帮我写一个lua读取本地txtzip文件自动上传到服务器

以下是一个基本的Lua脚本，它可以读取本地的txt和zip文件，并自动上传到服务器。你需要根据你的具体需求进行适当的修改和扩展。

local socket = require("socket")
local ltn12 = require("ltn12")
local mime = require("mime")

local function read_file(filename)
    local file = assert(io.open(filename, "rb"))
    local content = file:read("*all")
    file:close()
    return content
end

local function upload_file(filename, url)
    local file_content = read_file(filename)
    local request_body = {
        filename = filename,
        filedata = mime.b64(file_content)
    }

    local request_headers = {
        ["Content-Type"] = "application/x-www-form-urlencoded",
        ["Content-Length"] = tostring(#request_body)
    }

    local response_body = {}

    local res, code, response_headers = socket.http.request{
        url = url,
        method = "POST",
        headers = request_headers,
        source = ltn12.source.string(request_body),
        sink = ltn12.sink.table(response_body)
    }

    if code == 200 then
        print("File uploaded successfully.")
    else
        print("File upload failed.")
    end
end

-- 读取txt文件并上传
local txt_filename = "path/to/your/txt/file.txt"
local server_url = "http://yourserver.com/upload"
upload_file(txt_filename, server_url)

-- 读取zip文件并上传
local zip_filename = "path/to/your/zip/file.zip"
upload_file(zip_filename, server_url)

请确保你已经安装了相关的Lua库，例如luasocket和luafilesystem。你可以使用Lua的包管理器（如luarocks）来安装它们：

luarocks install luasocket
luarocks install luafilesystem

请将path/to/your/txt/file.txt和path/to/your/zip/file.zip替换为你实际的文件路径，将http://yourserver.com/upload替换为你的服务器上传URL