I want to convert long filenames/path to short filenames (8.3).
I'm developing a script that calls a command line tool that only accepts short filenames.
So i need to convert
C:\Ruby193\bin\test\New Text Document.txt
to
C:\Ruby193\bin\test\NEWTEX~1.TXT
So far i found How to get long filename from ARGV which uses WIN32API to convert short to long filenames (the opposite of what I want to achieve).
Is there any way to get the short filename in Ruby?
You can do this using FFI; there's actually an example that covers your exact scenario in their wiki under the heading "Convert a path to 8.3 style pathname":
require 'ffi'
module Win
extend FFI::Library
ffi_lib 'kernel32'
ffi_convention :stdcall
attach_function :path_to_8_3, :GetShortPathNameA, [:pointer, :pointer, :uint], :uint
end
out = FFI::MemoryPointer.new 256 # bytes
Win.path_to_8_3("c:\\program files", out, out.length)
p out.get_string # be careful, the path/file you convert to 8.3 must exist or this will be empty
This ruby code uses getShortPathName and don't need additional modules to be installed.
def get_short_win32_filename(long_name)
require 'win32api'
win_func = Win32API.new("kernel32","GetShortPathName","PPL"," L")
buf = 0.chr * 256
buf[0..long_name.length-1] = long_name
win_func.call(long_name, buf, buf.length)
return buf.split(0.chr).first
end
The windows function you require is GetShortPathName. You could use that in the same manner as described in your linked post.
EDIT: sample usage of GetShortPathName (just as a quick example) - shortname will contain "C:\LONGFO~1\LONGFI~1.TXT" and returned value is 24.
TCHAR* longname = "C:\\long folder name\\long file name.txt";
TCHAR* shortname = new TCHAR[256];
GetShortPathName(longname,shortname,256);
Related
I'm having difficulty to Encrypt large files (bigger than available memory) using GPGME in Ruby.
#!/usr/bin/ruby
require 'gpgme'
def gpgfile(localfile)
crypto = GPGME::Crypto.new
filebasename = File.basename(localfile)
filecripted = crypto.encrypt File.read(localfile), :recipients => "info#address.com", :always_trust => true
File.open("#{localfile}.gpg", 'w') { |file| file.write(filecripted) }
end
gpgpfile("/home/largefile.data")
In this case I got an error of memory allocation:
"read: failed to allocate memory (NoMemoryError)"
Someone can explain me how to read the source file chunk by chunk (of 100Mb for example) and write them passing by the crypting?
The most obvious problem is that you're reading the entire file into memory with File.read(localfile). The Crypto#encrypt method will take an IO object as its input, so instead of File.read(localfile) (which returns the contents of the file as a string) you can pass it a File object. Likewise, you can give an IO object as the :output option, letting you write the output directly to a file instead of in memory:
def gpgfile(localfile)
infile = File.open(localfile, 'r')
outfile = File.open("#{localfile}.gpg", 'w')
crypto = GPGME::Crypto.new
crypto.encrypt(infile, recipients: "info#address.com",
output: outfile,
always_trust: true)
ensure
infile.close
outfile.close
end
I've never used ruby-gpgme, so I'm not 100% sure this will solve your problem since it depends a bit on what ruby-gpgme does behind the scenes, but from the docs and the source I've peeked at it seems like a sanely-built gem so I'm guessing this will do the trick.
I am practicing Ruby, and I am trying to copy contents from file "from" to file "to". can you tell me where I did it wrong?
thanks !
from = "1.txt"
to = "2.txt"
data = open(from).read
out = open(to, 'w')
out.write(data)
out.close
data.close
Maybe I am missing the point, but I think writing it like so is more 'ruby'
from = "1.txt"
to = "2.txt"
contents = File.open(from, 'r').read
File.open(to, 'w').write(contents)
Personally, however, I like to use the Operating systems terminal to do File operations like so. Here is an example on linux.
from = "1.txt"
to = "2.txt"
system("cp #{from} #{to}")
And for Windows I believe you would use..
from = "1.txt"
to = "2.txt"
system("copy #{from} #{to}")
Finally, if you were needing the output of the command for some sort of logging or other reason, I would use backticks.
