The times when this worked were when I used ZIPCrypto compression. It's with AES-256 that it fails. How to get around this please?
I previously had success using the following Python code to open a password protected .zip file created with 7-Zip:
import zipfile
zip_file = zipfile.ZipFile('crack_me.zip')
output_verbose = 1 # increase that for long password list
with open('passwords.txt', 'rb') as password_list:
for index, line in enumerate(password_list):
try:
pwd = line.strip(b'\r\n')
zip_file.extractall(pwd=pwd)
except RuntimeError as e:
print(e)
if index % output_verbose == 0:
print('{}. The {} word not matched.'.format(index + 1, pwd))
else:
print('{}. Wow ! found the password: {}'.format(index + 1, pwd))
break
zip_file.close()
However, for no understandable reason, it has only worked a couple of times out of many attempts. Most times is gives "That compression method is not supported" for the exception.
I've tried deleting, renaming, re-creating the .zip file but no success. It makes no sense to me that it works occasionally.
I tried to simplify the issue as below:
import zipfile
zip_file = zipfile.ZipFile('crack_me.zip')
try:
zip_file.extractall(pwd=b"password")
print('Opened')
except RuntimeError as e:
print(e)
But I get the same exception. I've tried variations of pwd such as bytes("password", "utf-8) and others.
The provided password opens the .zip file when opening with 7-Zip.
What is going on here please?
Try using pyzipper. Here is a sample script:
import pyzipper
password = 'abc'
with pyzipper.AESZipFile('yourdocument.zip', 'r', compression=pyzipper.ZIP_DEFLATED, encryption=pyzipper.WZ_AES) as extracted_zip:
extracted_zip.extractall(pwd=str.encode(password))
In order for the zipfile library to work on a password protected zip you need to have ticked the "Zip legacy encryption" option when setting up the password.
Related
Have a peculiar issue that I can't seem to fix on my own..
I'm attempting to FTP a list of files in a directory over to an iSeries IFS using Python's ftplib library.
Note, the files are in a single subdirectory down from the python script.
Below is an excerpt of the code that is giving me trouble:
from ftplib import FTP
import os
localpath = os.getcwd() + '/Files/'
def putFiles():
hostname = 'host.name.com'
username = 'myuser'
password = 'mypassword'
myftp = FTP(hostname)
myftp.login(username, password)
myftp.cwd('/STUFF/HERE/')
for file in os.listdir(localpath):
if file.endswith('.csv'):
try:
file = localpath + file
print 'Attempting to move ' + file
myftp.storbinary("STOR " + file, open(file, 'rb'))
except Exception as e:
print(e)
The specific error that I am getting throw is:
Attempting to move /home/doug/Files/FILE.csv
426-Unable to open or create file /home/doug/Files to receive data.
426 Data transfer ended.
What I've done so far to troubleshoot:
Initially I thought this was a permissions issue on the directory containing my files. I used chmod 777 /home/doug/Files and re-ran my script, but the same exception occured.
Next I assumed there was an issue between my machine and the iSeries. I validated that I could indeed put files by using ftp. I was successfully able to put the file on the iSeries IFS using the shell FTP.
Thanks!
Solution
from ftplib import FTP
import os
localpath = os.getcwd() + '/Files/'
def putFiles():
hostname = 'host.name.com'
username = 'myuser'
password = 'mypassword'
myftp = FTP(hostname)
myftp.login(username, password)
myftp.cwd('/STUFF/HERE/')
for csv in os.listdir(localpath):
if csv.endswith('.csv'):
try:
myftp.storbinary("STOR " + csv, open(localpath + csv, 'rb'))
except Exception as e:
print(e)
As written, your code is trying to execute the following FTP command:
STOR /home/doug/Files/FILE.csv
Meaning it is trying to create /home/doug/Files/FILE.csv on the IFS. Is this what you want? I suspect that it isn't, given that you bothered to change the remote directory to /STUFF/HERE/.
If you are trying to issue the command
STOR FILE.csv
then you have to be careful how you deal with the Python variable that you've named file. In general, it's not recommended that you reassign a variable that is the target of a for loop, precisely because this type of confusion can occur. Choose a different variable name for localpath + file, and use that in your open(..., 'rb').
