PyQt4 signals and slots

Question

I am writing my first Python app with PyQt4. I have a MainWindow and a Dialog class, which is a part of MainWindow class:

self.loginDialog = LoginDialog();

I use slots and signals. Here's a connection made in MainWindow:

QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL("aa(str)"), self.login)

And I try to emit signal inside the Dialog class (I'm sure it is emitted):

self.emit(QtCore.SIGNAL("aa"), "jacek")

Unfortunately, slot is not invoked. I tried with no arguments as well, different styles of emitting signal. No errors, no warnings in the code. What might be the problem?

score 27 · Answer 1 · edited Jul 26 '14 at 18:27

There are some concepts to be clarified

[QT signal & slot] VS [Python signal & slot]

All the predefined signals & slots provided by pyqt are implemented by QT's c++ code. Whenever you want to have a customized signal & slot in Python, it is a python signal & slot. Hence there are four cases to emits a signal to a slot:

from a QT signal to a QT slot
from a QT signal to a Python slot
from a Python signal to a QT slot
from a Python signal to a Python slot

The code below shows how to connect for these four different scnarios

    import sys
    from PyQt4.QtCore import *
    from PyQt4.QtGui import *

    class Foo(QtCore.QObject):

        def __init__(self, parent=None):
            super(Foo, self).__init__(parent)
            dial = QDial()
            self.spinbox = QSpinbox()

            # --------------------------------------
            # QT signal & QT slot
            # --------------------------------------

            # option 1: more efficient 
            self.connect(self.spinbox, SIGNAL("valueChanged(int)"), 
                dial, SLOT("setValue(int)"))
            # option 2:
            self.connect(self.spinbox, SIGNAL("valueChanged(int)"), 
                dial.setValue)


            # --------------------------------------
            # QT signal & Python slot
            # --------------------------------------

            self.connect(self.spinbox, SIGNAL("valueChanged(int)"), 
                self.myValChanged)


            # --------------------------------------
            # Python signal & Qt slot
            # --------------------------------------

            # connect option 1: more efficient
            self.connect(self, SIGNAL("mysignal"), dial, 
                SLOT("setValue(int)"))

            # connect option 2:
            self.connect(self, SIGNAL("mysignal"), dial.setValue)

            # emit
            param = 100
            self.emit(SIGNAL("mysignal"), param)


            # --------------------------------------
            # Python signal & Python slot
            # --------------------------------------

            # connect
            self.connect(self, SIGNAL("mysignal"), self.myValChanged)

            # emit
            param = 100
            self.emit(SIGNAL("mysignal"), param)


    def myValChanged(self):
        print "New spin val entered {0}".format(self.spinbox.value())

Conclusion is --

Signal signature for Python signal differentiate from that of QT signal in that it doesn't have the parenthesis and can be passed any python data types when you emit it. The Python signal is created when you emit it.

For slot, there are three forms of signatures.

s.connect(w, SIGNAL("signalSignature"), functionName)
s.connect(w,SIGNAL("signalSignature"), instance.methodName)
s.connect(w,SIGNAL("signalSignature"), instance, SLOT("slotSignature"))

Number 1 & 2 are available for Python slot, while number 2 & 3 are available for QT slot. It is clear that besides QT predefined slot, any python callable function/methods is qulified to be a Python slot.

These points are made in Summerfield's article on Signals and Slots.

[Old style qt signal & slot] VS [new style qt singal & slot]

Well, all the description above is based on the old style pyqt signal & slot. As @Idan K suggested there is an alternative new-style to do the things, especially for the Python signal. Refer to here for more.

IMHO, awesome explanation! One of the best I was able to find so far! — Eugene Sajine, Aug 30 '12 at 17:31

score 12 · Accepted Answer · answered Jan 12 '10 at 11:44

12

You don't use the same signal, when emitting and connecting.

QtCore.SIGNAL("aa(str)") is not the same as QtCore.SIGNAL("aa"). Signals must have the same signature. By the way, if you are defining your own signals, don't define parametres. Just write SIGNAL('aa'), because defining parametres is a thing from C++ and Python version of Qt doesn't need this.

So it should look like this:

QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL("aa"), self.login)

and if you pass any parametres in emit, your login method must accept those parametres. Check, if this helps :-)

answered Jan 12 '10 at 11:44

gruszczy

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That was my mistake in pasting the code. I cut it to be more readable. I didin't do it in the real code. Thanks for a hint ;) – Jacek Jan 12 '10 at 12:49
Could you post some more of your code? Maybe the whole snippet. Then I can try to help you. – gruszczy Jan 12 '10 at 14:42
I would be grateful. The code is avaliable here: http://github.com/jbajor/Pwitter . Code I have problems with: ptwitt_win.py, line 83-98, ptwitter_login.py, line 54-58. – Jacek Jan 12 '10 at 16:25
I assume, you are sure, that code enters login method in ptwitter_login.py. Try connecting this way: self.connect(self.loginDialog, QtCore.SIGNAL('aa'), self.login) Remember to remove brackets after aa. This kind of connecting always worked for me. – gruszczy Jan 12 '10 at 16:47
Well, I have no further idea then. You must have some error somewhere deeper. – gruszczy Jan 13 '10 at 12:34

score 3 · Answer 3 · edited Jul 26 '14 at 20:05

3

What @bialix suggested should have worked, but try an alternative way of connecting:

class Foo(QtCore.QObject):
    mysignal = QtCore.pyqtSignal(str, name='mysignal')

    def connect_to_signal(self):
        # you can use this syntax instead of the 'old' one
        self.mysignal.connect(self.myslot)

