how to import files from command line in python - python

I am using python and I am supposed to read a file from command line for further processing. My input file has a binary that should be read for further processing. Here is my input file sub.py:
CODE = " \x55\x48\x8b\x05\xb8\x13\x00\x00"
and my main file which should read this is like the following:
import pyvex
import archinfo
import fileinput
import sys
filename = sys.argv[-1]
f = open(sys.argv[-1],"r")
CODE = f.read()
f.close()
print CODE
#CODE = b"\x55\x48\x8b\x05\xb8\x13\x00\x00"
# translate an AMD64 basic block (of nops) at 0x400400 into VEX
irsb = pyvex.IRSB(CODE, 0x1000, archinfo.ArchAMD64())
# pretty-print the basic block
irsb.pp()
# this is the IR Expression of the jump target of the unconditional exit at the end of the basic block
print irsb.next
# this is the type of the unconditional exit (i.e., a call, ret, syscall, etc)
print irsb.jumpkind
# you can also pretty-print it
irsb.next.pp()
# iterate through each statement and print all the statements
for stmt in irsb.statements:
stmt.pp()
# pretty-print the IR expression representing the data, and the *type* of that IR expression written by every store statement
import pyvex
for stmt in irsb.statements:
if isinstance(stmt, pyvex.IRStmt.Store):
print "Data:",
stmt.data.pp()
print ""
print "Type:",
print stmt.data.result_type
print ""
# pretty-print the condition and jump target of every conditional exit from the basic block
for stmt in irsb.statements:
if isinstance(stmt, pyvex.IRStmt.Exit):
print "Condition:",
stmt.guard.pp()
print ""
print "Target:",
stmt.dst.pp()
print ""
# these are the types of every temp in the IRSB
print irsb.tyenv.types
# here is one way to get the type of temp 0
print irsb.tyenv.types[0]
The problem is that when I run "python maincode.py sub.py' it reads the code as content of the file but its output is completely different from when I directly add CODE into the statement irsb = pyvex.IRSB(CODE, 0x1000, archinfo.ArchAMD64()). Does anyone know what is the problem and how can I solve it? I even use importing from inputfile but it does not read a text.

Have you considered the __import__ way?
You could do
mod = __import__(sys.argv[-1])
print mod.CODE
and just pass the filename without the .py extension as your command line argument:
python maincode.py sub
EDIT: Apparently using __import__ is discouraged. Instead though you can use importlib module:
import sys,importlib
mod = importlib.import_module(sys.argv[-1])
print mod.CODE
..and it should work the same as using __import__
If you need to pass a path to the module, one way is if in each of the directories you added an empty file named
__init__.py
That will allow python to interpret the directories as module namespaces, and you can then pass the path in its module form: python maincode.py path.to.subfolder.sub
If for some reason you cannot or don't want to add the directories as namespaces, and don't want to add the init.py files everywhere, you could also use imp.find_module. Your maincode.py would instead look like this:
import sys, imp
mod = imp.find_module("sub","/path/to/subfolder/")
print mod.code
You'll have to write code which breaks apart your command line input into the module part "sub" and the folder path "/path/to/subfolder/" though. Does that make sense? Once its ready you'll call it like you expect, python maincode.py /path/to/subfolder/sub/

you're reading the code as text, while when reading as file you're likely reading as binary
you probably need to convert binary to text of vice-versa to make this work
Binary to String/Text in Python

