Python kann Prozess nicht mit Process.start () unter Windows starten. PySide Signale

Methoden-Header:

@engine.async @guiLoop def for_every_client(self): 

PySide-Signal mit Methode verbinden:

 self.ui.pushButton.clicked.connect(lambda: self.for_every_client(self)) 

Stacktarce – http://pastebin.com/R1ZKcdPy

oder hier:

 Traceback (most recent call last): File "app2.py", line 437, in <lambda> self.ui.pushButton.clicked.connect(lambda: self.for_every_client(self)) File "c:\Python27\lib\site-packages\async_gui\engine.py", line 79, in wrapper gen = func(*args, **kwargs) File "C:\visa\visa_production\libs\guiLoop.py", line 70, in __call__ _loop_in_the_gui(gui_element, generator, self.start_in_gui) File "C:\visa\visa_production\libs\guiLoop.py", line 44, in _loop_in_the_gui wait_time = next(generator) File "app2.py", line 530, in for_every_client p.start() File "c:\Python27\lib\multiprocessing\process.py", line 130, in start self._popen = Popen(self) File "c:\Python27\lib\multiprocessing\forking.py", line 277, in __init__ dump(process_obj, to_child, HIGHEST_PROTOCOL) File "c:\Python27\lib\multiprocessing\forking.py", line 199, in dump ForkingPickler(file, protocol).dump(obj) File "c:\Python27\lib\pickle.py", line 224, in dump self.save(obj) File "c:\Python27\lib\pickle.py", line 331, in save self.save_reduce(obj=obj, *rv) File "c:\Python27\lib\pickle.py", line 425, in save_reduce save(state) File "c:\Python27\lib\pickle.py", line 286, in save f(self, obj) # Call unbound method with explicit self File "c:\Python27\lib\pickle.py", line 655, in save_dict self._batch_setitems(obj.iteritems()) File "c:\Python27\lib\pickle.py", line 687, in _batch_setitems save(v) File "c:\Python27\lib\pickle.py", line 286, in save f(self, obj) # Call unbound method with explicit self File "c:\Python27\lib\pickle.py", line 554, in save_tuple save(element) File "c:\Python27\lib\pickle.py", line 331, in save self.save_reduce(obj=obj, *rv) File "c:\Python27\lib\pickle.py", line 425, in save_reduce save(state) File "c:\Python27\lib\pickle.py", line 286, in save f(self, obj) # Call unbound method with explicit self File "c:\Python27\lib\pickle.py", line 655, in save_dict self._batch_setitems(obj.iteritems()) File "c:\Python27\lib\pickle.py", line 687, in _batch_setitems save(v) File "c:\Python27\lib\pickle.py", line 331, in save self.save_reduce(obj=obj, *rv) File "c:\Python27\lib\pickle.py", line 425, in save_reduce save(state) File "c:\Python27\lib\pickle.py", line 286, in save f(self, obj) # Call unbound method with explicit self File "c:\Python27\lib\pickle.py", line 655, in save_dict self._batch_setitems(obj.iteritems()) File "c:\Python27\lib\pickle.py", line 687, in _batch_setitems save(v) File "c:\Python27\lib\pickle.py", line 331, in save self.save_reduce(obj=obj, *rv) File "c:\Python27\lib\pickle.py", line 396, in save_reduce save(cls) File "c:\Python27\lib\pickle.py", line 286, in save f(self, obj) # Call unbound method with explicit self File "c:\Python27\lib\pickle.py", line 754, in save_global (obj, module, name)) pickle.PicklingError: Can't pickle <type 'PySide.QtCore.SignalInstance'>: it's n ot found as PySide.QtCore.SignalInstance Traceback (most recent call last): File "<string>", line 1, in <module> File "c:\Python27\lib\multiprocessing\forking.py", line 381, in main self = load(from_parent) File "c:\Python27\lib\pickle.py", line 1384, in load return Unpickler(file).load() File "c:\Python27\lib\pickle.py", line 864, in load dispatch[key](self) File "c:\Python27\lib\pickle.py", line 886, in load_eof raise EOFError EOFError 

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  • One Solution collect form web for “Python kann Prozess nicht mit Process.start () unter Windows starten. PySide Signale”

     class Window SignalProgressBar = QtCore.pyqtSignal() def __init__(self): self.SignalProgressBar.connect(lambda: self.for_every_client(self)) 

    Verbindungssignal mit Methode:

      w = Window() w.SignalProgressBar.emit() 
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