How to take multiple screenshots through Selenium in Python? - python-2.7

GECKODRIVER_PATH = 'F:/geckodriver.exe'
firefox_options = Options()
firefox_options .add_argument("-headless")
driver = webdriver.Firefox(executable_path=CHROMEDRIVER_PATH, firefox_options = firefox_options )
test = []
test.append('http://google.com')
test.append('http://stackoverflow.com')
for x in test:
print x
driver.get(x)
driver.set_page_load_timeout(20)
filename = str(x)+'.png'
driver.save_screenshot( filename )
driver.close()
Now, how can I take multiple screenshots and save them in the different filename? As you can see I am trying to save the filename according to domain URL but failed.
See the error below:
http://google.com
http://card.com
Traceback (most recent call last):
File "F:\AutoRecon-master\test.py", line 125, in <module>
driver.get(x)
File "C:\Python27\lib\site-packages\selenium\webdriver\remote\webdriver.py", line 326, in get
self.execute(Command.GET, {'url': url})
File "C:\Python27\lib\site-packages\selenium\webdriver\remote\webdriver.py", line 314, in execute
self.error_handler.check_response(response)
File "C:\Python27\lib\site-packages\selenium\webdriver\remote\errorhandler.py", line 242, in check_response
raise exception_class(message, screen, stacktrace)
selenium.common.exceptions.SessionNotCreatedException: Message: Tried to run command without establishing a connection
Can anyone please tell me what is the exact problem is? Will be a big help.

Try to move driver.close() out of loop:
for x in test:
print x
driver.get(x)
driver.set_page_load_timeout(20)
filename = str(x)+'.png'
driver.save_screenshot( filename )
driver.close()
Also note that x is already a string, so there is no need in str(x)
P.S. I'm not sure that http://stackoverflow.com.png filename is acceptable, you might need to use:
filename = x.split('//')[-1] + '.png'

Tried to run command without establishing a connection
you are closing the browser within your for loop... so the 2nd time through the loop it fails with the error above (since the browser is closed, the connection to geckodriver has already been terminated).
other issues:
you are setting the page_load_timeout after you have already fetched the page, so it is not doing anything useful.
using CHROMEDRIVER_PATH as the name for Geckodriver is just confusing. Chromedriver is not used at all here.

Related

Saving files in Python using "with" method

I am wanting to create a file and save it to json format. Every example I find specifies the 'open' method. I am using Python 2.7 on Windows. Please help me understand why the 'open' is necessary for a file I am saving for the first time.
I have read every tutorial I could find and researched this issue but with no luck still. I do not want to create the file outside of my program and then have my program overwrite it.
Here is my code:
def savefile():
filename = filedialog.asksaveasfilename(initialdir =
"./Documents/WorkingDirectory/",title = "Save file",filetypes = (("JSON
files","*.json"), ("All files", "*.")))
with open(filename, 'r+') as currentfile:
data = currentfile.read()
print (data)
Here is this error I get:
Exception in Tkinter callback Traceback (most recent call last):
File "C:\Python27\lib\lib-tk\Tkinter.py", line 1542, in call
return self.func(*args) File "C:\Users\CurrentUser\Desktop\newproject.py", line 174, in savefile
with open(filename, 'r+') as currentfile: IOError: [Errno 2] No such file or directory:
u'C:/Users/CurrentUser/Documents/WorkingDirectory/test.json'
Ok, I figured it out! The problem was the mode "r+". Since I am creating the file, there is no need for read and write, just write. So I changed the mode to 'w' and that fixed it. I also added the '.json' so it would be automatically added after the filename.
def savefile():
filename = filedialog.asksaveasfilename(initialdir =
"./Documents/WorkingDirectory/",title = "Save file",filetypes = (("JSON
files","*.json"), ("All files", "*.")))
with open(filename + ".json", 'w') as currentfile:
line1 = currentfile.write(stringone)
line2 = currentfile.write(stringtwo)
print (line1,line2)

cx_Freeze: 'The system cannot find the file specified' error during build [win10] [PyQt4] [python2.7]

