I'm doing a basic coding project for class. It's the password locker from "Automate the Boring Stuff".
I have everything coded properly, compiled without errors and I've made the bat file and it's in the right spot.
This is the code (minus the dashes in front of the hashtags)
#! python3
#pw.py This is an insecure Password Locker.
PASSWORDS = {'yahooEmail':'12345', 'googleEmail':'23456','facebook':'34567', 'luggage':'45678'}
import sys, pyperclip
if len(sys.argv) < 2:
print('Usage: py pw.py [account] - copy account password')
sys.exit()
account = sys.argv[1] # first command line arg is the account name
if account in PASSWORDS:
pyperclip.copy(PASSWORDS[acount])
print('Password for ' + account + ' copied to clipboard.')
else:
print('There is no account named ' + account +'.')
So it should, of course, just allow me to type in the key to see the value, and copy/past new values as needed.
But when I run it I get this:
Traceback (most recent call last):
File "C:\Users\shawn\Desktop\Python homework\pw.py", line 6, in <module>
import sys, pyperclip
ModuleNotFoundError: No module named 'pyperclip'
Related
I need to check the change of one parameter, if it has changed - I need to restart the script.
code:
import subprocess, os.path, time
from engine.db_manager import DbManager
DB = DbManager(os.path.abspath(os.path.join('../webserver/db.sqlite3')))
tmbot = None
telegram_config = DB.get_config('telegram')
old_telegram_token = ''
vkbot = None
vk_config = DB.get_config('vk')
old_vk_token = ''
while True:
telegram_config = DB.get_config('telegram')
if old_telegram_token != telegram_config['token']:
if vkbot != None:
tmbot.terminate()
tmbot = subprocess.Popen(['python', 'tm/tm_bot.py'])
old_telegram_token = telegram_config['token']
print('telegram token was updated')
vk_config = DB.get_config('vk')
if old_vk_token != vk_config['token']:
if vkbot != None:
vkbot.terminate()
vkbot = subprocess.Popen(['python', 'vk/vk_bot.py'])
old_vk_token = vk_config['token']
print('vk token was updated')
time.sleep(30)
I get errors:
enter image description here
While there might be subtle differences between unix and windows, the straight-up answer is: you can use PYTHONPATH environment variable to let python know where to look for libraries.
However, if you use venv, I'd recommend activating it first, or calling the relevant binary instead of setting the environment variable.
Consider this scenario: you have a venv at /tmp/so_demo/venv, and you try to run this file:
$ cat /tmp/so_demo/myscript.py
import requests
print("great success!")
Running the system python interpreter will not find the requests module, and will yield the following error:
$ python3 /tmp/so_demo/myscript.py
Traceback (most recent call last):
File "/tmp/so_demo/myscript.py", line 1, in <module>
import requests
ModuleNotFoundError: No module named 'requests'
As I have installed requests in the venv, if I provide the path to python, it will know where to look:
$ PYTHONPATH='/tmp/so_demo/venv/lib/python3.8/site-packages/' python3 /tmp/so_demo/myscript.py
great success!
but using the binary inside the venv is better, as you don't need to be familiar with the internal venv directory paths, and is much more portable (notice that the path I provided earlier depicts the minor version):
$ /tmp/so_demo/venv/bin/python /tmp/so_demo/myscript.py
great success!
I am trying to built a script that creates a playlist on a user's spotify profile. To learn spotipy I decided to try the examples they have on the documentation page.
The code I run is:
import sys
import spotipy
import spotipy.util as util
token = util.prompt_for_user_token('idxxx',
'user-library-read',
client_id='axxx',
client_secret='Bxxx',
redirect_uri='http://localhost')
scope = 'user-library-read'
if len(sys.argv) > 1:
username = sys.argv[1]
else:
print("Usage: %s username" % (sys.argv[0],))
sys.exit()
token = util.prompt_for_user_token(username, scope)
if token:
sp = spotipy.Spotify(auth=token)
results = sp.current_user_saved_tracks()
for item in results['items']:
track = item['track']
print(track['name'] + ' - ' + track['artists'][0]['name'])
else:
print("Can't get token for", username)
The problem occurs when i run the code. I get redirected on my redirect uri and after i paste it back on the terminal i get this:
Traceback (most recent call last):
File "spot01.py", line 9, in <module>
redirect_uri='http://localhost')
File "/home/user/.local/lib/python3.6/site-packages/spotipy/util.py", line 92, in prompt_for_user_token
token = sp_oauth.get_access_token(code, as_dict=False)
File "/home/user/.local/lib/python3.6/site-packages/spotipy/oauth2.py", line 434, in get_access_token
raise SpotifyOauthError(response.reason)
spotipy.oauth2.SpotifyOauthError: Bad Request
I tried to access the oauth2.py from the File Manager and the terminal but it says that this repository does not exist. Also i tried to install spotipy through the github page they have where the neccessary files exist but still nothing.