Incidentally, it looks like you're using Python 2, since there is a bare print statement with no parentheses. I'm sure you're aware that Python 3 is recommended by now, but if you do stick to Python 2, it's recommended that you avoid using file as a variable name, because it actually means something in Python 2 (it's the name of a type; specifically, the return type of the open function).
I'm trying to control exceptions when reading files, but I have a problem. I'm new to Python, and I am not yet able to control how I can catch an exception and still continue reading text from the files I am accessing. This is my code:
import errno
import sys
class Read:
#FIXME do immutables this 2 const
ROUTE = "d:\Profiles\user\Desktop\\"
EXT = ".txt"
def setFileReaded(self, fileToRead):
content = ""
try:
infile = open(self.ROUTE+fileToRead+self.EXT)
except FileNotFoundError as error:
if error.errno == errno.ENOENT:
print ("File not found, please check the name and try again")
else:
raise
sys.exit()
with infile:
content = infile.read()
infile.close()
return content
And from another class I tell it:
read = Read()
print(read.setFileReaded("verbs"))
print(read.setFileReaded("object"))
print(read.setFileReaded("sites"))
print(read.setFileReaded("texts"))
Buy only print this one:
turn on
connect
plug
File not found, please check the name and try again
And no continue with the next files. How can the program still reading all files?
It's a little difficult to understand exactly what you're asking here, but I'll try and provide some pointers.
sys.exit() will terminate the Python script gracefully. In your code, this is called when the FileNotFoundError exception is caught. Nothing further will be ran after this, because your script will terminate. So none of the other files will be read.
Another thing to point out is that you close the file after reading it, which is not needed when you open it like this:
with open('myfile.txt') as f:
content = f.read()
The file will be closed automatically after the with block.
I am trying to make a simple bruteforcer for rar files. My code is...
import rarfile
file = input("Password List Directory: ")
rarFile = input("Rar File: ")
passwordList = open(file,"r")
for i in passwordList:
try :
rarfile.read(rarFile, psw=i)
print('[+] Password Found: '+i)
except Exception as e:
print('[-] '+i+' is not a password ')
passwordList.close()
I think this has to do with my use of the module, because when I input a password list that I am 10000% sure contains the password to the rarFile, it prints the exception.
The real problem here is that you are catching all exceptions, not just the one you want. So use except rarfile.PasswordRequired: That will show you that the error is not a missing password. Instead there is no function read in the rarfile module.
Have a look at some Documentation. Rar encryption is per file, not per archive.
You need to create a object from the RarFile class and try the password on each file in the archive. (or just the first if you know that is encrypted)
import rarfile
file = input("Password List Directory: ")
rarFilename = input("Rar File: ")
rf = rarfile.RarFile(rarFilename)
passwordList = open(file,"r")
first_file = next(rf.infolist)
for i in passwordList:
password = i.rstrip()
try:
rf.open(first_file, psw=password)
print(password, "found")
except rarfile.PasswordRequired:
print(password,"is not a password")
When you open and read lines from a file, the "new line" character is kept
at the end of the line. This needs to be stripped from each line.
for i in passwordList:
password = i.rstrip()
try :
rarfile.read(rarFile, psw=password)
print('[+] Password Found: '+password)
I am going to start of by showing the code I have thus far:
def err(em):
print(em)
exit
def rF(f):
s = ""
try:
fh = open(f, 'r')
except IOError:
e = "Could not open the file: " + f
err(e)
try:
with fh as ff:
next(ff)
for l in ff:
if ">" in l:
next(ff)
else:
s += l.replace('\n','').replace('\t','').replace('\r','')
except:
e = "Unknown Exception"
err(e)
fh.close()
return s
For some reason the python shell (I am using 3.2.2) freezes up whenever I tried to read a file by typing:
rF("mycobacterium_bovis.fasta")
The conditionals in the rF function are to prevent reading each line that starts with a ">" token. These lines aren't DNA/RNA code (which is what I am trying to read from these files) and should be ignored.
I hope anyone can help me out with this, I don't see my error.
As per the usual, MANY thanks in advance!
EDIT:
*The problem persists!*
This is the code I now use, I removed the error handling which was a fancy addition anyway, still the shell freezes whenever attempting to read a file. This is my code now:
def rF(f):
s = ""
try:
fh = open(f, 'r')
except IOError:
print("Err")
try:
with fh as ff:
next(ff)
for l in ff:
if ">" in l:
next(ff)
else:
s += l.replace('\n','').replace('\t','').replace('\r','')
except:
print("Err")
fh.close()
return s
You didn't ever define e.
So you'll get a NameError that is being hidden by the naked except:.
This is why it is good and healthy to specify the exception, e.g.:
try:
print(e)
except NameError as e:
print(e)
In cases like yours, though, when you don't necessarily know what the exception will be you should at least use this method of displaying information about the error:
import sys
try:
print(e)
except: # catch *all* exceptions
e = sys.exc_info()[1]
print(e)
Which, using the original code you posted, would have printed the following:
name 'e' is not defined
Edit based on updated information:
Concatenating a string like that is going to be quite slow if you have a large file.
Consider instead writing the filtered information to another file, e.g.:
def rF(f):
with open(f,'r') as fin, open('outfile','w') as fou:
next(fin)
for l in fin:
if ">" in l:
next(fin)
else:
fou.write(l.replace('\n','').replace('\t','').replace('\r',''))
I have tested that the above code works on a FASTA file based on the format specification listed here: http://en.wikipedia.org/wiki/FASTA_format using Python 3.2.2 [GCC 4.6.1] on linux2.
A couple of recommendations:
Start small. Get a simple piece working then add a step.
Add print() statements at trouble spots.
Also, consider including more information about the contents of the file you're attempting to parse. That may make it easier for us to help.
In my app, I write to an excel file. After writing, the user is able to view the file by opening it. But if the user forgets to close the file before any further writing, a warning message should appear. So I need a way to check this file is open before the writing process. Could you supply me with some python code to do this task?
If all you care about is the current process, an easy way is to use the file object attribute "closed"
f = open('file.py')
if f.closed:
print 'file is closed'
This will not detect if the file is open by other processes!
source: http://docs.python.org/2.4/lib/bltin-file-objects.html
I assume that you're writing to the file, then closing it (so the user can open it in Excel), and then, before re-opening it for append/write operations, you want to check that the file isn't still open in Excel?
This is how you could do that:
while True: # repeat until the try statement succeeds
try:
myfile = open("myfile.csv", "r+") # or "a+", whatever you need
break # exit the loop
except IOError:
input("Could not open file! Please close Excel. Press Enter to retry.")
# restart the loop
with myfile:
do_stuff()
For windows only
None of the other provided examples would work for me when dealing with this specific issue with excel on windows 10. The only other option I could think of was to try and rename the file or directory containing the file temporarily, then rename it back.
import os
try:
os.rename('file.xls', 'tempfile.xls')
os.rename('tempfile.xls', 'file.xls')
except OSError:
print('File is still open.')
You could use with open("path") as file: so that it automatically closes, else if it's open in another process you can maybe try
as in Tims example you should use except IOError to not ignore any other problem with your code :)
try:
with open("path", "r") as file: # or just open
# Code here
except IOError:
# raise error or print
Using
try:
with open("path", "r") as file:#or just open
may cause some troubles when file is opened by some other processes (i.e. user opened it manually).
You can solve your poblem using win32com library.
Below code checks if any excel files are opened and if none of them matches the name of your particular one, openes a new one.
import win32com.client as win32
xl = win32.gencache.EnsureDispatch('Excel.Application')
my_workbook = "wb_name.xls"
xlPath="my_wb_path//" + my_workbook
if xl.Workbooks.Count > 0:
# if none of opened workbooks matches the name, openes my_workbook
if not any(i.Name == my_workbook for i in xl.Workbooks):
xl.Workbooks.Open(Filename=xlPath)
xl.Visible = True
#no workbooks found, opening
else:
xl.Workbooks.Open(Filename=xlPath)
xl.Visible = True
'xl.Visible = True is not necessary, used just for convenience'
Hope this will help
Try this method if the above methods corrupt your excel file.
This function attempts to rename the file with its own name. If the file has already been opened, the edit will be reject by the os and an OSError exception will be raised. It does not touch the inner code so it will not corrupt your excel files. LMK if it worked for you.
def check_file_status(self):
try:
os.rename("file1.xlsx", "file1.xlsx")
print("File is closed.")
except OSError:
print("File is opened.")
if myfile.closed == False:
print("File is still open ################")
Just use this function. It will close any already opened excel file
import os
def close():
try:
os.system('TASKKILL /F /IM excel.exe')
except Exception:
print("KU")
close()