I'm trying to create a .exe file from a python script (which use PyQt4 GUI and matplotlib). I'm using cx_Freeze version 5.1.1 for 64-bit windows with the following setup.py:
import cx_Freeze
import sys
import matplotlib
base = "Win32GUI"
includes = ["atexit"]
buildOptions = dict(
#create_shared_zip=False,
#append_script_to_exe=True,
includes=includes
)
executables = [cx_Freeze.Executable(script = "main.py", base = base)] # icon = "chart32.jpg")]
cx_Freeze.setup(
name= "1ChPlotGUI",
options = dict(build_exe=buildOptions), # {"build_exe": {"packages": ["matplotlib"], "include_files":["chart32.jpg"]}},
version = "0.01",
description = "1 Channel Plotting app with GUI",
executables = executables
)
after running
python setup.py build
in the cmd from the position of
C:\Users\Us.Er\Pyth-examples\Qt\UI-examples\ChannelplotGUI-to-exe
I have something like this:
running build
running build_exe
copying c:\users\Us.Er\appdata\local\enthought\canopy\user\lib\site-
packages\cx_Freeze\bases\Win32GUI.exe -> build\exe.win-amd64-2.7\main.exe
copying
c:\users\Us.Er\appdata\local\enthought\canopy\user\scripts\python27.dll ->
build\exe.win-amd64-2.7\python27.dll
Traceback (most recent call last):
File "setup.py", line 23, in <module>
executables = executables
File "c:\users\Us.Er\appdata\local\enthought\canopy\user\lib\site-packages\cx_Freeze\dist.py", line 349, in setup
distutils.core.setup(**attrs)
File "C:\Users\Us.Er\AppData\Local\Enthought\Canopy\App\appdata\canopy-1.7.4.3348.win-x86_64\lib\distutils\core.py", line 151, in setup
dist.run_commands()
File "C:\Users\Us.Er\AppData\Local\Enthought\Canopy\App\appdata\canopy-1.7.4.3348.win-x86_64\lib\distutils\dist.py", line 953, in run_commands
self.run_command(cmd)
File "C:\Users\Us.Er\AppData\Local\Enthought\Canopy\App\appdata\canopy-1.7.4.3348.win-x86_64\lib\distutils\dist.py", line 972, in run_command
cmd_obj.run()
File "C:\Users\Us.Er\AppData\Local\Enthought\Canopy\App\appdata\canopy-1.7.4.3348.win-x86_64\lib\distutils\command\build.py", line 127, in run
self.run_command(cmd_name)
File "C:\Users\Us.Er\AppData\Local\Enthought\Canopy\App\appdata\canopy-1.7.4.3348.win-x86_64\lib\distutils\cmd.py", line 326, in run_command
self.distribution.run_command(command)
File "C:\Users\Us.Er\AppData\Local\Enthought\Canopy\App\appdata\canopy-1.7.4.3348.win-x86_64\lib\distutils\dist.py", line 972, in run_command
cmd_obj.run()
File "c:\users\Us.Er\appdata\local\enthought\canopy\user\lib\site-packages\cx_Freeze\dist.py", line 219, in run
freezer.Freeze()
File "c:\users\Us.Er\appdata\local\enthought\canopy\user\lib\site-packages\cx_Freeze\freezer.py", line 626, in Freeze
self._FreezeExecutable(executable)
File "c:\users\Us.Er\appdata\local\enthought\canopy\user\lib\site-packages\cx_Freeze\freezer.py", line 232, in _FreezeExecutable
self._AddVersionResource(exe)
File "c:\users\Us.Er\appdata\local\enthought\canopy\user\lib\site-packages\cx_Freeze\freezer.py", line 172, in _AddVersionResource
stamp(fileName, versionInfo)
File "c:\users\Us.Er\appdata\local\enthought\canopy\user\lib\site-packages\win32\lib\win32verstamp.py", line 159, in stamp
h = BeginUpdateResource(pathname, 0)
pywintypes.error: (2, 'BeginUpdateResource', 'The system cannot find the file specified.')
What is the possible solution for above problem?
EDIT
To be clear: I don't want to add any icon for now. I am happy with just a simple, working .exe
I have added the target in this way:
executables = [cx_Freeze.Executable(script = "main.py", base = base, targetName="main.exe")]
I have tried to add
targetDir = "C:\Users\Us.Er\Pyth-examples\Qt\UI-examples\ChannelplotGUI-to-exe"
anyway, the return is:
TypeError: __init__() got an unexpected keyword argument 'targetDir'
