Lua实战之table.remove
Description
local tbl1 = {1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 6}
local tbl2 = {1, 2, 3, 3, 5, 6}
删除 tbl1 中存在于 tbl2 中的元素，使删除之后 tbl1 的值为：{2, 3, 4, 5, 5}
Implemention
local LogManager = require("LogManager")
local tbl1 = {1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 6}
local tbl2 = {1, 2, 3, 3, 5, 6}
for _, v in pairs(tbl2) do
for i = #tbl1, 1, -1 do -- 倒着删
if v == tbl1[i] then
table.remove(tbl1, i)
break -- 跳出内层循环
end
end
end
LogManager:dump(tbl1, "tbl1")
--[[
- "tbl1" = {
- 1 = 2
- 2 = 3
- 3 = 4
- 4 = 5
- 5 = 5
- }
]]