Remove specific entry from Lua table

Question

I am inserting into a table like this

Admin = {}

table.insert(Admins, {id = playerId, Count = 0})

And that works fine.

How do I remove that specific admin from that table now?

The following does not work, and Im sure its because ID is stored in an array that's inside of the table, but how would I access that then?

table.remove(Admins, playerId)

Basically, I want to remove from the table Admins, where the ID == playerId that I input.

Answer 1

There are two approaches to remove an entry from the table, both are acceptable ways:

1. myTable[index] = nil
Removes an entry from given index, but adds a hole in the table by maintaining the indices

local Admins = {}
table.insert(Admins, {id = 10, Count = 0})
table.insert(Admins, {id = 20, Count = 1})
table.insert(Admins, {id = 30, Count = 2})
table.insert(Admins, {id = 40, Count = 3})


local function removebyKey(tab, val)
    for i, v in ipairs (tab) do 
        if (v.id == val) then
          tab[i] = nil
        end
    end
end
-- Before
-- [1] = {['Count'] = 0, ['id'] = 10},
-- [2] = {['Count'] = 1, ['id'] = 20},
-- [3] = {['Count'] = 2, ['id'] = 30},
-- [4] = {['Count'] = 3, ['id'] = 40}}
removebyKey(Admins, 20)
-- After
-- [1] = {['Count'] = 0, ['id'] = 10},
-- [3] = {['Count'] = 2, ['id'] = 30},
-- [4] = {['Count'] = 3, ['id'] = 40}

2. table.remove(myTable, index)
Removes the entry from given index and renumbering the indices

local function getIndex(tab, val)
    local index = nil
    for i, v in ipairs (tab) do 
        if (v.id == val) then
          index = i 
        end
    end
    return index
end
local idx = getIndex(Admins, 20) -- id = 20 found at idx = 2
if idx == nil then 
    print("Key does not exist")
else
    table.remove(Admins, idx) -- remove Table[2] and shift remaining entries
end
-- Before is same as above
-- After entry is removed. Table indices are changed
-- [1] = {['id'] = 10, ['Count'] = 0},
-- [2] = {['id'] = 30, ['Count'] = 2},
-- [3] = {['id'] = 40, ['Count'] = 3}