The main problem is as before - the error with last lines as:
h = BeginUpdateResource(pathname, 0)
pywintypes.error: (2, 'BeginUpdateResource', 'The system cannot find the file specified.')
In my case the issue was solved by changing
'build_exe': 'build_folder'
to
'build_exe': './/build_folder'
This is because open(pathname) would work on the pathname='build_folder\\executable.exe', allowing the code to advance during the check at the begginning of the stamp method, but BeginUpdateResource(pathname, 0) would only work on './/build_folder\\executable.exe'
Solution
options = {
'build_exe': {
'build_exe': './/build'
}
}
Note: .//build here, build is the folder name where you want to build your project.
Thanks to help of user jpeg I managed to successfully freeze the script.
To eliminate the problem with path I have added a line print(pathname) in win32verstamp.py before line 159.
The print statement worked fine showing the relative path to the newly made .exe file.
Despite showing the correct path, the error was still present. I went to the stamp() definition in win32verstamp.py and found a try - except block, and have inserted print(pathname) over there.
The part is:
def stamp(pathname, options):
# For some reason, the API functions report success if the file is open
# but doesnt work! Try and open the file for writing, just to see if it is
# likely the stamp will work!
#print("Current path is " + pathname)
try:
f = open(pathname, "a+b")
f.close()
print("Possible to open" + pathname) #<---line added
except IOError, why:
print "WARNING: File %s could not be opened - %s" % (pathname, why)r code here
....
Since that the freeze was possible. (Not sure why tho, I'm happy to add some info if explanation is known)
I had the same issue tying to freeze my PyQt5 app (using python 3.7 under windows 10) with the following message :
h = BeginUpdateResource(pathname, 0)
pywintypes.error: (2, 'BeginUpdateResource', 'Le fichier spécifié est introuvable.')
So I choose to use cx_freeze Sample provided as a first step.
(This samples are availables after cx_freeze installation under MyVENVpath\Lib\site-packages\cx_Freeze\samples)
The very basic PyQt5 exemple was working, but my case was way more complex as I used several sub Python class I have developed.
I honestly don't really understand why but I was able to make it work So here is all modifications I have performed :
In setup.py identify sub python files
add my personnal packages and class like this :
include = ["PyQt5", "PySerial", "myclass1", "myclass2", "myclass3", "PandasModel"]
packages = ["PySerial", "myclass1", "myclass2", "myclass3", "PandasModel", "pandas","math"]
Use relative path in setup.py
executables = [cx_Freeze.Executable(os.getcwd() + r'\..\..\MyFileName.py', base=base)]
include_files = [(os.getcwd() + r'\..\..\CalibrationTool.ui', r'Inputs\GUI\CalibrationTool.ui')
The reason for "...." is to jump two directories higher level, as my python.exe is in sub folders VENV\Script\
Add python Files containing "Myclass1" at same level as setup.py.
Even if in my application, class files are in a subfolder Input\Packages\Myclass1File.py, I had to copy-paste it at the same level as setup.py. This way the include and package line of setup.py work and was able to generate the executable.
Last but not least,the terminal command
I have no idea why but the only command who did suceed was:
to use full path to the python.exe and full path to the setup.py with the "build" at the end.
cmd:>C:\Folder1\Folder2\MyVENV\Scripts\python.exe D:\Folder1\Folder2\setup.py build
This finaly created the executable in a build folder under C:\Folder1\Folder2\MyVENV\Scripts\