Any ideas?
Thanks a lot.
I solved the problem by downloading the required files from here https://github.com/plamere/spotipy/tree/master/spotipy
Then I changed some things inside of each .py file and i run every code I wanted to inside there. It must be another more fancy solution but this one worked for me.
Oh before running any command I typed this set of commands on the terminal:
$ bash
$ export SPOTIPY_CLIENT_ID='xxx'
$ export SPOTIPY_CLIENT_SECRET='xxx'
$ export SPOTIPY_REDIRECT_URI='http://localhost/'
where xxx you put your credentials and at the refirect uri your localhost or your github profile link.
I'm creating a really basic program that simulates a terminal with Python3.6, its name is Prosser(The origin is that "Prosser" sounds like "Processer" and Prosser is a command processer).
A problem that I'm having is with command import, this is, all the commands are stored in a folder called "lib" in the root folder of Prosser and inside it can have folders and files, if a folder is in the "lib" dir it can't be named as folder anymore, its name now is package(But this doesn't care for now).
The interface of the program is just a input writed:
Prosser:/home/pedro/documents/projects/prosser/-!
and the user can type a command before the text, like a normal terminal:
Prosser:/home/pedro/documents/projects/prosser/-! console.write Hello World
let's say that inside the "lib" folder exists one folder called "console" and inside it has a file called "write.py" that has the code:
class exec:
def main(args):
print(args)
As you can see the first 2 lines is like a important structure for command execution: The class "exec" is the main class for the command execution and the def "main" is the main and first function that the terminal will read and execute also pass the arguments that the user defined, after that the command will be responsible to catch any error and do what it will be created to do.
At this moment, everything is ok, but now comes the true help that I need of U guys, the command import.
Like I writed the user can type any command, and in the example above I typed a "console.write Hello World" and exists one folder called "console" and one file "write.py". The point is that the packages can be defined by a "dot", this is:
-! write Hello World
Above I only typed "write" and this says that the file is only inside the "lib" folder, it doesn't has a package to storage and separate it, so it is a Freedom command(A command that doesn't has packages or nodes).
-! console.write Hello World
Now I typed above "console.write" and this says that the file has a package or node to storage and separate it, this means that it is a Tied command(A command that has packages or nodes).
With that, a file is separated from the package(s) with a dot, the more dots you put, more folders will be navigated to find the file and proceed to the next execution.
Code
Finnaly the code. With the import statement I tryied this:
import os
import form
curDir = os.path.dirname(os.path.abspath(__file__)) # Returns the current direrctory open
while __name__ == "__main__":
c = input('Prosser:%s-! ' % (os.getcwd())).split() # Think the user typed "task.kill"
path = c[0].split('.') # Returns a list like: ["task", "kill"]
try:
args = c[1:] # Try get the command arguments
format = form.formater(args) # Just a text formatation like create a new line with "$n"
except:
args = None # If no arguments the args will just turn the None var type
pass
if os.path.exists(curDir + "/lib/" + "/".join(path) + ".py"): # If curDir+/lib/task/kill.py exists
module = __import__("lib." + '.'.join(path)) # Returns "lib.task.kill"
module.exec.main(args) # Execute the "exec" class and the def "**main**"
else:
pathlast = path[-1] # Get the last item, in this case "kill"
path.remove(path[-1]) # Remove the last item, "kill"
print('Error: Unknow command: ' + '.'.join(path) + '. >>' + pathlast + '<<') # Returns an error message like: "Error: Unknow command: task. >>kill<<"
# Reset the input interface
The problem is that when the line "module = __import__("lib." + '.'.join(path))" is executed the console prints the error:
Traceback (most recent call last):
File "/home/pedro/documents/projects/prosser/main.py", line 18, in <module>
module.exec.main(path) # Execute the "exec" class and the def "**main**"
AttributeError: module 'lib' has no attribute 'exec'
I also tried to use:
module = \_\_import\_\_(curDir + "lib." + '.'.join(path))
But it gets the same error. I think it's lighter for now. I'd like if someone help me or find some replacement of the code. :)
I think you have error here:
You have diffirent path here:
if os.path.exists(curDir + "/lib/" + "/".join(path) + ".py")
And another here, you dont have curDir:
module = __import__("lib." + '.'.join(path)) # Returns "lib.task.kill"
You should use os.path.join to build paths like this:
module = __import__(os.path.join(curdir, 'lib', path + '.py'))
I'm currently completing an assignment which requires me to create a script to crack a password from a PDF file, I already have a list which contains the password within, I am having issues when prompt to enter the path to the file and met with an Name not define error, please mind I am a novice to coding.
