I am learning F# and Deedle. I am trying to extract the contents of this TGZ File using SharpZipLib. I downloaded the TGZ to my local drive. I think I am close because out1 works, but out2 errs. I am sure the code could be written better with pipe forwarding or composition, but it first needs to work. Does anyone have any ideas?
open ICSharpCode.SharpZipLib.GZip
open ICSharpCode.SharpZipLib.Tar
let extract path filename =
let fullpath = Path.Combine(path, filename)
let inSt = File.OpenRead(fullpath)
let gSt = new GZipInputStream(inSt)
let tar = TarArchive.CreateOutputTarArchive(gSt)
let out1 = tar.ListContents
let out2 = tar.ExtractContents(path)
out2
extract "C:/Downloads/" "dragontail-12.4.1.tgz"
This is the error:
Error: System.NullReferenceException: Object reference not set to an instance of an object.
at ICSharpCode.SharpZipLib.Tar.TarArchive.ExtractContents(String destinationDirectory, Boolean allowParentTraversal) in /_/src/ICSharpCode.SharpZipLib/Tar/TarArchive.cs:line 620
at ICSharpCode.SharpZipLib.Tar.TarArchive.ExtractContents(String destinationDirectory) in /_/src/ICSharpCode.SharpZipLib/Tar/TarArchive.cs:line 600
at FSI_0104.extract(String path, String filename)
at <StartupCode$FSI_0104>.$FSI_0104.main#()
Does this help?
https://stackoverflow.com/a/52200001/1594263
Looks like you should be using CreateInputTarArchive(). I modified your example to use CreateInputTarArchive(), and it worked for me.
BTW you're just assigning a function to out1, you're not actually calling ListContents().
I'm not a SharpZipLib expert, but this works for me:
use tar = TarArchive.CreateInputTarArchive(gSt, System.Text.Encoding.UTF8)
tar.ExtractContents(path, true)
I think you have to explicitly allow for path traversal to unzip into an absolute path. See SharpZipLib unit tests in this file.
Related
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()
In order to create a Json provider I need to pass a literal with the path. There are several people working on the project from different locations, and the paths are different in each case. (Actually only the beginning of each path). I tried to create a literal with pattern matching but the compiler does not accept it. Is there another way to do this?
My failed attempt is below:
open FSharp.Data
[<Literal>]
let bitbucketRoot = // Error message: This is not a valid constant expression
let computerName = Environment.MachineName
match computerName with
| "DESKTOP-G3OF32U" -> "C:\\Users\\Fernando"
| "HPW8" -> #"H:\Dropbox\"
| _ -> failwith "Unknown computer"
[<Literal>] // Error message: This is not a valid constant expression
let projDataPath = bitbucketRoot + #"Bitbucket\VSProjects\Fractal10\Fractal10\data\"
[<Literal>] // Error message: This is not a valid constant expression
let jsonPath = projDataPath + "fractal.json"
type PathInfo = JsonProvider<Sample=jsonPath>
I would advise that you store it in source control and make it a path relative to your project root, assuming you are working out of a common source control repository.
Either that, or host the sample on a public URL. (I wouldn't actually recommend this because including it in your source repository allows versioning and doesn't publicly expose your data)
You cannot create a conditional literal as the other comments point it out. However this is a fairly frequent use case and the way to deal with it is as follows:
#r #"..\packages\FSharp.Data\lib\net40\FSharp.Data.dll"
open FSharp.Data
open System
open System.IO
[<Literal>]
let JsonSource = __SOURCE_DIRECTORY__ + #"\test.json"
type JSonType = JsonProvider<JsonSource>
let json1 = JSonType.GetSamples()
let anotherPath = #"C:\tmp"
let anotherJson = anotherPath + #"\test.json"
let json2 = JSonType.Load(anotherJson)
The __SOURCE_DIRECTORY__ directive will point to the project root (just display it in the REPL) and then you can add the filename to it and make that a literal. If you check in this file into a git repo, then everyone who checks it out can have it in a relative path, and you can refer it when generating the type. When actually using the type or referring to the full file you can just use the .Load() method to load any file, and this doesn't have to be a literal.
There is actually a second way, which could work for you depending on the circumstances, compile a sample, and distribute it as a .dll. You can refer to this and use it directly without having access to the actual file. Please see the Using the JSON Provider in a Library section at the end of the documentation.
I have not tried referring to the json in a config file, it might also be possible.
The following code runs without a hitch:
On the other hand, I get an access-denied error with this:
The destination is in my personal folder and I have full control. The directory is not read-only. Anyway, in either of those cases, the first code sample should not run either! I appreciate the help ...
In the second sample, you have two problems:
There are back slashes instead of forward slashes, so some of them may get interpreted as escape sequences.
You completely ignore the first parameter of write and specify what I assume is a folder as destination. You can't open a file stream on a folder, no wonder you get access denied.
This should work:
let write filename (ms:MemoryStream) =
let path = System.IO.Path.Combine( "C:/Users/<whatever>/signal_processor", filename )
use fs = new FileStream( path, FileMode.Create )
ms.WriteTo(fs)
I am doing my project on incremental deep drawing using ABAQUS.
I am trying to import a text file of loop program into abaqus script so that there is no need of entering amplitude values manually.
But I am getting an error when trying to import the data using the following code
f = open('data_x', 'r')
values=f.read()
values=f.readline()
Error:
data_x is not defined
Error NameError: name 'data_x' is not defined points that you are using data_x as a name in your code, not as a string (with quotes).
This means that in your code, you probably have something like
f = open(data_x)
Python is trying to figure out which value is associated with data_x, which is a Python name, not a string. Since it's not defined before getting to that line, you are getting an error.
If you want to store the name of a file and then open a file, write
data_x = 'data_x.txt'
f = open(data_x)
You could also directly write
f = open('data_x.txt')
Whichever solution you adopt, make sure that a correct path to the file is passed to the function open, so that it could find the file.
the following is a code sample that takes a list of file names and zips them into a single archive. The problem I'm having is that I'd like for the file described by filname be in the top level of the zip archive (i.e. when the archive is opened, "clientName....xml" is the first thing you see, instead of the folder "XML").
let filename = sprintf "C:\\XML\\ClientName_%s.xml" (System.DateTime.Now.ToString("ddMMyyyy"))
use fs = new FileStream(filename, FileMode.Create)
let xmlSerializer = XmlSerializer(typeof<log>)
xmlSerializer.Serialize(fs,logObj)
fs.Close()
use zipfile = new ZipFile()
let basePath = path.Replace("/", "\\")
for fileObj in files do
let relativeFilePath = basePath + (fileObj.Filename).Replace("/", "\\")
printfn "%s" relativeFilePath
zipfile.AddFile(relativeFilePath) |> ignore
()
zipfile.AddFile(filename) |> ignore
let zipFileName = sprintf "C:\\XML\\Compliance_%s.zip" (System.DateTime.Now.ToString("ddMMyyyy"))
zipfile.Save(zipFileName)
Where does the ZipFile type come from? I don't think this is a standard .NET class... I tried searching and found this library http://dotnetzip.codeplex.com/ which has a class matching to your sample :-)
The mentioned library also has AddFile overload that takes two string - the source file name and a relative file name in the ZIP file. This seems exactly like what you're looking for. I guess the call would be something like zipfile.AddFile(absolutePath, "/")...