are still polled regularly but the highest priority are polled more frequently.
Sorry I don't have more details but I don't really have any source code or anything to build off of, I'm kinda at a blank slate right now.
Also I'm working Lua, but I'm just looking for a rough pseudo-code version of ideas, I'd like to work out the exact implementation my self. For simplicity sake, polling the database can be represented by a method call: getDatabaseInfo(itemIndex);
All answers appreciated, thank you for your time!

Answer 1

local items = { -- all your items are here
{name = "item #1", update_period = 1 },
{name = "item #2", update_period = 1.5}, -- updates for item#2 will happen 1.5 times less frequent than updates for item#1
{name = "item #3", update_period = 2 }, -- updates for item#3 will happen 2 times less frequent than updates for item#1
{name = "item #4", update_period = 3 }, -- updates for item#4 will happen 3 times less frequent than updates for item#1
}
local function update_one_item()
local min_ctr, itemIndex = math.huge
for idx, item in ipairs(items) do
item.current_counter = item.current_counter or 0
if item.current_counter < min_ctr then
min_ctr = item.current_counter
itemIndex = idx
end
end
for idx, item in ipairs(items) do
item.current_counter = item.current_counter - min_ctr
end
items[itemIndex].current_counter = items[itemIndex].current_counter + items[itemIndex].update_period
getDatabaseInfo(itemIndex);
print(items[itemIndex].name.." updated")
end
for _ = 1, 20 do
update_one_item()
end
Output:
item #1 updated
item #2 updated
item #3 updated
item #4 updated
item #1 updated
item #2 updated
item #1 updated
item #3 updated
item #1 updated
item #2 updated
item #4 updated
item #1 updated
item #3 updated
item #2 updated
item #1 updated
item #1 updated
item #2 updated
item #3 updated
item #4 updated
item #1 updated

Answer 2

Depending on how often you wish for each value to be updated, a loop like this could work.
Things with primary priority are updated every time
Things with secondary priority are updated every 2 cycles
Things with tertiary priority are updated every 3 cycles
Things with quaternary priority are updated every 4 cycles
Code:
for i = 1,math.huge do
for ii = 1,#items do do
if getPriority(ii) == 1 then
items[ii] = getDatabaseInfo(ii)
elseif i%2 == 0 and getPriority(ii) == 2 then
items[ii] = getDatabaseInfo(ii)
elseif i%3 == 0 and getPriority(ii) == 3 then
items[ii] = getDatabaseInfo(ii)
elseif i%4 == 0 and GetPriority(ii) == 4 then
items[ii] = getDatabaseInfo(ii)
end
end
sleep(interval)
end
The way this works is it does a check for if the current iteration number is divisible by 1,2,3, or 4.
One could add a 3rd nested loop which loops over all priorities, so you can have limitless amounts of priorities, but I'll leave that as an exercise to the reader :)
As noted by Piglet in the comments, you can use this shortcut to make the code more efficient:
if i % getPriority(ii) == 0 then ...