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[ltddev]

Hi, I'm trying to build and package a script and its .pyui file into a standalone .ipa file in Xcode. I have Apple developer provisioning profile and have successfully been using the Pythonista v1.5 template to create standalone apps where any UI code is done in the main script file.

I want to create a standalone app from UI built with the UI Editor. I am able to build and export my target .ipa and install it on a jailbroken device. I have my Python script named Script.py and my UI file named Script.pyui. No matter how I define a path for the the pyui file, the call to ui.load_view() consistently fails and I see the following error message on my device:

"Error (line nn): UI file could not be loaded" The nn will be the actual line number that the ui.load_view() call is in my script.

I presume this error message is coming from the ui module itself. Any suggestions on how to get around this problem to load the UI from a file in a standalone app? The script and UI file load and run fine within PYthonista itself.

I have tried the following, without luck for a file actually named Script.pyui and called from the script called Script.py:

ui.load_view()

ui.load_view("Script")

ui.load_view(".\Script")

ui.load_view("Script.pyui")

ui.load_view(".\Script.pyui")

Since I never saw any note or documentation to suggest this is not officially supported -- to package and run UI code from a loaded UI file in a standalone app, I figure there must be a way to do this but it is just not documented anywhere.

Can anyone tell me how to accomplish this?

[ccc]

Sorry... I fixed the script above. Please try again with this new formulation (double backslashes).

[techteej]

[ccc]

Here is a verbose version of ui.load_view() that might help.

It takes the real, absolute path of this script and adds 'ui' to it.<br>
It prints out what that new filepath is and if that file exists or not.<br>
It tries to ui.load_view() it and if it fails it repeats the process with all / converted to \.

def ui_load_view_verbose(pyui_filepath = None):
    pyui_filepath = pyui_filepath or os.path.realpath(os.path.abspath(__file__)) + 'ui'
    fmt = 'Attempting to load_view({}) -- File {}found.'
    print(fmt.format(pyui_filepath, '' if os.path.isfile(pyui_filepath) else 'not '))
    try:
        return ui.load_view(pyui_filepath)
    except ValueError as e:
        print(e)
        return None if '\\' in pyui_filepath else ui_load_view_verbose(pyui_filepath.replace('/', '\\'))

# replace your current call to ui.load_view() with the following:
v = ui_load_view_verbose()  # with no parameters!!
if not v:
    print('Bummer!  ui_load_view_verbose() did not work.')
    sys.exit()

If this fails, please send us the printed output.

[JonB]

One problem with using __file__ is if os.chdir has been called. __file__ only has the relative path from the time you import it. If you chdir after import, you can't find it again.
I struggled with this problem when making a plugin for my javascript editor script... The calling script had imported the module, then changed directory. In retrospect, I probably could have done something in init, but here's what I ended up with. So possibly using inspect to locate things might be another route.

origPath= os.path.dirname(inspect.stack()[-1][1])
fullScriptPath=  os.path.join(origPath,os.path.splitext(__file__)[0])

[ccc]

Does sys.argv[0] suffer from the same problem as __file__?

[techteej]

[JonB]

Try this.. Which will display all files ending with pyui...

def findpyui():
    import os
    startingpath = os.path.expanduser('~')    
    # other things to try....
    #startingpath = '.'  
    #starting path = '..'  # i.e in pythonista, start above documents, to see into library,etc
    print startingpath
    for root, dirs, files in os.walk(startingpath):
        for name in files:
            if name.endswith('pyui'):
                print os.path.join(os.path.abspath(root),name)
findpyui()

[techteej]

The solution above worked. Thanks.

[ccc]

What was the path of the .pyiu file??