#A nice one liner
`cp 1.txt 2.txt`
Here is the system and backtick methods documentation.
http://ruby-doc.org/core-1.9.3/Kernel.html
You can't perform data.close — data.class would show you that you have a String, and .close is not a valid String method. By opening from the way you chose to, you lost the File reference after using it with your read. One way to fix that would be:
from = "1.txt"
to = "2.txt"
infile = open(from) # Retain the File reference
data = infile.read # Use it to do the read
out = open(to, 'w')
out.write(data)
out.close
infile.close # And finally, close it
I worked today in a simple script to checksum files in all available hashlib algorithms (md5, sha1.....) I wrote it and debug it with Python2, but when I decided to port it to Python 3 it just won't work. The funny thing is that it works for small files, but not for big files. I thought there was a problem with the way I was buffering the file, but the error message is what makes me think it is something related to the way I am doing the hexdigest (I think) Here is a copy of my entire script, so feel free to copy it, use it and help me figure out what the problem is with it. The error I get when checksuming a 250 MB file is
"'utf-8' codec can't decode byte 0xf3 in position 10: invalid continuation byte"
I google it, but can't find anything that fixes it. Also if you see better ways to optimize it, please let me know. My main goal is to make work 100% in Python 3. Thanks
#!/usr/local/bin/python33
import hashlib
import argparse
def hashFile(algorithm = "md5", filepaths=[], blockSize=4096):
algorithmType = getattr(hashlib, algorithm.lower())() #Default: hashlib.md5()
#Open file and extract data in chunks
for path in filepaths:
try:
with open(path) as f:
while True:
dataChunk = f.read(blockSize)
if not dataChunk:
break
algorithmType.update(dataChunk.encode())
yield algorithmType.hexdigest()
except Exception as e:
print (e)
def main():
#DEFINE ARGUMENTS
parser = argparse.ArgumentParser()
parser.add_argument('filepaths', nargs="+", help='Specified the path of the file(s) to hash')
parser.add_argument('-a', '--algorithm', action='store', dest='algorithm', default="md5",
help='Specifies what algorithm to use ("md5", "sha1", "sha224", "sha384", "sha512")')
arguments = parser.parse_args()
algo = arguments.algorithm
if algo.lower() in ("md5", "sha1", "sha224", "sha384", "sha512"):
Here is the code that works in Python 2, I will just put it in case you want to use it without having to modigy the one above.
#!/usr/bin/python
import hashlib
import argparse
def hashFile(algorithm = "md5", filepaths=[], blockSize=4096):
'''
Hashes a file. In oder to reduce the amount of memory used by the script, it hashes the file in chunks instead of putting
the whole file in memory
'''
algorithmType = hashlib.new(algorithm) #getattr(hashlib, algorithm.lower())() #Default: hashlib.md5()
#Open file and extract data in chunks
for path in filepaths:
try:
with open(path, mode = 'rb') as f:
while True:
dataChunk = f.read(blockSize)
if not dataChunk:
break
algorithmType.update(dataChunk)
yield algorithmType.hexdigest()
except Exception as e:
print e
def main():
#DEFINE ARGUMENTS
parser = argparse.ArgumentParser()
parser.add_argument('filepaths', nargs="+", help='Specified the path of the file(s) to hash')
parser.add_argument('-a', '--algorithm', action='store', dest='algorithm', default="md5",
help='Specifies what algorithm to use ("md5", "sha1", "sha224", "sha384", "sha512")')
arguments = parser.parse_args()
#Call generator function to yield hash value
algo = arguments.algorithm
if algo.lower() in ("md5", "sha1", "sha224", "sha384", "sha512"):
for hashValue in hashFile(algo, arguments.filepaths):
print hashValue
else:
print "Algorithm {0} is not available in this script".format(algorithm)
if __name__ == "__main__":
main()
I haven't tried it in Python 3, but I get the same error in Python 2.7.5 for binary files (the only difference is that mine is with the ascii codec). Instead of encoding the data chunks, open the file directly in binary mode:
with open(path, 'rb') as f:
while True:
dataChunk = f.read(blockSize)
if not dataChunk:
break
algorithmType.update(dataChunk)
yield algorithmType.hexdigest()
Apart from that, I'd use the method hashlib.new instead of getattr, and hashlib.algorithms_available to check if the argument is valid.
