防止 PyQt 使插槽中发生的异常静音

问题描述

据我所知，如果在 PyQt 下的插槽中发生异常，则该异常会打印到屏幕上，但不会冒泡.这在我的测试策略中产生了问题，因为如果插槽中发生异常，我不会看到测试失败.

As far as I can see, if an exception occurs in a slot under PyQt, the exception is printed to screen, but not bubbled. This creates a problem in my testing strategy, because if an exception occurs in a slot, I will not see the test fail.

这是一个例子:

import sys
from PyQt4 import QtGui, QtCore

class Test(QtGui.QPushButton):
    def __init__(self, parent=None):
        QtGui.QWidget.__init__(self, parent)
        self.setText("hello")
        self.connect(self, QtCore.SIGNAL("clicked()"), self.buttonClicked)

    def buttonClicked(self):
        print "clicked"
        raise Exception("wow")

app=QtGui.QApplication(sys.argv)
t=Test()
t.show()
try:
    app.exec_()
except:
    print "exiting"

注意异常如何永远不会退出程序.

Note how the exception never quits the program.

有没有办法解决这个问题?

Is there a way to work around this problem?

推荐答案

可以创建一个装饰器，包装 PyQt 的新信号/槽装饰器，并为所有槽提供异常处理.还可以覆盖 QApplication::notify 以捕获未捕获的 C++ 异常.

Can create a decorator that wraps PyQt' new signal/slot decorators and provides exception handling for all slots. Can also override QApplication::notify to catch uncaught C++ exceptions.

import sys
import traceback
import types
from functools import wraps
from PyQt4 import QtGui, QtCore

def MyPyQtSlot(*args):
    if len(args) == 0 or isinstance(args[0], types.FunctionType):
        args = []
    @QtCore.pyqtSlot(*args)
    def slotdecorator(func):
        @wraps(func)
        def wrapper(*args, **kwargs):
            try:
                func(*args)
            except:
                print "Uncaught Exception in slot"
                traceback.print_exc()
        return wrapper

    return slotdecorator

class Test(QtGui.QPushButton):
    def __init__(self, parent=None):
        QtGui.QWidget.__init__(self, parent)
        self.setText("hello")
        self.clicked.connect(self.buttonClicked)

    @MyPyQtSlot("bool")
    def buttonClicked(self, checked):
        print "clicked"
        raise Exception("wow")

class MyApp(QtGui.QApplication):
    def notify(self, obj, event):
        isex = False
        try:
            return QtGui.QApplication.notify(self, obj, event)
        except Exception:
            isex = True
            print "Unexpected Error"
            print traceback.format_exception(*sys.exc_info())
            return False
        finally:
            if isex:
                self.quit()

app = MyApp(sys.argv)

t=Test()
t.show()
try:
    app.exec_()
except:
    print "exiting"

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