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Call LUA function from GO #124
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The usual approach is to return a table from your Lua script containing all the exported functions of your "module". E.g.: local _M = {}
function _M.foo()
return 42
end
return _M and then something like: l := lua.NewState()
lua.LoadString(l, myLuaScript)
l.ProtectedCall(0, 1, 0) // Run the script, leaves the module at the top of stack
l.Field(-1, "foo") // Lookup the function 'foo' in the module
l.ProtectedCall(0, 0, 0) // Call the function 'foo' from the module |
or you could do: function hello()
return 42
end l := lua.NewState()
lua.LoadString(l, myLuaScript)
l.Global("hello")
l.ProtectedCall(0, 1, 0)
myInt, ok := l.ToInteger(-1) |
I tried to get a function result but instead of 42 it returns 0:
Any hints, what I do wrong? |
Please, change the call The demo file: package main
import (
"fmt"
"github.com/Shopify/go-lua"
)
func main2() {
l := lua.NewState()
lua.OpenLibraries(l)
// lua.LoadString(l, "function hello()\n return 42\nend\n")
lua.DoString(l, "function hello()\n return 42\nend\n")
l.Global("hello")
l.ProtectedCall(0, 1, 0)
myInt, errRet := l.ToInteger(-1)
fmt.Printf("RC = %v\n", errRet)
fmt.Printf("Lua says %d\n", myInt)
} The output: $ go run .
RC = true
Lua says 42 If you wanna test the error, comment the statement $ go run .
RC = false
Lua says 0 And please, always save the error and use it. If the The End |
Perfect! THANKS :-) |
There is one more thing about calling a Lua function in an external script. This code works fine: lua.DoFile(l, "script2.lua")
l.Global("main")
l.PushString("Hello World")
l.ProtectedCall(1, 1, 0)
res, err = l.ToString(-1)
fmt.Printf("RC = %v\n", err)
fmt.Printf("Lua says %s\n", res) script2.lua: function main(s)
return s.." from script file!"
end Output:
While this one throws an error in json.lua: lua.DoFile(l, "script3.lua")
l.Global("main")
l.ProtectedCall(0, 1, 0)
res, err = l.ToString(-1)
fmt.Printf("RC = %v\n", err)
fmt.Printf("Lua says %s\n", res) script3.lua: local json = require("json")
function main()
local data = {
name = "John Doe",
age = 30,
occupation = "Software Engineer"
}
local d = json.encode(data)
return d
end Output:
json.lua is located in the same folder as the script. |
Hello @hschneider, sorry for the delay. But the reported error is not in the shared code. It's inside the json package. The error message report the line 102, as the error point.
Is this package proprietary or it can be shared? The package was suceffuly imported? There's a It seems the package only a converter of a Lua table into a JSON string representation. As I wrote befone, check if the conversion has an error status and check it before return the data. You should agree with a convention about the error. How to proceed if it fail! |
json.lua here: If I call Maybe the problem is, that the requirement is located in the same folder as the script? |
@hschneider Sorry for the delay! Let me tell the bad news my friend! The package You can see by ourself at go-lua/string.go. If you wish to use this package local function encode_string(val)
return '"' .. val:gsub('[%z\1-\31\\"]', escape_char) .. '"'
end
|
OK - thanks you so much for looking into it! |
I implemented gsub/gmatch/match here tluyben@151c0f8#diff-2b61500b8c4093b9a24b8e54a62ef0276046f9df9a0f67abea61a96771e9ab2fR260 but I did not robustly test them; they work for all my cases, so maybe it'll help someone. |
Hi Folks,
I'm wondering if anybody was able to call a Lua function from GO, instead of just run all the file or a buffered script.
I'm aware of need to load all the file (and who any code outside a function will be executed), but I'm thinking about a way of load it and after it was done, We be able of call any function of the Lua script, any time with the desired arguments.
Without the need of recompile and than increasing the speed.
Does anyone here did it?
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