Error using OAuth2 to connect to dropbox in Python

On my Raspberry Pi running raspbian jessie I tried to go through the OAuth2 flow to connect a program to my dropbox using the dropbox SDK for Python which I installed via pip.
For a test, I copied the code from the documentation (and defined the app-key and secret, of course):
from dropbox import DropboxOAuth2FlowNoRedirect
auth_flow = DropboxOAuth2FlowNoRedirect(APP_KEY, APP_SECRET)
authorize_url = auth_flow.start()
print "1. Go to: " + authorize_url
print "2. Click \"Allow\" (you might have to log in first)."
print "3. Copy the authorization code."
auth_code = raw_input("Enter the authorization code here: ").strip()
try:
access_token, user_id = auth_flow.finish(auth_code)
except Exception, e:
print('Error: %s' % (e,))
return
dbx = Dropbox(access_token)
I was able to get the URL and to click allow. When I then entered the authorization code however, it printed the following error:
Error: 'str' object has no attribute 'copy'
Using format_exc from the traceback-module, I got the following information:
Traceback (most recent call last):
File "test.py", line 18, in <module>
access_token, user_id = auth_flow.finish(auth_code)
File "/usr/local/lib/python2.7/dist-packages/dropbox/oauth.py", line 180, in finish
return self._finish(code, None)
File "/usr/local/lib/python2.7/dist-packages/dropbox/oauth.py", line 50, in _finish
url = self.build_url(Dropbox.HOST_API, '/oauth2/token')
File "/usr/local/lib/python2.7/dist-packages/dropbox/oauth.py", line 111, in build_url
return "https://%s%s" % (self._host, self.build_path(target, params))
File "/usr/local/lib/python2.7/dist-packages/dropbox/oauth.py", line 89, in build_path
params = params.copy()
AttributeError: 'str' object has no attribute 'copy'
It seems the build_path method expects a dict 'params' and receives a string instead. Any ideas?
Thanks to smarx for his comment. The error is a known issue and will be fixed in version 3.42 of the SDK. source

Web2py application: How to reference .yaml file in controller?