Related

Loop an existing script

I'm using a script from a third party I can't modify or show (let's call it original.py) which takes a file and produces some calculations. At the end it ouputs a result (using the print statment).
Since I have many files I decided to make a second script that gets all wanted files and runs them through the original.py
1st get list of all files to run
2nd run each file through the original.py
3rd obtain results from each file
I have the 1st and 2nd step. However, the end result only saves the calculations from the last file it read.
import sys
import original
import glob
import os
fn=str(sys.argv[1])
for filename in sys.argv[1:]:
print(filename)
ficheiros = [f for f in glob.glob(fn)]
for ficheiro in ficheiros:
original.file = bytes(ficheiro,'utf-8')
original.function()
To summarize:
Knowing I can't change the original script (which is made with a print statement) how can I obtain the results for each loop? Is there a better way than using a for loop?.
The first script can be invoked with python original.py
It requires the file to be changed manually inside the script in the original.file line.
This script outputs the result in the console and I redirect it with: python original.py > result.txt
At the moment when I try to run my script, it reads all the correct files in the folder but only returns the results for the last file.
#
(I tried to reformulate the question hopefully it's easier to understand)
#
The problem is due to a mistake in the ````ficheiros = [f for f in glob.glob(fn)]`````it's only reading one file, hence only outputting one result.
Thanks for the time.sleep() trick in the comments.
Solved:
I changed the initial part to:
fn=str(sys.argv[1])
ficheiros= []
for filename in sys.argv[1:]:
ficheiros.append(filename)
#print(filename)
and now it correctly reads all the files and it outputs all the results
Depending on your operating system there are different ways to take what is printed to the console and append it to a file.
For example on Linux, you could run this file that calls original.py for every file python yourfile.py >> outputfile.txt, which will then effectively save everything that is printed into outputfile.txt.
The syntax is similar for Windows.
I'm not quite sure what you're asking, but you could try one of these:
Either redirecting all output to a file for later use, by running the script like so: python secondscript.py > outfilename.txt
Or, and this might or might not work for you, redefining the print command to a function that outputs the result how you want, eg:
def print(x):
with open('outfile.txt','w') as f:
f.write('example: ' + x)
If you choose the second option, I recommend saving the old print function (oldprint = print) so you can restore and use the regular print later.
I don't know if I got exactly what you want. You have a first script named original.py which takes some arguments and returns things in the form of print statements and you would like to grab these prints statements in your scripts to do things?
If so, a solution could be the subprocess module:
Let's say that this is original.py:
print("Hi, I'm original.py")
print("print me!")
And this is main.py:
import subprocess
script_path = "original.py"
print("Executing ", script_path)
process = subprocess.Popen(["python3", script_path], stdout=subprocess.PIPE)
for line in process.stdout:
print(line.decode("utf8"))
You can easily add more arguments in the Popen call like ["arg1", "arg2",] etc.
Output:
Executing original.py
Hi, I'm original.py
print me!
and you can grab the lines in the main.py to do what you want with them.

Print output to a file using another Python module

I have two python modules: buildContent.py which contains code that results in output i want. buildRun.py which i run in order to redirect the output to a file.
I'm trying to save the output from buildContent.py to a file and I did something like this in the buildRun.py:
import buildContent
import sys
with open('out.xhtml', 'w') as f:
sys.stdout = f
print buildContent
I can see my output in the console but the file result is:
<module 'buildContent' from 'here's my path to the file'>
what to do?
the redirection is working properly.
if you replace your print statement with a string you will see that it has
worked.
The reason for that output is that you are not calling any functions within buildcontent, merely importing it.
The solution is to run the buildContent file from within the above where your print statement should be.
see this question for an example
Instead of printing buildContent, just execute that module with the required parameters. Not sure of the content of buildContent but something like this should work:
buildContent(data)
This way the code inside buildContent will run on the "data" and print the results (if the print statements are given in the module). If you did not include print statements in buildContent, collect the output into a variable and print that variable. Something like this:
var = buildContent(data)
print var
If you do not need any data atall to run buildContent, just run:
buildContent()

Python String Query

I recently made a Twitter-bot that takes a specified .txt file and tweets it out, line by line. A lot of the features I built into the program to troubleshoot some formatting issues actually allows the program to work with pretty much any text file.
I would like build in a feature where I can "import" a .txt file to use.
I put that in quotes because the program runs in the command line at them moment.
I figured there are two ways I can tackle this problem but need some guidance on each:
A) I begin the program with a prompt asking which file the user want to use. This is stored as a string (lets say variable string) and the code looks like this-
file = open(string,'r')
There are two main issues with. The first is I'm unsure how to keep the program from crashing if the program specified is misspelled or does not exist. The second is that it won't mesh with future development (eventually I'd like to build app functionality around this program)
B) Somehow specify the desired file somehow in the command line. While the program will still occasionally crash, it isn't as inconvenient to the user. Also, this would lend itself to future development, as it'll be easier to pass a value in through the command line than an internal prompt.
Any ideas?
For the first part of the question, exception handling is the way to go . Though for the second part you can also use a module called argparse.
import argparse
# creating command line argument with the name file_name
parser = argparse.ArgumentParser()
parser.add_argument("file_name", help="Enter file name")
args = parser.parse_args()
# reading the file
with open(args.file_name,'r') as f:
file = f.read()
You can read more about the argparse module on its documentation page.
Regarding A), you may want to investigate
try:
with open(fname) as f:
blah = f.read()
except Exception as ex:
# handle error
and for B) you can, e.g.
import sys
fname = sys.argv[1]
You could also combine the both to make sure that the user has passed an argument:
#!/usr/bin/env python
# encoding: utf-8
import sys
def tweet_me(text):
# your bot goes here
print text
if __name__ == '__main__':
try:
fname = sys.argv[1]
with open(fname) as f:
blah = f.read()
tweet_me(blah)
except Exception as ex:
print ex
print "Please call this as %s <name-of-textfile>" % sys.argv[0]
Just in case someone wonders about the # encoding: utf-8 line. This allows the source code to contain utf-8 characters. Otherwise only ASCII is allowed, which would be ok for this script. So the line is not necessary. I was, however, testing the script on itself (python x.py x.py) and, as a little test, added a utf-8 comment (# รค). In real life, you will have to care a lot more for character encoding of your input...
Beware, however, that just catching any Exception that may arise from the whole program is not considered good coding style. While Python encourages to assume the best and try it, it might be wise to catch expectable errors right where they happen. For example , accessing a file which does not exist will raise an IOError. You may end up with something like:
except IndexError as ex:
print "Syntax: %s <text-file>" % sys.argv[0]
except IOError as ex:
print "Please provide an existing (and accessible) text-file."
except Exception as ex:
print "uncaught Exception:", type(ex)

