Hello I want to get the file id from files on windows with python. When I searched I could only find how to do it in other languages. Does anybody know how I can achieve this in python?
As far as I have looked and researched, there is no such file id available. But instead, you can have the creation date on Windows and Mac, and the last modified on Linux. These two are usually sufficient to find unique files, even if they are renamed, altered, or whatever.
Here's how to do it, along with the source SO thread I found the solution.
import os
import platform
def creation_date(path_to_file):
"""
Try to get the date that a file was created, falling back to when it was
last modified if that isn't possible.
See http://stackoverflow.com/a/39501288/1709587 for explanation.
"""
if platform.system() == 'Windows':
return os.path.getctime(path_to_file)
else:
stat = os.stat(path_to_file)
try:
return stat.st_birthtime
except AttributeError:
# We're probably on Linux. No easy way to get creation dates here,
# so we'll settle for when its content was last modified.
return stat.st_mtime
import os
path_to_file = r"path_to_your_file"
file_id = os.stat(path_to_file, follow_symlinks=False).st_ino
print(hex(file_id))
to check the result from the commandline:
c:\> fsutil file queryfileid path_to_your_file
so in Python you can also use
print(os.popen(fr"fsutil file queryfileid path_to_your_file").read())
or when you have hardlinks:
print(os.popen(fr"fsutil hardlink list path_to_your_file").read())
to find the filename with an id:
print(os.popen(fr'fsutil file queryFileNameById c:\ the_file_id').read())
Related
Well I searched a lot and found different ways to open program in python,
For example:-
import os
os.startfile(path) # I have to give a whole path that is not possible to give a full path for every program/software in my case.
The second one that I'm currently using
import os
os.system(fileName+'.exe')
In second example problem is:-
If I want to open calculator so its .exe file name is calc.exe and this happen for any other programs too (And i dont know about all the .exe file names of every program).
And assume If I wrote every program name hard coded so, what if user installed any new program. (my program wont able to open that program?)
If there is no other way to open programs in python so Is that possible to get the list of all install program in user's computer.
and there .exe file names (like:- calculator is calc.exe you got the point).
If you want to take a look at code
Note: I want generic solution.
There's always:
from subprocess import call
call(["calc.exe"])
This should allow you to use a dict or list or set to hold your program names and call them at will. This is covered also in this answer by David Cournapeau and chobok.
You can try with os.walk :
import os
exe_list=[]
for root, dirs, files in os.walk("."):
#print (dirs)
for j in dirs:
for i in files:
if i.endswith('.exe'):
#p=os.getcwd()+'/'+j+'/'+i
p=root+'/'+j+'/'+i
#print(p)
exe_list.append(p)
for i in exe_list :
print('index : {} file :{}'.format(exe_list.index(i),i.split('/')[-1]))
ip=int(input('Enter index of file :'))
print('executing {}...'.format(exe_list[ip]))
os.system(exe_list[ip])
os.getcwd()+'/'+i prepends the path of file to the exe file starting from root.
exe_list.index(i),i.split('/')[-1] fetches just the filename.exe
exe_list stores the whole path of an exe file at each index
Can be done with winapps
First install winapps by typing:
pip install winapps
After that use the library:
# This will give you list of installed applications along with some information
import winapps
for app in winapps.list_installed():
print(app)
If you want to search for an app you can simple do:
application = 'chrome'
for app in winapps.search_installed(application):
print(app)
I need to read an msi file and make some queries to it. But it looks like despite it is a standard lib for python, it has poor documentation.
To make queries I have to know database schema and I can't find any examples or methods to get it from the file.
Here is my code I'm trying to make work:
import msilib
path = "C:\\Users\\Paul\\Desktop\\my.msi" #I cannot share msi
dbobject = msilib.OpenDatabase(path, msilib.MSIDBOPEN_READONLY)
view = dbobject.OpenView("SELECT FileName FROM File")
rec = view.Execute(None)
r = v.Fetch()
And the rec variable is None. But I can open the MSI file with InstEd tool and see that File is present in the tables list and there are a lot of records there.
What I'm doing wrong?
Your code is suspect, as the last line will throw a NameError in your sample. So let's ignore that line.
The real problem is that view.Execute returns nothing of use. Under the hoods, the MsiViewExecute function only returns success or failure. After you call that, you then need to call view.Fetch, which may be what your last line intended to do.
I'm attempting to use a Python module (python-dvdvideo to be exact) to clone an ISO image. The provided class works fine if I pass it a filepath to an ISO file that is already on my computer, but it throws an exception if I attempt to pass it the drive letter of my CDROM drive instead.
After quickly inspecting the library's code, I determined that the class is expecting either a regular file or a block special device file, as shown here:
def __init__(self, filename):
s = os.stat(filename)
if stat.S_ISREG(s.st_mode):
f = self.File(filename)
elif stat.S_ISBLK(s.st_mode):
f = DvdCssFile(filename)
else:
raise RuntimeError
This leads me to my question: Is there a way to treat a Windows CDROM drive as either of these? I'm vaguely familiar with how Linux works in this regard (it treats a CDROM drive as a block device file under /dev/*), but not with how Windows sees drives.