file = raw_input('Path: ')
wordlist = 'wordlist.txt'
word =open(wordlist, 'r')
allpass = word.readlines()
word.close()
for password in allpass:
password = password.strip()
print ("Testing password: "+password)
instance = dispatch('pdf.Application')
try:
instance.Workbooks.Open(file, False, True, None, password)
print ("Password Cracked: "+password)
break
except:
pass
When the program is running, it attempts the first password from the list then proceeds to crash.
python Comsec.py
Path: /home/hall/Desktop/comsec121/examAnswers.pdf
Testing password: 123456
Traceback (most recent call last):
File "Comsec.py", line 11, in <module>
instance = dispatch('pdf.Application')
NameError: name 'dispatch' is not defined
Please excuse my formatting on this website, I am trying my best to help you understand my issue!
Thanks in advance!
This error means that inside your Python script there is no object or function with the name dispatch. You would get that same error if you tried:
instance = this_is_a_function_that_has_not_been_defined('pdf.Application')
Typically, this function should be defined in a Python module. To access it, at the top of your code you should have some import statement like this:
from aBeautifulModuleForPDFs import dispatch
That module will be providing you the missing dispatch function. Checking in Google, I suggest you to try with module pywin32. Install it (run pip install pywin32 in a terminal) and add this line at the beginning of the code:
from win32com.client import Dispatch as dispatch
I recognize there are a decent amount of ValueError questions on here, but it seems none are specifically related to psychopy or my issue. I am coding an experiment from scratch on psychopy (no builder involved). Yesterday, my script was running totally fine. Today I tried running it without adding anything new or taking anything away and it's suddenly giving me this error:
File "/Users/vpam/Documents/fMRI_binding/VSTMbindingpaige.py", line 53, in <module>
script, filename = argv
ValueError: need more than 1 value to unpack
These are lines 52 and 53, apparently something in 53 (the last one) is making this happen, but I can't imagine what since it was working just fine yesterday. Anyone know why it's doing that? (I am running the oldest version of python in order to be able to include corrective audio feedback, but I have been running it on that with success):
from sys import argv
script, filename = argv
This is what I'm calling the filename (in the script it is above those other lines)
from sys import argv
script, filename = argv
from psychopy import gui
myDlg = gui.Dlg(title="Dr. S's experiment")
myDlg.addField('Subject ID','PJP')
ok_data = myDlg.show()
if myDlg.OK:
print(ok_data)
else:
print('user cancelled')
[sID]=myDlg.data
# Data file name stem = absolute path + name; later add .psyexp, .csv, .log, etc
data_file = sID + '_VSTMbinding.txt'
f = open(data_file,'a') #name file here
f.write(sID)
print myDlg.data
It looks like you're using Python2. Python3 gives a more detailed information in it's error message. The problem is that argv only contains a single value and you're trying to unpack it into two variables. argv contains the command line variables -- if this was running yesterday "without any changes" as you suggest, it's because you were providing a filename as a command-line argument.
py2.py
#!/usr/bin/env python
from sys import argv
script, filename = argv
print("Script: {0}\nFilename: {1}".format(script, filename))
py3.py
#!/usr/bin/env python3
from sys import argv
script, filename = argv
print("Script: {0}\nFilename: {1}".format(script, filename))
Running py2.py:
$ charlie on laptop in ~
❯❯ ./py2.py
Traceback (most recent call last):
File "./py2.py", line 4, in <module>
script, filename = argv
ValueError: need more than 1 value to unpack
$ charlie on laptop in ~
❯❯ ./py2.py filename
Script: ./py2.py
Filename: filename
Running py3.py:
$ charlie on laptop in ~
❯❯ ./py3.py
Traceback (most recent call last):
File "./py3.py", line 4, in <module>
script, filename = argv
ValueError: not enough values to unpack (expected 2, got 1)
$ charlie on laptop in ~
❯❯ ./py3.py filename
Script: ./py3.py
Filename: filename