Question

我要插入这样的表

Admin = {}

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

那很好。

我现在如何从该表中删除该特定管理员？

以下内容不起作用，由于ID存储在表内部的数组中，因此不能确定，但是我该如何访问呢？

table.remove(Admins, playerId)

基本上，我想从表Admins中删除我输入的ID == playerId。

Answer 1

有两种方法可以从表中删除条目，两种方法都是可以接受的：

1。 myTable [index] = nil
从给定索引中删除一个条目，但通过维护索引在表中添加一个空洞

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）
从给定索引中删除条目并重新编号索引

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}

从Lua表中删除特定条目

1 个答案: