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Encrypt folder or zip file using python

So I am trying to encrypt a directory using python and I'm not sure what the best way to do that is. I am easily able to turn the folder into a zip file, but from there I have tried looking up how to encrypt it with AES, but couldn't get that to work and I have also tried encrypting using 7zip to archive the folder, but also couldn't get that to work, so if anybody has another solution to encrypt a directory or could point me in the right direction on how to use one of the previous methods that would be helpful. (I'm on windows if that has any significance)

I still recommend 7-zip.
let's say you want to name the zip folder as myzip.zip
Import subprocess
zp = subprocess.call(['7z', 'a', 'your password', '-y', 'myzip.zip'] + ['your file'])
An alternative way:
Import pyminzip
level=4 #level of compression
pyminizip.compress("your file", "myzip.zip", "your password", level)

Using 7-Zip through the subprocess module works. Here are some issues I encountered and had to resolve:
You need to specify the path to 7zip separate from the cmd variable in the Popen subprocess, and build the command with variables rather than a solid string:
appPath="C:\Program Files\\7-Zip"
zApp="7z.exe"
zAction='a'
zPass='-pPASSWORD'
zAnswer='-y'
zDir=directoryToZip
progDir=os.path.join(appPath,zApp)
cmd = [zApp, zAction, zipFileName, zPass, zAnswer, zDir]
subprocess.Popen(cmd, executable=progDir, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
That will create a zip file (in the location with the name in the zipFileName variable), including the contents (directories and files) inside the "directoryToZip" path
progDir has to be specified for separate from the application you are calling as part of the Open command (this is the executable path), and the command string needed to be built out as variables to deal with the windows backslash escaping setup.

Wildcard or * for matching a datetime python 2.7

I am trying to match the following string and not having any luck. Below you will find my attempt.
LOG FORMAT:
riskserver.2014-04-07-08:45:01.log
I think I only will need the year month and date. So I was attempting a wildcard * which python 2.7 does not seem to like.
cmd = 'tail -n10000 /opt/rubedo/log/riskserver.'+nowFormat+*'
Help is very much appreciated here. Thanks, I hope I explained this well, and some can understand.
I am using subprocess with grep involved.
tail: cannot open `/opt/rubedo/log/riskserver.2014-04-08' for reading: No such file or directory
grep: not: No such file or directory
EDIT:
now = datetime.datetime.now().strftime("%H:%M:%S")
nowFormat = datetime.datetime.now().strftime("%Y\-%m\-%d")

glob is the splat (*) , eg ls *.txt gets processed by the linux shell into ls f1.txt f2.txt f3.txt f4.txt ...
so that ls actually recieves a list of files that match not the matching string. that is what they mean in the comments
nowFormat = "2014-04-07"
cmd = 'tail -n10000 /opt/rubedo/log/riskserver.'+nowFormat+'*'
os.system(cmd) #this will execute it through your linux shell you should see the output, allthough this call will not give you access to the output in python
or in python
import glob
fnames = glob.glob('/opt/rubedo/log/riskserver.'+nowFormat+'*')
print fnames

os.listdir can't see my directory

I am working on a python script that installs an 802.1x certificate on a Windows 8.1 machine. This script works fine on Windows 8 and Windows XP (haven't tried it on other machines).
I have isolated the issue. It has to do with clearing out the folder
"C:\Windows\system32\config\systemprofile\AppData\LocalLow\Microsoft\CryptURLCache\Content"
The problem is that I am using the module os and the command listdir on this folder to delete each file in it. However, listdir errors, saying the folder does not exist, when it does indeed exist.
The issue seems to be that os.listdir cannot see the LocalLow folder. If I make a two line script:
import os
os.listdir("C:\Windows\System32\config\systemprofile\AppData")
It shows the following result:
['Local', 'Roaming']
As you can see, LocalLow is missing.
I thought it might be a permissions issue, but I am having serious trouble figuring out what a next step might be. I am running the process as an administrator from the command line, and it simply doesn't see the folder.
Thanks in advance!
Edit: changing the string to r"C:\Windows\System32\config\systemprofile\AppData", "C:\Windows\System32\config\systemprofile\AppData", or C:/Windows/System32/config/systemprofile/AppData" all produce identical results
Edit: Another unusual wrinkle in this issue: If I manually create a new directory in that location I am unable to see it through os.listdir either. In addition, I cannot browse to the LocalLow or my New Folder through the "Save As.." command in Notepad++
I'm starting to think this is a bug in Windows 8.1 preview.

I encountered this issue recently.
I found it's caused by Windows file system redirector
and you can check out following python snippet
import ctypes
class disable_file_system_redirection:
_disable = ctypes.windll.kernel32.Wow64DisableWow64FsRedirection
_revert = ctypes.windll.kernel32.Wow64RevertWow64FsRedirection
def __enter__(self):
self.old_value = ctypes.c_long()
self.success = self._disable(ctypes.byref(self.old_value))
def __exit__(self, type, value, traceback):
if self.success:
self._revert(self.old_value)
#Example usage
import os
path = 'C:\\Windows\\System32\\config\\systemprofile\\AppData'
print os.listdir(path)
with disable_file_system_redirection():
print os.listdir(path)
print os.listdir(path)
ref : http://code.activestate.com/recipes/578035-disable-file-system-redirector/

You must have escape sequences in your path. You should use a raw string for file/directory paths:
# By putting the 'r' at the start, I make this string a raw string
# Raw strings do not process escape sequences
r"C:\path\to\file"
or put the slashes the other way:
"C:/path/to/file"
or escape the slashes:
# You probably won't want this method because it makes your paths huge
# I just listed it because it *does* work
"C:\\path\\to\\file"

I'm curious as to how you are able to list the contents with those two lines. You are using escape sequences \W, \S, \c, \s, \A in your code. Try escaping the back slash like this:
import os
os.listdir('C:\\Windows\\System32\\config\\systemprofile\\AppData')

File Manipulation: Scripting Question

I have a script which connects to database and gets all records which statisfy the query. These record results are files present on a server, so now I have a text file which has all file names in it.
I want a script which would know:
What is the size of each file in the output.txt file?
What is the total size of all the files present in that text file?
Update:
I would like to know how can I achieve my task using Perl programming language, any inputs would be highly appreciated.
Note: I do not have any specific language constraint, it could be either Perl or Python scripting language which I can run from the Unix prompt. Currently I am using the bash shell and have sh and py script. How can this be done?
My scripts:
#!/usr/bin/ksh
export ORACLE_HOME=database specific details
export PATH=$ORACLE_HOME/bin:path information
sqlplus database server information<<EOF
SET HEADING OFF
SET ECHO OFF
SET PAGESIZE 0
SET LINESIZE 1000
SPOOL output.txt
select * from my table_name;
SPOOL OFF
EOF
I know du -h would be the command which I should be using but I am not sure how should my script be, I have tried something in python. I am totally new to Python and it's my first time effort.
Here it is:
import os
folderpath='folder_path'
file=open('output file which has all listing of query result','r')
for line in file:
filename=line.strip()
filename=filename.replace(' ', '\ ')
fullpath=folderpath+filename
# print (fullpath)
os.system('du -h '+fullpath)
File names in the output text file for example are like: 007_009_Bond Is Here_009_Yippie.doc
Any guidance would be highly appreciated.
Update:
How can I move all the files which are present in output.txt file to some other folder location using Perl ?
After doing step1, how can I delete all the files which are present in output.txt file ?
Any suggestions would be highly appreciated.

In perl, the -s filetest operator is probaby what you want.
use strict;
use warnings;
use File::Copy;
my $folderpath = 'the_path';
my $destination = 'path/to/destination/directory';
open my $IN, '<', 'path/to/infile';
my $total;
while (<$IN>) {
chomp;
my $size = -s "$folderpath/$_";
print "$_ => $size\n";
$total += $size;
move("$folderpath/$_", "$destination/$_") or die "Error when moving: $!";
}
print "Total => $total\n";
Note that -s gives size in bytes not blocks like du.
On further investigation, perl's -s is equivalent to du -b. You should probably read the man pages on your specific du to make sure that you are actually measuring what you intend to measure.
If you really want the du values, change the assignment to $size above to:
my ($size) = split(' ', `du "$folderpath/$_"`);

Eyeballing, you can make YOUR script work this way:
1) Delete the line filename=filename.replace(' ', '\ ') Escaping is more complicated than that, and you should just quote the full path or use a Python library to escape it based on the specific OS;
2) You are probably missing a delimiter between the path and the file name;
3) You need single quotes around the full path in the call to os.system.
This works for me:
#!/usr/bin/python
import os
folderpath='/Users/andrew/bin'
file=open('ft.txt','r')
for line in file:
filename=line.strip()
fullpath=folderpath+"/"+filename
os.system('du -h '+"'"+fullpath+"'")
The file "ft.txt" has file names with no path and the path part is '/Users/andrew/bin'. Some of the files have names that would need to be escaped, but that is taken care of with the single quotes around the file name.
That will run du -h on each file in the .txt file, but does not give you the total. This is fairly easy in Perl or Python.
Here is a Python script (based on yours) to do that:
#!/usr/bin/python
import os
folderpath='/Users/andrew/bin/testdir'
file=open('/Users/andrew/bin/testdir/ft.txt','r')
blocks=0
i=0
template='%d total files in %d blocks using %d KB\n'
for line in file:
i+=1
filename=line.strip()
fullpath=folderpath+"/"+filename
if(os.path.exists(fullpath)):
info=os.stat(fullpath)
blocks+=info.st_blocks
print `info.st_blocks`+"\t"+fullpath
else:
print '"'+fullpath+"'"+" not found"
print `blocks`+"\tTotal"
print " "+template % (i,blocks,blocks*512/1024)
Notice that you do not have to quote or escape the file name this time; Python does it for you. This calculates file sizes using allocation blocks; the same way that du does it. If I run du -ahc against the same files that I have listed in ft.txt I get the same number (well kinda; du reports it as 25M and I get the report as 24324 KB) but it reports the same number of blocks. (Side note: "blocks" are always assumed to be 512 bytes under Unix even though the actual block size on larger disc is always larger.)
Finally, you may want to consider making your script so that it can read a command line group of files rather than hard coding the file and the path in the script. Consider:
#!/usr/bin/python
import os, sys
total_blocks=0
total_files=0
template='%d total files in %d blocks using %d KB\n'
print
for arg in sys.argv[1:]:
print "processing: "+arg
blocks=0
i=0
file=open(arg,'r')
for line in file:
abspath=os.path.abspath(arg)
folderpath=os.path.dirname(abspath)
i+=1
filename=line.strip()
fullpath=folderpath+"/"+filename
if(os.path.exists(fullpath)):
info=os.stat(fullpath)
blocks+=info.st_blocks
print `info.st_blocks`+"\t"+fullpath
else:
print '"'+fullpath+"'"+" not found"
print "\t"+template % (i,blocks,blocks*512/1024)
total_blocks+=blocks
total_files+=i
print template % (total_files,total_blocks,total_blocks*512/1024)
You can then execute the script (after chmod +x [script_name].py) by ./script.py ft.txt and it will then use the path to the command line file as the assumed path to the files "ft.txt". You can process multiple files as well.

You can do it in your shell script itself.
You have all the files names in your spooled file output.txt, all you have to add at the end of existing script is:
< output.txt du -h
It will give size of each file and also a total at the end.

You can use the Python skeleton that you've sketched out and add os.path.getsize(fullpath) to get the size of individual file.
For example, if you wanted a dictionary with the file name and size you could:
dict((f, os.path.getsize(f)) for f in file)
Keep in mind that the result from os.path.getsize(...) is in bytes so you'll have to convert it to get other units if you want.
In general os.path is a key module for manipulating files and paths.

running a system command in a python script

I have been going through "A byte of Python" to learn the syntax and methods etc...
I have just started with a simple backup script (straight from the book):
#!/usr/bin/python
# Filename: backup_ver1.py
import os
import time
# 1. The files and directories to be backed up are specified in a list.
source = ['"C:\\My Documents"', 'C:\\Code']
# Notice we had to use double quotes inside the string for names with spaces in it.
# 2. The backup must be stored in a main backup directory
target_dir = 'E:\\Backup' # Remember to change this to what you will be using
# 3. The files are backed up into a zip file.
# 4. The name of the zip archive is the current date and time
target = target_dir + os.sep + time.strftime('%Y%m%d%H%M%S') + '.zip'
# 5. We use the zip command to put the files in a zip archive
zip_command = "zip -qr {0} {1}".format(target, ' '.join(source))
# Run the backup
if os.system(zip_command) == 0:
print('Successful backup to', target)
else:
print('Backup FAILED')
Right, it fails. If I run the zip command in the terminal it works fine. I think it fails because the zip_command is never actually run. And I don't know how to run it.
Simply typing out zip_command does not work. (I am using python 3.1)

Are you sure that the Python script is seeing the same environment you have access to when you enter the command manually in the shell? It could be that zip isn't on the path when Python launches the command.

It would help us if you could format your code as code; select the code parts, and click on the "Code Sample" button in the editor toolbar. The icon looks like "101/010" and if you hold the mouse pointer over it, the yellow "tool tip" box says "Code Sample <pre></pre> Ctrl+K"
I just tried it, and if you paste code in to the StackOverflow editor, lines with '#' will be bold. So the bold lines are comments. So far so good.
Your strings seem to contain backslash characters. You will need to double each backslash, like so:
target_dir = 'E:\\Backup'
This is because Python treats the backslash specially. It introduces a "backslash escape", which lets you put a quote inside a quoted string:
single_quote = '\''
You could also use a Python "raw string", which has much simpler rules for a backslash. A raw string is introduced by r" or r' and terminated by " or ' respectively. examples:
# both of these are legal
target_dir = r"E:\Backup"
target_dir = r'E:\Backup'

The next step I recommend is to modify your script to print the command string, and just look at the string and see if it seems correct.
Another thing you can try is to make a batch file that prints out the environment variables, and have Python run that, and see what the environment looks like. Especially PATH.
Here is a suggested example:
set
echo Trying to run zip...
zip
Put those in a batch file called C:\mytest.cmd, and then have your Python code run it:
result_code = os.system("C:\\mytest.cmd")
print('Result of running mytest was code', result_code)
If it works, you will see the environment variables printed out, then it will echo "Trying to run zip...", then if zip runs it will print a message with the version number of zip and how to run it.

zip command only work in linux not for windows.. thats why it make an error..