Python script that prints its source

Is it possible (not necessarly using python introspection) to print the source code of a script?
I want to execute a short python script that also print its source (so I can see which commands are executed).
The script is something like this:
command1()
#command2()
command3()
print some_variable_that_contain_src
The real application is that I want to run a script from IPython with the run -i magic and have as output the source (i.e. the commands executed). In this way I can check which commands are commented at every execution. Moreover, if executed in a Notebook I leave a trace of which commands have been used.
Solution
Using korylprince solution I end up with this one-liner to be put at the beginning of the script:
with open(__file__) as f: print '\n'.join(f.read().split('\n')[1:])
This will print the script source except the first line (that would be only noise). It's also easy to modify the slicing in order to print a different "slice" of the script.
If you want to print the whole file instead, the one-liner simplifies to:
with open(__file__) as f: print f.read()
As long as you're not doing anything crazy with packages, put this at the top of your script
with open(__file__) as f:
print f.read()
Which will read in the current file and print it out.
For python 3 make sure to use instead
print(f.read())
For the most simple answer:
import my_module
print open(my_module.__file__).read()
I also tried using the inspect package.
import inspect
import my_module
source_list = inspect.getsourcelines(my_module)
Will give you a list of strings with the source code defined in it
for line in source_list[0]:
print line
Will print out the entire source code in a readable manner

Saving data in Python without a text file?

I have a python program that just needs to save one line of text (a path to a specific folder on the computer).
I've got it working to store it in a text file and read from it; however, I'd much prefer a solution where the python file is the only one.
And so, I ask: is there any way to save text in a python program even after its closed, without any new files being created?
EDIT: I'm using py2exe to make the program an .exe file afterwards: maybe the file could be stored in there, and so it's as though there is no text file?
You can save the file name in the Python script and modify it in the script itself, if you like. For example:
import re,sys
savefile = "widget.txt"
x = input("Save file name?:")
lines = list(open(sys.argv[0]))
out = open(sys.argv[0],"w")
for line in lines:
if re.match("^savefile",line):
line = 'savefile = "' + x + '"\n'
out.write(line)
This script reads itself into a list then opens itself again for writing and amends the line in which savefile is set. Each time the script is run, the change to the value of savefile will be persistent.
I wouldn't necessarily recommend this sort of self-modifying code as good practice, but I think this may be what you're looking for.
Seems like what you want to do would better be solved using the Windows Registry - I am assuming that since you mentioned you'll be creating an exe from your script.
This following snippet tries to read a string from the registry and if it doesn't find it (such as when the program is started for the first time) it will create this string. No files, no mess... except that there will be a registry entry lying around. If you remove the software from the computer, you should also remove the key from the registry. Also be sure to change the MyCompany and MyProgram and My String designators to something more meaningful.
See the Python _winreg API for details.
import _winreg as wr
key_location = r'Software\MyCompany\MyProgram'
try:
key = wr.OpenKey(wr.HKEY_CURRENT_USER, key_location, 0, wr.KEY_ALL_ACCESS)
value = wr.QueryValueEx(key, 'My String')
print('Found value:', value)
except:
print('Creating value.')
key = wr.CreateKey(wr.HKEY_CURRENT_USER, key_location)
wr.SetValueEx(key, 'My String', 0, wr.REG_SZ, 'This is what I want to save!')
wr.CloseKey(key)
Note that the _winreg module is called winreg in Python 3.
Why don't you just put it at the beginning of the code. E.g. start your code:
import ... #import statements should always go first
path = 'what you want to save'
And now you have path saved as a string

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