Moho, run script and get ScriptInterface instance - lua

I need to run Moho lua script which created new document, import some files etc., by command line.
The problem is: to call this function requered ScriptInterface instance. It may called by mouse clicking, by menu selection in Moho ui (hand made calling) and sending ScriptingInterface instance as function parameter, like script:Run(moho). If I trying to call this function from commandline this instance not set as parameter.
So the question is - can I take ScriptInterface instance from some global vars or from somewhere else ??

You can prepare a special .moho file, which includes a layer with embed lua script. If you run command line opening Moho with this special file in it, it will execute the layer script (because it is executed on every new frame enter) and script will receive ScriptInterface instance as moho argument given to its LayerScript function.

Related

Lua wait for log file to be closed

I have a script that is executed via a function LuaExportStop() that is called when a program is stopped.
The same function is used by other scripts. One of those other scripts writes and closes a log file when that function is called. I want to import that log file, parse the contents and write the results to another file or possibly to a MySql DB down the road. I have accomplished the code to do this with a static copy of the file in question.
I believe the problem I am having is that both scripts execute with the same function at the same time so an error is thrown when my script tries to access a file that is open and/or being written.
My question is can I have my script "wait" until the file has been closed to execute? How would this be implemented around this code?
luaExportStop()
local filePath = "h:\\somepath\\stats.log"
newFile = io.open(filePath, "w")
dofile("this is the log file")
newFile:write(pp(stats))
newFile:close()
end

Find name and location of Lua executeable

I need to find the name of the Lua executeable from within a Lua script as it sets up a task for later execution.
Using arg I can find out the name, however this becomes un reliable if options are used. For example, if no arguments are used running within a script print( arg[-1]) would print lua53. However if options are used they would be printed instead, such as -i, and to get the exe I would have change the line to print( arg[-2]).
What method will reliably get the name of the lua binary?
Try this
i=0
repeat i=i-1 until arg[i]==nil
i=i+1
print(i,arg[i])

Assign bash output to variable in .lua script

I would like to assign output of a bash command to a variable in .lua script. Is it possible?
For instance, something similar to:
var = `ps uax | grep myprocess`
Yes, you need to use io.popen for this.
io.popen (prog [, mode])
Starts program prog in a separated process and returns a file handle that you can use to read data from this program (if mode is "r", the default) or to write data to this program (if mode is "w").
This function is system dependent and is not available on all platforms.
Also see How to execute an external command?.
io.popen calls a command but returns a file object so you can read the output of the command, if the second argument is 'r', but you can also pass input to a command with a second argument of 'w'. Unfortunately, you don't get a io.popen2, and you don't get the return code.

Luac don't work with one script

I have embedded lua and I want precompile my script. For that, I call the main of luac (with argc the number of file is 1). My problem is on the function doargs of luac. I don't understand the use of the variable i. Because when I use one script. The result of i after the doargs function is 1. And in the main function we have argc -= i after. And so argc = 0 and I have a error "no file". Any idea ?
luac is meant to be a command line utility for compiling .lua files. This expected usage is the reason why you're getting an error.
When you start an executable the OS passes the name of the program as its first argument (argv[0]). The luac main function assumes it is being called by the OS, so it expects that there will always be at least one argument and its argv[0] will be the name of the executable.
For this reason doargs starts its for loop at 1 and will always ignore that first (0th) argument. It returns how many options it has processed, which is also the offset of the first filename in the argv array. The main function uses this to know where the list of files starts.
If you really want to use the main function to precompile your scripts, then supply an extra dummy argument at the beginning of your argument array and list your files after that. Preferably you should use luac from the command line and supply an output file where the precompiled script will be stored like this:
luac -o outputFile script.lua
Alternatively, take a look at chapter 8 of Programming in Lua (Compilation, Execution, and Errors) for a pure Lua solution, or the the luaL_dofile, luaL_dostring, lua_dump, and lua_load functions in the Reference Manual for a C API solution.

Calling back to a main file from a dofile in lua

Say i have two files:
One is called mainFile.lua:
function altDoFile(name)
dofile(debug.getinfo(1).source:sub(debug.getinfo(1).source:find(".*\\")):sub(2)..name)
end
altDoFile("libs/caller.lua")
function callBack()
print "called back"
end
doCallback()
The other called caller.lua, located in a libs folder:
function doCallback()
print "performing call back"
_G["callBack"]()
end
The output of running the first file is then:
"performing call back"
Then nothing more, i'm missing a line!
Why is callBack never getting executed? is this intended behavior, and how do i get around it?
The fact that the function is getting called from string is important, so that can't be changed.
UPDATE:
I have tested it further, and the _G["callBack"] does resolve to a function (type()) but it still does not get called
Why not just use dofile?
It seems that the purpose of altDoFile is to replace the running script's filename with the script you want to call thereby creating an absolute path. In this case the path for caller.lua is a relative path so you shouldn't need to change anything for Lua to load the file.
Refactoring your code to this:
dofile("libs/caller.lua")
function callBack()
print "called back"
end
doCallback()
Seems to give the result you are looking for:
$ lua mainFile.lua
performing call back
called back
Just as a side note, altDoFile throws an error if the path does not contain a \ character. Windows uses the backslash for path names, but other operating systems like Linux and MacOS do not.
In my case running your script on Linux throws an error because string.find returns nill instead of an index.
lua: mainFile.lua:2: bad argument #1 to 'sub' (number expected, got nil)
If you need to know the working path of the main script, why not pass it as a command line argument:
C:\LuaFiles> lua mainFile.lua C:/LuaFiles
Then in Lua:
local working_path = arg[1] or '.'
dofile(working_path..'/libs/caller.lua')
If you just want to be able to walk back up one directory, you can also modify the loader
package.path = ";../?.lua" .. package.path;
So then you could run your file by doing:
require("caller")
dofile "../Untitled/SensorLib.lua" --use backpath librarys
Best Regards
K.

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