I am trying to save compressed strings to a file and load them later for use in the game. I kept getting "in 'finish': buffer error" errors when loading the data back up for use. I came up with this:
require "zlib"
def deflate(string)
zipper = Zlib::Deflate.new
data = zipper.deflate(string, Zlib::FINISH)
end
def inflate(string)
zstream = Zlib::Inflate.new
buf = zstream.inflate(string)
zstream.finish
zstream.close
buf
end
setting = ["nothing","nada","nope"]
taggedskills = ["nothing","nada","nope","nuhuh"]
File.open('testzip.txt','wb') do |w|
w.write(deflate("hello world")+"\n")
w.write(deflate("goodbye world")+"\n")
w.write(deflate("etc")+"\n")
w.write(deflate("etc")+"\n")
w.write(deflate("Setting: name "+setting[0]+" set"+(setting[1].class == String ? "str" : "num")+" "+setting[1].to_s)+"\n")
w.write(deflate("Taggedskill: "+taggedskills[0]+" "+taggedskills[1]+" "+taggedskills[2]+" "+taggedskills[3])+"\n")
w.write(deflate("etc")+"\n")
end
File.open('testzip.txt','rb') do |file|
file.each do |line|
p inflate(line)
end
end
It was throwing errors at the "Taggedskill:" point. I don't know what it is, but trying to change it to "Skilltag:", "Skillt:", etc. continues to throw a buffer error, while things like "Setting:" or "Thing:" work fine, while changing the setting line to "Taggedskill:" continues to work fine. What is going on here?
In testzip.txt, you are storing newline separated binary blobs. However, binary blobs may contain newlines by themselves, so when you open testzip.txt and split it by line, you may end up splitting one binary blob that inflate would understand, into two binary blobs that it does not understand.
Try to run wc -l testzip.txt after you get the error. You'll see the file contains one more line, than the number of lines you are putting in.
What you need to do, is compress the whole file at once, not line by line.
I'm currently experimenting with creating my own gem in Ruby. The gem requires some static resources (say an icon in ICO format). Where do I put such resources within my gem directory tree and how to I access them from code?
Also, parts of my extension are native C code and I would like the C-parts to have access to the resources too.
You can put resources anywhere you want, except in the lib directory. Since it will be will be part of Ruby's load path, the only files that should be there are the ones that you want people to require.
For example, I usually store translated text in the i18n/ directory. For icons, I'd just put them in resources/icons/.
As for how to access these resources... I ran into this problem enough that I wrote a little gem just to avoid repetition.
Basically, I was doing this all the time:
def Your::Gem.root
# Current file is /home/you/code/your/lib/your/gem.rb
File.expand_path '../..', File.dirname(__FILE__)
end
Your::Gem.root
# => /home/you/code/your/
I wrapped this up into a nice DSL, added some additional convenience stuff and ended up with this:
class Your::Gem < Jewel::Gem
root '../..'
end
root = Your::Gem.root
# => /home/you/code/your/
# No more joins!
path = root.resources.icons 'your.ico'
# => /home/you/code/your/resources/icons/your.ico
As for accessing your resources in C, path is just a Pathname. You can pass it to a C function as a string, open the file and just do what you need to do. You can even return an object to the Ruby world:
VALUE your_ico_new(VALUE klass, VALUE path) {
char * ico_file = NULL;
struct your_ico * ico = NULL;
ico_file = StringValueCStr(path);
ico = your_ico_load_from_file(ico_file); /* Implement this */
return Data_Wrap_Struct(your_ico_class, your_ico_mark, your_ico_free, ico);
}
Now you can access it from Ruby:
ico = Your::Ico.new path