        # but this will also work
        self.connect(self, QtCore.SIGNAL('mysignal(QString)'), self.myslot) 

        self.mysignal.emit("hello")

    def myslot(self, param):
        print "received %s" % param

For a more detailed explanation of how signals/slots work in PyQt I'd suggest going through it's documentation, specifically this section.

edited Jul 26 '14 at 20:05

eric

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answered Jan 12 '10 at 20:55

Idan K

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Does not work. Thanks for input though. I would be grateful for any new ideas. – Jacek Jan 12 '10 at 21:07
Try creating a separate script with the code I've written and call connect_to_signal on a Foo object, do you see any output? – Idan K Jan 12 '10 at 21:11

score 3 · Answer 4 · answered Jan 18 '10 at 03:45

I checked your code and it looks like the problem is in the way how you're connecting your signal

you emit the signal in Ui_Dialog class

self.emit(QtCore.SIGNAL("aa()"))
you connect to the signal in Ui_MainWindow's setupUi method by calling

QtCore.QObject.connect(self.loginDialog.ui, QtCore.SIGNAL("aa()"), self.login)

notice first parameter is changed to self.loginDialog.ui; your original connect call was using self.loginDialog which is of the LoginDialog type, whereas signal is emitted by the Ui_Dialog class which is ui property of the LoginDialog. After this change login method of the Ui_MainWindow got called

hope this helps, regards

score 1 · Answer 5 · answered Jan 12 '10 at 12:23

1

As noted by gruszczy you have to use the same QtCore.SIGNAL('xxx') to connect signal and to emit it. Also I think you should use Qt types in the arguments list of signal function. E.g.:

QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL("aa(QString&)"), self.login)

And then emit with:

self.emit(QtCore.SIGNAL("aa(QString&)"), "jacek")

Sometimes it makes sense to define signal only once as global variable and use it elsewhere:

MYSIGNAL = QtCore.SIGNAL("aa(QString&)")
...
QtCore.QObject.connect(self.loginDialog, MYSIGNAL, self.login)
...
self.emit(MYSIGNAL, "jacek")

answered Jan 12 '10 at 12:23

bialix

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Still, nothing happens. I also tried to emit with no arguments, I failed again. Thanks for a hint. – Jacek Jan 12 '10 at 13:04
Which version of PyQt you're using? – bialix Jan 12 '10 at 15:25
PyQt4 263682 (returned by QtCore.PYQT_VERSION) – Jacek Jan 12 '10 at 16:33
1

QtCore.PYQT_VERSION_STR will show you human-readable string. But it seems you're using 4.6.2. I dunno why it does not work for you. – bialix Jan 12 '10 at 20:24

score 0 · Answer 6 · answered Jan 12 '10 at 10:33

0

I haven't used PyQT4 but take a look at here.

answered Jan 12 '10 at 10:33

erelender

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Error: QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL('aa()'), self, QtCore.SIGNAL('login()')) RuntimeError: underlying C/C++ object has been deleted Thanks anyway ;) – Jacek Jan 12 '10 at 12:47
Should be QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL('aa()'), self, QtCore.SLOT('login()')) – bialix Jan 12 '10 at 15:27
You're right Bialix. Unfortunately I still get: QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL('aa()'), self, QtCore.SLOT('login()')) RuntimeError: underlying C/C++ object has been deleted – Jacek Jan 12 '10 at 16:26

Andy M · Answer 7 · 2010-01-12T12:25:16.410

0

Looks like you miss the "SLOT" part in your connect call.

Here is an example :

QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL("NotifyMySignal(int)"), QtCore.SLOT("onNotifyMySignal(int)"));

then

self.emit(QtCore.SIGNAL('NotifyMySignal(1)'));

Hope this helps !

edited Jan 12 '10 at 12:25

answered Jan 12 '10 at 10:39

Andy M

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SLOT is not necessary in PyQt4. Just callable is enough. – bialix Jan 12 '10 at 12:19
Unfortunataly, it does not work. QtCore.QObject.connect(self.loginDialog, QtCore.SIGNAL('aa'), QtCore.SIGNAL('login')) TypeError: QObject.connect(): arguments did not match any overloaded call: overload 1: argument 3 has unexpected type 'str' overload 2: argument 3 has unexpected type 'str' overload 3: argument 2 has unexpected type 'str' Thanks anyway :) – Jacek Jan 12 '10 at 12:51

PyQt4 signals and slots

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