I am working on automating program that requires choosing file path.
After I click the browse button using code:
dlg.child_window(auto_id='btnBrwsBinFile').click()
browse window opens and freeze execution of code and I cannot control the popup window.
I have tried also different approach. I have edited text filed using code:
dlg['Edit'].set_text(path)
but then program do not see the path, it treats the field like it was never edited, like it was empty
I would like to ask if someone solved this issue before.
try:
dlg.child_window(title='btnBrwsBinFile').click_input()
pywinauto documentation：https://pywinauto.readthedocs.io/en/latest/code/pywinauto.findwindows.html
You can use UISpy.exe to get the title, class name
Related
I'm using the latest version (to date) of pywinauto; and also using PyInspect (uia) to identify controls.
I'm automating controls on an application, and part of the process is to check a few boxes on a window that pops up after triggering the window to appear from a menu selection (like Edit->Settings).
The problem is, pywinauto doesn't seem to be able to detect the new window. I see successfully opens; and can see the window and elements as a sub (child?) window of the application in PyInspect.
I've tried wait methods, thinking the automation is occurring too fast- but to no avail. something like:
mysettings = app['app-name'].child_window(title_re="my target settings window", class_name="#32770").wait('exists', timeout=10)
this will just timeout. And if I print control identifiers, "my target settings window" is never included.
app['app-name'].print_control_identifiers()
I also tried set_focus on top_window.. that didn't work either. My conclusion is that pywinauto is having trouble detect that it is there. Any thoughts on this?
I was able to resolve this issue of pywinauto detecting the child window, and the issue that immediately followed: accessing the child window.
First, I was able to get pywinauto to detect the new window by defining backend='uia' in the application definition, like this:
app = application.Application(backend='uia')
I previously just had:
app = application.Application()
My next issue was accessing elements on the child window. I could not access the window directly, as I may have anticipated:
app['my app']['child window']['textbox'].set_edit_text("hello world")
Instead, this code worked:
app['my app'].child_window(title='child window name').Edit1.set_edit_text("hello world")
While I have resolved my issue, I have noticed that after defining backend='uia', the process now executes much slower than before. If anyone who stumbles across this has any feedback in that regard (or optimizing my efforts above), please contribute.
Thanks
I am using pywinauto for entering the button in an application installed on windows 7.
from pywinauto.application import Application
from pywinauto.keyboard import send_keys, KeySequenceError
app = Application().start()
send_keys("{ENTER}")
The above code opens the app and does the action ENTER. The problem is, it wrongly selected another button. Is there any way to specify the name of the button and hit the enter key using pywinautoin python.
Thanks
It's worth reading the Getting Started Guide for core concept. In a few words you can do something like
app.window(title="Main Window Title").child_window(title="Button name", class_name="Button").click_input()
# or .click() which may not work in some specific cases
depending on
what [w.window_text() for w in app.windows()] returns and
what app.window(title="Main Window Title").dump_tree() prints.
Also it's important to distinguish Application(backend="win32") and Application(backend="uia").
In order to print stuff on campus, I need to login to the webpage and submit the printing job manually on a browser, but I wanted to submit a job locally. So I wrote a small python script that uses Selenium to automate print job submission. I've verified that it works, but there's one thing that bugs me. Even after I select a file to print, the file dialog stays there until the actual code runs to the end.
The structure of the code is like the following.
1. Enter information on terminal (username, pw, which printer to use, how many copies, etc.).
2. Call tkinter.Tk().withdraw() to select a file (after selecting a file, "Submitting a print job..." is printed, as shown on the screenshot)
3. Do the actual Selenium job using information I collected above to submit a job
How can I make that file dialog disappear as soon as I select the file? Is it an ascyncio problem?
I noticed that many other people were experiencing a similar issue. While many had solved the issue by adding Tk().update() before or after askopenfilename() line, my problem was only gone when Tk().update() line was added both before and after the askopenfilename() line. FYI, I'm running the script on macOS with python 3.7.
I hope find an answer. I have created an application with Python and PyQt4, and I have made the buttons checkable to select a file or folder from the treeview, and all button are auto-exclusive. The problem is when I choose another tab and check a button from there, the button in the first tab is still checked and I can't call setchecked(False). Any idea how to solve this problem ?
archive with the code and the ui used : sorry i don't know how to insert the code here .
enter link description here
So my App needs to be able to open a single webpage(and it must be from the internet and not saved) in it, and specifically I'd like to use the Tkinter GUI toolkit since it's the one i'm most comfortable with. On top of that though, I'd like to be able to generate events in the window(say a mouse click) but without actually using the mouse. What's a good method to go about this?
EDIT: I suppose to clarify this a bit, I need a way to load a webpage, or maybe even a specific java applet into a tkinter widget or window. Or if not that perhaps another method to do this where I can generate mouse and keyboard events without using either the mouse of the keyboard.
If you want it to be opened inside your GUI use Bryans suggestion, if you just want to open a webpage you can use:
import webbrowser
webbrowser.open("page.html")
Tkinter does not have a widget that can render a web page.
So i found this module named pywebview
pip install pywebview
sample code:-
import webview
webview.create_window('duckduckgo', 'https://www.duckduckgo.com')
webview.start() #this will open the webpage in a new window
You should use pywebview it is very easy only code three lines .
I used it but in my case it didn't work everywhere. Comment and let me know if it works for you.
The best option that works everywhere is PyQt's QtWebview module. You might run into one problem that is to rename the window, so here is the solution
web.setWindowTitle(title)
You can use all the functions as it is but just replace window or self with web like the above code.