While trying to do something similar I found this thread useful. Based on the information there (and also here) I created this, which shows you the basics:
import os
driveName = "D"
# Get Window raw block device name from logical drive
# Adapted from https://stackoverflow.com/a/6523306/1209004
deviceName = "\\\\.\\" + driveName + ":"
# Open as file object
# Adapted from https://stackoverflow.com/q/7135398/1209004
d = os.fdopen(os.open(deviceName, os.O_RDONLY|os.O_BINARY), 'rb+')
# Read data
data = d.read()
# Close file object
d.close()
# Write data to an output file
fOut = open('data.bin','wb')
fOut.write(data)
fOut.close()
One thing I noted is that compared against dedicated imaging tools like IsoBuster the data read in this way may be incomplete. Also, it doesn't appear to work to access data sessions on an 'enhanced' audio CD. So use with caution.
I'm trying to code a simple application that must read all currently open files within a certain directory.
More specificly, I want to get a list of files open anywhere inside my Documents folder,
but I don't want only the processes' IDs or process name, I want the full path of the open file.
The thing is I haven't quite found anything to do that.
I couldn't do it neither in linux shell (using ps and lsof commands) nor using python's psutil library. None of these is giving me the information I need, which is only the path of currently open files in a dir.
Any advice?
P.S: I'm tagging this as python question (besides os related tags) because it would be a plus if it could be done using some python library.
This seems to work (on Linux):
import subprocess
import shlex
cmd = shlex.split('lsof -F n +d .')
try:
output = subprocess.check_output(cmd).splitlines()
except subprocess.CalledProcessError as err:
output = err.output.splitlines()
output = [line[3:] for line in output if line.startswith('n./')]
# Out[3]: ['file.tmp']
it reads open files from current directory, non-recursively.
For recursive search, use +D option. Keep in mind, that it is vulnerable to race condition - when you get your ouput, situation might have changed already. It is always best to try to do something (open file), and check for failure, e.g. open file and catch exception or check for null FILE value in C.
Currently on a project that my client needs the reference file path to
remain in linux format. For example
A.ma , referencing objects from --> //linux/project/scene/B.ma
B.ma , referencing objects from --> //linux/project/scene/C.ma
Most of our Maya license here however are on Windows. I can run a
Python script that convert all the paths windows paths and save the
file. For example
Z:\project\scene\B.ma
However I'm trying to figure out a way to do this without converting
or altering the original file.... I'll try to explain what I'm trying to do.
Run the script to open the file.
The script checks for the linux formatted reference path, and all
child path down the hierarchy.
Maps all paths to their appropriate windows formatted paths.
Giving the animators the ability to "save" files normally without running a separate save script.
Is this possible to achieve this with Python script? Or will I need a
fully-compiled plug in to get this to work?
Any suggestion is greatly appreciated.
edit: Thank you for your input.
A little more clarification. The projects were set up for us by a remote company and part of the requirement is that we have to keep the path as is. They come as absolute path and we have no choice in that matter.
We match the mount //linux/ on our Fedora workstations. That same drive is mapped to Z:\ on our windows workstations. We only have 2 Maya license for Linux tho which is why I'm trying to do this.
Here is a solution. First step is to create a dict that keeps track of linux/windows references (don't forget to import the re module for regexp):
>>> def windows_path(path):
return path.replace('//linux', 'Z:').replace('/', '\\')
>>> reg = re.compile('(\w+\.ma) , referencing objects from --> (.*)')
>>> d = {}
>>> for line in open('D:\\temp\\Toto.txt'):
match = reg.match(line)
if match:
file_name = match.groups()[0]
linux_path = match.groups()[1]
d[file_name] = (linux_path, windows_path(linux_path))
>>> d
{'B.ma': ('//linux/project/scene/C.ma', 'Z:\\project\\scene\\C.ma'),
'A.ma': ('//linux/project/scene/B.ma', 'Z:\\project\\scene\\B.ma')}
Then you just need to loop on this dict to ask for file save:
>>> for file_name in d.keys():
s = raw_input('do you want to save file %s ? ' % file_name)
if s.lower() in ('y', 'yes'):
# TODO: save your file thanks to d[file][0] for linux path,
# d[file][1] for windows path
print '-> file %s was saved' % file_name
else:
print '-> file %s was not saved' % file_name
do you want to save file B.ma ? n
-> file B.ma was not saved
do you want to save file A.ma ? yes
-> file A.ma was saved
Many Windows applications will interpret paths with two leading "/"s as UNC paths. I don't know if Maya is one of those, but try it out. If Maya can understand paths like "//servername/share/foo", then all you need to do is set up a SMB server named "linux", and the paths will work as they are. I would guess that this is actually what your client does, since the path "//linux" would not make sense in a Linux-only environment.
You can use environment variables to do this. Maya will expand environment vars present in a file path, you could use Maya.env to set them up properly for each platform.
What you are looking for is the dirmap mel command. It is completely non-intrusive to your files as you just define a mapping from your linux paths to windows and/or vice versa. Maya will internally apply the mapping to resolve the paths, without changing them when saving the file.
To setup dirmap, you need to run a MEL script which issues the respective commands on maya startup. UserSetup.mel could be one place to put it.
For more details, see the official documentation - this particular link points to maya 2012, the command is available in Maya 7.0 and earlier as well though:
http://download.autodesk.com/global/docs/maya2012/en_us/Commands/dirmap.html