I have string variable which represents the full path of some file, like:
x = "/home/user/.local/share/app/some_file" on Linux
or
x = "C:\\Program Files\\app\\some_file" on Windows
I'm wondering if there is some programmatic way, better then splitting string manually to get to directory path
How do I return directory path (path without filename) in Lua, without loading additional library like LFS, as I'm using Lua extension from other application?
In plain Lua, there is no better way. Lua has nothing working on paths. You'll have to use pattern matching. This is all in the line of the mentality of offering tools to do much, but refusing to include functions that can be replaced with one-liners:
-- onelined version ;)
-- getPath=function(str,sep)sep=sep or'/'return str:match("(.*"..sep..")")end
getPath=function(str,sep)
sep=sep or'/'
return str:match("(.*"..sep..")")
end
x = "/home/user/.local/share/app/some_file"
y = "C:\\Program Files\\app\\some_file"
print(getPath(x))
print(getPath(y,"\\"))
Here is a platform independent and simpler solution based on jpjacobs solution:
function getPath(str)
return str:match("(.*[/\\])")
end
x = "/home/user/.local/share/app/some_file"
y = "C:\\Program Files\\app\\some_file"
print(getPath(x)) -- prints: /home/user/.local/share/app/
print(getPath(y)) -- prints: C:\Program Files\app\
For something like this, you can just write your own code. But there are also libraries in pure Lua that do this, like lua-path or Penlight.
Related
In LibreOffice, It is possible to run python scripts like this:
sURL = "vnd.sun.star.script:file.function?language=Python&location=document"
oScript = scriptProv.getScript(sURL)
x = oScript.Invoke(args, Array(), Array())
In that example 'file' is a filename, and 'function' is the name of a function in that file.
Is it possible to embed script in that URL? sURL="vnd.." & scriptblock & "?language.."
(It seems like the kind of thing that might be possible with the correct URL, or might not be possible if just not supported).
We can use Python's eval() function. Here is an example inspired by JohnSUN's explanation in the discussion. Note: xray() uses XrayTool to show output, but you could replace that line with any output method of your choosing, such as writing to a file.
def runArbitraryCode(*args):
url = args[0]
codeString = url.split("&codeToRun=")[1]
x = eval(codeString)
xray(x)
Now enter this formula in Calc and Ctrl+click on it.
=HYPERLINK("vnd.sun.star.script:misc_examples.py$runArbitraryCode?language=Python&location=user&codeToRun=5+1")
Result: 6
Obligatory caveat: Running eval() on an unknown string is about the worst idea imaginable in terms of security. So hopefully you're the one controlling the URL and not some black hat hacker!
I'm using Lua in Scite on Windows, but hopefully this is a general Lua question.
Let's say I want to write a temporary string content to a temporary file in Lua - which I want to be eventually read by another program, - and I tried using io.tmpfile():
mytmpfile = assert( io.tmpfile() )
mytmpfile:write( MYTMPTEXT )
mytmpfile:seek("set", 0) -- back to start
print("mytmpfile" .. mytmpfile .. "<<<")
mytmpfile:close()
I like io.tmpfile() because it is noted in https://www.lua.org/pil/21.3.html :
The tmpfile function returns a handle for a temporary file, open in read/write mode. That file is automatically removed (deleted) when your program ends.
However, when I try to print mytmpfile, I get:
C:\Users\ME/sciteLuaFunctions.lua:956: attempt to concatenate a FILE* value (global 'mytmpfile')
>Lua: error occurred while processing command
I got the explanation for that here Re: path for io.tmpfile() ?:
how do I get the path used to generate the temp file created by io.tmpfile()
You can't. The whole point of tmpfile is to give you a file handle without
giving you the file name to avoid race conditions.
And indeed, on some OSes, the file has no name.
So, it will not be possible for me to use the filename of the tmpfile in a command line that should be ran by the OS, as in:
f = io.popen("python myprog.py " .. mytmpfile)
So my questions are:
Would it be somehow possible to specify this tmpfile file handle as the input argument for the externally ran program/script, say in io.popen - instead of using the (non-existing) tmpfile filename?
If above is not possible, what is the next best option (in terms of not having to maintain it, i.e. not having to remember to delete the file) for opening a temporary file in Lua?
You can get a temp filename with os.tmpname.
local n = os.tmpname()
local f = io.open(n, 'w+b')
f:write(....)
f:close()
os.remove(n)
If your purpose is sending some data to a python script, you can also use 'w' mode in popen.
--lua
local f = io.popen(prog, 'w')
f:write(....)
#python
import sys
data = sys.stdin.readline()
Here's a screenshot:
https://drive.google.com/open?id=0B4Dqo44FM648VTVKVzBKSERldGc
I would like to have fun contributing to the development of an open source game but this problem is preventing me from doing this.
My LuaRocks install is configured for Lua 5.3 and is set to use MinGW's build tools. My OS is Windows 10 64 bit.
My Lua 5.3.2 binaries were provided by this website (lua-5.3.2_Win64_bin.zip).
If you have an idea for a possible cause please post it.
You have to resolve environment variables manually:
os.getenv (varname)
Returns the value of the process environment variable varname, or nil
if the variable is not defined.
So, in your case you need to obtain the %APPDATA% path first:
os.getenv("APPDATA")
Returns: C:\Users\USERNAME\AppData\Roaming
When searching for packages to load, Lua uses a pre-defined list of folders to search in. This list is defined in package.path which is a semicolon-delimited (;) string. To add a new folder, you must append it's path to package.path:
package.path = package.path .. ";" .. NEWFOLDER .. "\\?.lua"
Finally, your complete solution is this:
package.path = package.path .. ";" .. os.getenv("APPDATA") .. "\\luarocks\\share\\lua\\5.3\\?.lua"
local s = require("socket")
"Lua does not expand environment variable references in package.path, so %APPDATA% won't work. You need the real path. The LuaRocks installation script should have said that: "Note that the %APPDATA% element in the paths above is user specific and it MUST be replaced by its actual value." - siffiejoe
Hi i'm a noob and asking this newbie question, please forgive me.
