Is there a global variable in Lua that contains the path to the file currently being interpreted? Something like Python's __file__ variable?

I ran a quick for k, v in pairs(_G) do print(k) end in the interpreter to see if I could find anything. Only the following variables were listed?

string xpcall package tostring print os unpack require getfenv setmetatable next assert tonumber io rawequal collectgarbage getmetatable module rawset math debug pcall table newproxy type coroutine _G select gcinfo pairs rawget loadstring ipairs _VERSION dofile setfenv load error loadfile

Any suggestions on how to get the path the file currently being executed?

有帮助吗?

解决方案

The debug library has a getinfo method you can call, which can return, amongst other things, the source file for a function.

local info = debug.getinfo(1,'S');
print(info.source);

That would return the name of the source file (which will begin with an @ symbol, indicating it is a filename) of the function at the first level of the call stack. By passing 1 you are asking for information about the current function. If you passed in 0 it would return =[C] as it would be returning information about the getinfo function itself.

For more detailed information check out the Programming in Lua reference on the official Lua website: http://www.lua.org/pil/23.1.html

其他提示

In Lua 5.2, when a script is loaded via require, it receives as arguments the module name given to require and the filename that require used to open the script:

$ cat a.lua
require"b"
$ cat b.lua
print("in b",...)
$ lua a.lua
in b    b   ./b.lua

In Lua 5.1, only the module name is passed, not the filename.

In response to the answer by lhf:
Being new to Lua, I was initially confused what ... meant. Turns out it's a vararg just like with ANSI C: https://www.lua.org/manual/5.3/manual.html#3.4. In my experience with lua 5.3, using

local packageName, packagePath = ...

got me the package name as when used in a require and the absolute file path of the package.

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