I have an app running online under web2py. Now, i am adding names.yml file which i need to call in my controller file (default.py) on web2py server. where should I keep the .yml/.yaml files. Currently I have kept them in views with default/names.yml but when I call it in default.py like:
dicttagger = DictionaryTagger([ 'default/names.yml', 'default/surname.yml'])
i get no such file error.
Also tried below:
dicttagger = DictionaryTagger([ 'views/default/names.yml', 'views/default/surname.yml'])
same error
class snapshot as under:
class DictionaryTagger(object):
def __init__(self, dictionary_paths):
files = [open(path, 'r') for path in dictionary_paths]
dictionaries = [yaml.load(dict_file) for dict_file in files]
map(lambda x: x.close(), files)
Any suggestions as how to do this or am I making mistake of using yaml/yml file in we2py and it doesn't work in web2py app hosted online?
question 2
thank you. it resolved an error but I am not sure how to add nltk.download() into my hosted app. I keep getting the below error. Can you pls have a look:
Traceback
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11.
12.
13.
Traceback (most recent call last):
File "/home/prakashsukhwal/web2py/gluon/restricted.py", line 220, in restricted
exec ccode in environment
File "/home/prakashsukhwal/web2py/applications/Sensiva/controllers/default.py", line 4, in
nltk.download()
File "/usr/local/lib/python2.7/dist-packages/nltk/downloader.py", line 644, in download
self._interactive_download()
File "/usr/local/lib/python2.7/dist-packages/nltk/downloader.py", line 958, in _interactive_download
DownloaderShell(self).run()
File "/usr/local/lib/python2.7/dist-packages/nltk/downloader.py", line 981, in run
user_input = raw_input('Downloader> ').strip()
EOFError: EOF when reading a line
Error snapshot help
(EOF when reading a line)
inspect attributes
Frames
File /home/prakashsukhwal/web2py/gluon/restricted.py in restricted at line 220 code arguments variables
File /home/prakashsukhwal/web2py/applications/Sensiva/controllers/default.py in at line 4 code arguments variables
File /usr/local/lib/python2.7/dist-packages/nltk/downloader.py in download at line 644 code arguments variables
File /usr/local/lib/python2.7/dist-packages/nltk/downloader.py in _interactive_download at line 958 code arguments variables
File /usr/local/lib/python2.7/dist-packages/nltk/downloader.py in run at line 981 code arguments variables
Function argument list
(self=)
Code listing
def run(self):
print 'NLTK Downloader'
while True:
self._simple_interactive_menu(
'd) Download', 'l) List', ' u) Update', 'c) Config', 'h) Help', 'q) Quit')
user_input = raw_input('Downloader> ').strip()
if not user_input: print; continue
command = user_input.lower().split()[0]
args = user_input.split()[1:]
try:
Variables
user_input undefined
builtinraw_input
).strip undefined
Context
You can store the files wherever you want, but if you're using the Python open function, you'll need to give it full paths, not paths relative to the web2py application folder. Instead, try:
import os
dicttagger = DictionaryTagger([os.path.join(request.folder, 'views',
'default', 'names.yml'),
...])

pywinauto batch file running error

Im a biologist and new to pywinauto, i wrote a code to open an input file in HYPHY application using pywinauto, when i run my code line by line in command line it works fine but when i run the code as a batch file it gives the following error.
Traceback (most recent call last):
File "C:\Users\Masyh\Desktop\autowin_test.py", line 8, in <module>
w_handle = pywinauto.findwindows.find_windows(title=u' Please select a batch file to run:', class_name='#32770')[0]
IndexError: list index out of range
the code is:
import pywinauto
pwa_app = pywinauto.application.Application()
w_handle = pywinauto.findwindows.find_windows(title=u'HYPHY Console', class_name='HYPHY')[0]
window = pwa_app.window_(handle=w_handle)
window.SetFocus()
window.MenuItem(u'&File->&Open->Open &Batch File\tCtrl+O').Click()
w_handle = pywinauto.findwindows.find_windows(title=u' Please select a batch file to run:', class_name='#32770')[0]
window = pwa_app.window_(handle=w_handle)
window.SetFocus()
ctrl = window['Edit']
ctrl.Click()
ctrl.TypeKeys('brown.nuc')
ctrl=window['&open']
ctrl.Click()
i guess the problem is that the window which gets the input(#'please select a batch file menue') is not open at the beginning and the first part of the code opens it but python looks for it from the beginning and cant find it.
i really appreciate any suggestions how to solve this.
It looks like the window does not exist when the checking is performed. You should wait for a some time for window is opened.
try the next construction:
a_check = lambda: pywinauto.findwindows.find_windows(title=u' Please select a batch file to run:', class_name='#32770')[0]
try:
w_handle = pywinauto.timings.WaitUntilPasses(timeout=10, retry_interval=1, a_check)
except:
print('Something went wrong')
Also, your problem can be caused by the window has extra attributes/state. For example, inisible.
Use allowed argumens from find_windows to handle such cases, here the arguments list with defaul values:
pywinauto.findwindows.find_windows(class_name=None,class_name_re=None, parent=None, process=None, title=None, title_re=None, top_level_only=True, visible_only=True, enabled_only=False, best_match=None, handle=None, ctrl_index=None, predicate_func=None, active_only=False, control_id=None)