I've installed successfully FSharpChart in my local directory
...
Added package 'MSDN.FSharpChart.dll.0.60.0' to folder 'C:\Users\Fagui\Documents\GitHub\Learning Fsharp\Expert in F\packages'
Added package 'MSDN.FSharpChart.dll.0.60.0' to 'packages.config'
Successfully installed 'MSDN.FSharpChart.dll 0.60.0' to Expert in F
now, if i do
#load "FSharpChart.fsx";;
^^^^^^^^^^^^^^^^^^^^^^^
stdin(4,1): error FS0078: Unable to find the file 'FSharpChart.fsx' in any of
C:\Users\Fagui\AppData\Local\Temp
additional info:
inside this folder, i see a nupkg file, and a lib directory
Inside the lib directory, there is a dll, and a pdf file,
but i don't see any .fsx file.
basically, F# has installed the package in the active folder for the current project, and F#interactive is in another folder ?? bit strange ?
should i install another time the package ? or what is the way around it ?
thanks
UPDATE:
i don't know why, but apparently when i installed the FSharpChart package, I only got the dll, no fsx file
i managed to load it doing
#I #"C:\Users\Fagui\Documents\GitHub\Learning Fsharp\Expert in F"
#r #"packages\MSDN.FSharpChart.dll.0.60\lib\MSDN.FSharpChart.dll";;
unfortunately, typing the script in F# interactive
open MSDN.FSharp.Charting
let rnd = System.Random()
let rand() = rnd.NextDouble()
let randomPoints = [for i in 0 .. 1000 -> 10.0 * rand(), 10.0 * rand()]
randomPoints |> FSharpChart.Point;;
doesn't yield any chart, but just returns a list
val rnd : Random
val rand : unit -> float
val randomPoints : (float * float) list =
[(9.765916457, 2.272289941); (0.8211438594, 1.625466995);
...
(7.783786034, 7.572208311); (6.497914692, 3.66987128); ...]
val it : ChartTypes.PointChart
this may be due to the fact that the library is not supported anymore, and that i should use a newer library like Thomas Petricek indicated.
So, i did manage to install FSharp.Charting instead
let rnd = System.Random()
let rand() = rnd.NextDouble()
let randomPoints = [for i in 0 .. 1000 -> 10.0 * rand(), 10.0 * rand()]
randomPoints |> Chart.Point;;
and it did work
There is a newer version of the FSharpChart.fsx library which is called F# Charting, so first of all, I would recommend using this newer library instead (the API is quite similar, but F# Charting has a number of improvements).
The documentation for F# Charting also has a detailed page on referencing the library.
Typically, when you reference the library using NuGet, you'll need to specify relative reference:
// On Mac OSX use packages/FSharp.Charting.Gtk.0.90.13/FSharp.Charting.Gtk.fsx
#load "packages/FSharp.Charting.0.90.13/FSharp.Charting.fsx"
Where 0.90.13 is the version of the library that you got from NuGet (you may need to check the folder name - the path references in #load are relative to the place where your script lives).
Yes, F# Interactive is independent of the current project.
Use:
#load #"C:\Users\Fagui\Documents\GitHub\Learning Fsharp\Expert in F\packages\FSharpChart.fsx";;
Also you can use the #I directive if you need to reference assemblies of a specific folder, see the reference.
I have a c# .net library I am looking to use within FSI/FSX. As part of the initialization of the .net lib, by default it expects and references a custom config file (MyAppConfig.xml) which loads various things before it can be used. When using it in c# it gets copied to the bin folder and the app by default expects it to be there and references it there unless there is a specific entry in the app.config to tell it otherwise. (I should add that it does it all by convention rather than injecting a path + filename, as per NLog, say)
I have an f# source file in a console app which will execute this initialization find, but I can't quite work out how to achieve this with FSI/FSX.
So my program.fs looks simply like
open System
open myApp
module Program =
[<EntryPoint>]
let Main(args) =
myApp.Initialization.Load() // references MyAppConfig.xml
Console.WriteLine("do my stuff!")
Console.ReadLine() |> ignore
0
If I try and do the same in FSI or using FSX, I have
#r #"E:\...path to MyApp...\MyApp.dll"
#I #"E:\...path to MyAppConfig.xml ..."
Environment.CurrentDirectory <- #"E:\...path to MyAppConfig.xml ..."
myApp.Initialization.Load() |> ignore // fails ... can't find MyAppConfig.xml
//do my stuff
I suspect that I've not got the paths quite right.
I'd be grateful of a steer
EDIT:
So I've managed to attach a debugger to the c# lib and see where it is looking for the config file - turns out it is "c:\Program Files\Microsoft F#\v4.0\" ( System.AppDomain.CurrentDomain.BaseDirectory) which again shows I've not quite understood how to tell FSI/FSX to use a particular path. If I copy the config file (MyAppConfig.xml) to that location it works fine.
Many thx
S
I'm not sure of the implications, but one possiblity might be temporarily changing the app base:
let origAppBase = AppDomain.CurrentDomain.BaseDirectory
AppDomain.CurrentDomain.SetData("APPBASE", "path_to_MyAppConfig.xml")
myApp.Initialization.Load() |> ignore
AppDomain.CurrentDomain.SetData("APPBASE", origAppBase) //restore original app base