如何通过lua CLI将参数传递给lua文件

Question

I have a LUA CLI which takes in the lua command,

Something like this (lua)> #

Now here inorder to execute the lua file I run the command

(lua)> # dofile("a.lua")

I want a command which will execute the file and also pass a argument to it.

Now here I want to pass a argument to the "a.lua" file which will take this argument and call one more lua file, and this 2nd lua file is called according to the argument, So, I need to parse this argument.

Please, can somebody tell me about the parsing commands that will be used in a.lua. I mean what are the functions to be used to parse it.

Please, can somebody tell me how to pass a argument to this file "a.lua".

Answer 1

Now here inorder to execute the lua file I run the command

This is generally not how you execute Lua files. Usually, if you have some Lua script, you execute it with this command: lua a.lua . You don't type lua and then use the interface there to execute it.

Using a proper command line to execute the script, you can pass string parameters to the file: lua a.lua someParam "Param with spaces" . The a.lua script can then fetch these parameters using the standard Lua ... mechanic:

local params = {...}
params[1] -- first parameter, if any.
params[2] -- second parameter, if any.
#params   -- number of parameters.

However, if you insist on trying to execute this using your method of invoking the interpreter (with lua ) and typing commands into it one by one, then you can do this:

> GlobalVariable = assert(loadfile(`a.lua`))
> GlobalVariable(--[[Insert parameters here]])

However, if you don't want to do it in two steps, with the intermediate global variable, you can do it in one:

> assert(loadfile(`a.lua`))(--[[Insert parameters here]])