Develop Reference

Python 2.7: Close a dialog without exiting the entire application

I'm running a script coded in python from a scripts menu in a desktop application. It's basically a giant macro that I wrote and added a GUI to. I'm pretty sure the GUI is a really old one that my desktop app uses called dialogKit from MIT.
GitHub still has it here.
The problem is the word "stop" at the very end of the dialog code.
I keep getting a "stop is undefined" message, which I understand, but I've tried everything to close the dialog and if I use exit(), sys.exit(), I don't get an error, but it also closes my entire desktop app.
I need to close the dialog and keep the software open.
The limited dialog documentation for what I'm using can be found here.
(you might have to click on the Dialog section. Their site uses frames.)
class MyDialog:
def __init__(self):
self.d = Dialog(self)
self.d.size = Point(300, 340)
self.d.Center()
self.d.title = "Halftone" #<----- Title of the dialogue
self.d.AddControl(STATICCONTROL, Rect(aIDENT, aIDENT, aIDENT, aIDENT), "frame", STYLE_FRAME)
# more controls and methods..
def on_ok(self, code):
return 1
def on_cancel(self, code):
print "blah"
def Run(self):
return self.d.Run()
d = MyDialog()
if d.Run()!= 1:
stop
I just need a way to change stop to something that 1) will prevent the script from running, and 2) close the dialog without quitting the entire application. This is the functionality of a typical "cancel" button, which is what I want.
Another option is the method called on_cancel(), which I also tried and could get the event itself to work, but still the entire application quits with any kind of exit().
The docs show a method called End(), which claims to terminate the dialog object, but I've tried and failed to get that to work either.

Okay, I'm posting an answer because I think I have a handle on your problem.
Try replacing stop with:
d.d.End()
If that works, you might want to try putting:
self.d.End()
inside of the on_cancel function in your class. That should close the dialogue without closing your program.

Python Maya: access result of a function when called in button flag

I am working on UI window with a browse button. I have a browsePath function that I am calling in a cmds.button command. How can I get the output of browsePath() in my main function?
def main():
cmds.button(label='Browse', command=browsePath)
def browsePath(*args):
path = cmds.fileDialog2(fm=2)
if path:
cmds.textField('txt', tx=path[0])
return path

I don't think so maya gui events return anything, you probaly need to relay on a variable outside. But since you already getting the path value to your textField you can always query the textField value where you want.

Python: Button widget in Tkinter

I have just now begun with GUI programming in Python 2.7 with Tkinter.
I want to have a button Browse, which when clicked opens the Windows File Explorer and returns the path of file selected to a variable. I wish to use that path later.
I am following the code given here. It outputs a window displaying 5 buttons, but the buttons do nothing. On clicking the first button, it doesn't open the selected file.
Likewise, on clicking the second button, the askopenfilename(self) function is called and it should return a filename. Like I mentioned, I need that filename later.
How to I get the value returned by the function into some variable for future use?

There is no point in using return inside a callback to a button. It won't return to anywhere. The way to make a callback save data is to store it in a global variable, or an instance variable if you use classes.
def fetchpath():
global filename
filename = tkFileDialog.askopenfilename(initialdir = 'E:')
FWIW (and unrelated to the question): you're making a very common mistake. In python, when you do foo=bar().baz(), foo takes the value in baz(). Thus, when you do this:
button = Button(...).pack()
button will take the value of pack() which always returns None. You should separate widget creation from widget layout if you expect to save an actual reference to the widget being created. Even if you're not, it's a good practice to separate the two.

pywinauto Wait and focus

I'm trying to automate a windows application using pywinauto.
I can select the menu and open the "open file window".
I need to wait for this window to appear and then set focus to that window and click some buttons.
For some reason is not working.
def open_file():
return pywinauto.findwindows.find_windows(best_match=u'Open File', class_name='#32770')[0]
pywinauto.timings.WaitUntilPasses(20, 0.5,open_file)
print('wait for window')
open_file.SetFocus()
When I try to run this it says that open_file doesn't have a SetFocus option.
I'm a beginner in pywinauto and I'm pretty sure that this is something easy to fix but I don't know how :/

open_file is a function. It has no such method. It is much simpler to use Application object to wait dialog.
OpenDialog = pwa_app.window(best_match=u'Open', class_name='#32770').wait('visible', timeout=20, retry_interval=0.5)
OpenDialog.set_focus()
Low-level functions like wait_until_passes are already encapsulated inside wait and wait_not methods of class WindowSpecification.
Even more simple code should work:
pwa_app.OpenDialog.wait('visible', timeout=20)
pwa_app.OpenDialog.set_focus()

How to open a popup window with a spinner in python + Gtk

I have a python Gtk application, with the GUI designed in Glade. It has a "scan" feature which will scan the network for a few seconds and then report its results to the user. During the scanning I want a popup window to appear stealing the focus from the parent until scanning is done.
I use a threading.Lock to synchronize the GUI and the scan thread, which makes the popup to last exactly the right time I want (see scanLock.acquire() ). It seems straightforward to me to implement something like a show() and hide() call before and after the scanLock.acquire(). I did use waitPopupShow and waitPopupHide instead of just calling the window.show() and window.hide() because I also may want to set the Label in the popup or start/stop the GtkSpinner. Here is some code from the GUI class:
def scan(self):
sT = scannerThread(self,self.STagList)
self.dataShare.threadsList.append(sT)
sT.start() # start scanning
self.waitPopupShow('Scanning... Please Wait')
self.scanLock.acquire() # blocks here until scan is finished
self.waitPopupHide()
def waitPopupShow(self, msg): # shows a GtkSpinner until the semaphore is cleared
self.waitDialogLabel.set_text(msg)
self.waitDialogBox.show_all()
self.waitDialog.show()
self.waitDialogSpinner.start()
def waitPopupHide(self):
# how to get the handle to the spinner and stop it?
self.waitDialogSpinner.stop()
self.waitDialog.hide()
def getAll(self):
# GUI
self.builder = Gtk.Builder()
self.builder.add_from_file(path to main GUI)
# ... getting stuff from a first glade file
# getting stuff from the waitDialog glade file
self.builder.add_from_file(path to waitDialog GUI)
self.waitDialog = self.builder.get_object("waitDialog") # GtkWindow
self.waitDialogBox = self.builder.get_object("waitDialogBox") # GtkBox
self.waitDialogLabel = self.builder.get_object("waitDialogLabel") # GtkLabel
self.waitDialogSpinner = self.builder.get_object("waitDialogSpinner") # GtkSpinner
self.waitDialog.hide()
I'm trying hardly since a couple of days to show a dialog with a label and a Gtk.Spinner. The best I obtain at the moment is to have the window showing up with no content. Please note that the self.waitDialog.hide() right after getting it with self.builder.get_object is needed because I set the property of the waitDialog Gtkwindow to Visibile. If I stop with the debugger before .hide() the waitDialog shows up perfectly. Afterwards its broken.
This is the waitDialog GUI file: http://pastebin.com/5enDQg3g
So my best guess is that I'm dooing something wrong, and I could find nothing on creating a new Gtk window over the main one, only basic examples and dialogs. A pointer to the documentation saying a bit about this would be a good starting point...