PyQt:qthread通过信号中断

我是pyqt的新手，所以我希望我所做的不会有什么奇怪的事情。我试图通过使用PyQt信号在QT线程之间创建交互。特别是，我想做的是从线程发出一个信号，这个信号应该会中断线程运行特定方法所做的事情。顺便说一下，我在做这件事时遇到了一些问题，所以我想知道我所做的是否合法

例如，现在我正试图做这样的事情：

import sys
import time
from PyQt4 import QtGui as qt
from PyQt4 import QtCore as qtcore
from PyQt4.QtCore import QThread
import threading

app = qt.QApplication(sys.argv)
class widget(qt.QWidget):

    def __init__(self, parent=None):
        qtcore.QObject.__init__(self)


    def appinit(self):
        self.mysignal = qtcore.SIGNAL("mysignal")
        thread = mythread(self)
        thread.start()
        time.sleep(5)
        print "before emit",str(threading.current_thread())
        self.emit(self.mysignal,"hello, I'm thread "+str(threading.current_thread()))
        print "after emit",str(threading.current_thread())



class mythread(QThread):
    def __init__(self,parent):
        qtcore.QThread.__init__(self, parent=app)
        self.parent=parent

    def run(self):
        self.mysignal = qtcore.SIGNAL("mysignal")
        self.connect(self.parent, self.mysignal, self.myfunc)
        for i in range(15):
            print "**",threading.current_thread(),i
            time.sleep(1)

    def myfunc(self, msg):
        print threading.current_thread(), msg, "Enter"
        time.sleep(5)
        print threading.current_thread(), msg, "Exit"


def main():
    mywidget = widget()
    mywidget.show()
    qtcore.QTimer.singleShot(0, mywidget.appinit)
    sys.exit(app.exec_())

main()

我得到的结果是：

** <_DummyThread(Dummy-1, started daemon 1968)> 0
** <_DummyThread(Dummy-1, started daemon 1968)> 1
** <_DummyThread(Dummy-1, started daemon 1968)> 2
** <_DummyThread(Dummy-1, started daemon 1968)> 3
** <_DummyThread(Dummy-1, started daemon 1968)> 4
before emit <_MainThread(MainThread, started 2928)>
<_MainThread(MainThread, started 2928)> hello, I'm thread <_MainThread(MainThread, started 2928)> Enter
** <_DummyThread(Dummy-1, started daemon 1968)> 5
** <_DummyThread(Dummy-1, started daemon 1968)> 6
** <_DummyThread(Dummy-1, started daemon 1968)> 7
** <_DummyThread(Dummy-1, started daemon 1968)> 8
** <_DummyThread(Dummy-1, started daemon 1968)> 9
<_MainThread(MainThread, started 2928)> hello, I'm thread <_MainThread(MainThread, started 2928)> Exit
after emit <_MainThread(MainThread, started 2928)>
** <_DummyThread(Dummy-1, started daemon 1968)> 10
** <_DummyThread(Dummy-1, started daemon 1968)> 11
** <_DummyThread(Dummy-1, started daemon 1968)> 12
** <_DummyThread(Dummy-1, started daemon 1968)> 13
** <_DummyThread(Dummy-1, started daemon 1968)> 14

**0
**  1
**  2
**  3
**  4
发射前
您好，我是thread Enter
**  5
**  6
**  7
**  8
**  9
你好，我是线程出口
后发
**  10
**  11
**  12
**  13
**  14

我本来希望mythread中断它的执行并运行myfunc。顺便说一句，mythread不会中断它的执行，函数myfunc实际上是由主线程运行的。我试着做相反的事情（生成的线程向主线程发送一个信号），结果它成功了

我想我还没有完全理解信号是如何工作的，以及是否有可能做我想做的事情。有什么线索吗？我在网上寻找解决方案，但没有结果

---

谢谢

好的，据我所知，您希望线程在进一步迭代之前进入5秒睡眠

import sys
import time
from PyQt4 import QtGui as qt
from PyQt4 import QtCore as qtcore
from PyQt4.QtCore import QThread
import threading

app = qt.QApplication(sys.argv)
class widget(qt.QWidget):

    def __init__(self, parent=None):
        qtcore.QObject.__init__(self)


    def appinit(self):
        self.mysignal = qtcore.SIGNAL("mysignal")
        thread = mythread(self,self.mysignal)
        thread.start()
        time.sleep(5)
        print "before emit",str(threading.current_thread())
        self.emit(self.mysignal,"hello, I'm thread "+str(threading.current_thread()))
        print "after emit",str(threading.current_thread())



class mythread(QThread):
    def __init__(self,parent,sig):
        qtcore.QThread.__init__(self, parent=app)
        self.parent=parent
        self.mysignal = sig
        self.stop_event=threading.Event()
        self.connect(self.parent, self.mysignal, self.setIt)

    def run(self):

        for i in range(15):
            if self.stop_event.isSet():
                self.myfunc()
            print "**",threading.current_thread(),i
            time.sleep(1)

    def setIt(self,msg):
        self.stop_event.set()
        self.msg = msg

    def myfunc(self):
        print threading.current_thread(),self.msg ,  "Enter"
        time.sleep(5)
        print threading.current_thread(),self.msg ,  "Exit"
        self.stop_event.clear()


def main():
    mywidget = widget()
    mywidget.show()
    qtcore.QTimer.singleShot(0, mywidget.appinit)
    sys.exit(app.exec_())

main()

输出：

** <_DummyThread(Dummy-1, started daemon 7312)> 0
** <_DummyThread(Dummy-1, started daemon 7312)> 1
** <_DummyThread(Dummy-1, started daemon 7312)> 2
** <_DummyThread(Dummy-1, started daemon 7312)> 3
** <_DummyThread(Dummy-1, started daemon 7312)> 4
before emit <_MainThread(MainThread, started 232)>
after emit <_MainThread(MainThread, started 232)>
<_DummyThread(Dummy-1, started daemon 7312)> hello, I'm thread <_MainThread(MainThread,     started 232)> Enter
<_DummyThread(Dummy-1, started daemon 7312)> hello, I'm thread <_MainThread(MainThread,     started 232)> Exit
** <_DummyThread(Dummy-1, started daemon 7312)> 5
** <_DummyThread(Dummy-1, started daemon 7312)> 6
** <_DummyThread(Dummy-1, started daemon 7312)> 7
** <_DummyThread(Dummy-1, started daemon 7312)> 8
** <_DummyThread(Dummy-1, started daemon 7312)> 9
** <_DummyThread(Dummy-1, started daemon 7312)> 10
** <_DummyThread(Dummy-1, started daemon 7312)> 11
** <_DummyThread(Dummy-1, started daemon 7312)> 12
** <_DummyThread(Dummy-1, started daemon 7312)> 13
** <_DummyThread(Dummy-1, started daemon 7312)> 14

**0
**  1
**  2
**  3
**  4
发射前
后发
您好，我是thread Enter
你好，我是线程出口
**  5
**  6
**  7
**  8
**  9
**  10
**  11
**  12
**  13
**  14

通过信号中断同步代码（如长时间运行的循环）不起作用：线程间的信号/插槽需要在接收线程中有一个事件循环（即，必须通过调用exec（）在run（）中输入）；要调用插槽，接收线程必须返回到事件循环；也就是说，如果你有一个长时间运行的循环，事件循环将被阻塞，插槽将永远不会被调用。我想我现在开始更好地理解了。顺便说一句，我将mythread.run（）中的for循环替换为app.exec\uz（）。结果是，函数myfunc仍然由主线程调用，因此我觉得有其他问题。