线程功能块 main 并非一直如此

问题描述

我正在开发一个项目，其中我有一个主类，它创建一个线程来执行一个很长的函数，而主类（我的 GUI）将继续可用于在执行这个长函数期间显示一些信息.所以我在 Stack Overflow 上找到了一种方法：

from PyQt5 import QtWidgets,QtCore,QtGui
from PyQt5.QtWidgets import QApplication
import threading
import sys
import time
from PyQt5.QtCore import QObject,pyqtSignal


class Ui_SaftMainWindow(object):
    def setupUi(self,SaftMainWindow):
        SaftMainWindow.setobjectName("SaftMainWindow")
        SaftMainWindow.resize(178,284)
        icon = QtGui.QIcon()
        icon.addpixmap(QtGui.Qpixmap(":/logo/img_logo-E.png"),QtGui.QIcon.normal,QtGui.QIcon.Off)
        SaftMainWindow.setwindowIcon(icon)
        self.gridLayout = QtWidgets.qgridLayout(SaftMainWindow)
        self.gridLayout.setobjectName("gridLayout")
        self.frame_all_pb = QtWidgets.qframe(SaftMainWindow)
        self.frame_all_pb.setMinimumSize(QtCore.QSize(160,0))
        self.frame_all_pb.setMaximumSize(QtCore.QSize(160,16777215))
        self.frame_all_pb.setobjectName("frame_all_pb")
        self.gridLayout_7 = QtWidgets.qgridLayout(self.frame_all_pb)
        self.gridLayout_7.setobjectName("gridLayout_7")
        self.pb_stop_simu = QtWidgets.QPushButton(self.frame_all_pb)
        self.pb_stop_simu.setText("STOP")
        self.gridLayout_7.addWidget(self.pb_stop_simu,5,1,1)
        self.pb_start_simu = QtWidgets.QPushButton(self.frame_all_pb)
        self.pb_start_simu.setobjectName("pb_start_simu")
        self.pb_start_simu.setText("START")
        self.gridLayout_7.addWidget(self.pb_start_simu,4,1)
        self.gridLayout.addWidget(self.frame_all_pb,1)

        QtCore.QMetaObject.connectSlotsByName(SaftMainWindow)


class Solver(QObject,object):
    data_changed = pyqtSignal(str,int)

    def __init__(self):
        # Call init function of QObject class for pyQt signal
        QObject.__init__(self)


    def init_parameters(self,input_files,**keys):
        print("Some initialisation")

    def time_step(self):
        # Function with Fortran call
        print("Fortran function")

    def transient_solve(self,arg):
        print("starting")
        t = threading.currentThread()
        i = 1
        while getattr(t,"do_run",True):
            print("working on " + str(i))
            i += 1
            time.sleep(5)
        print("Stopping as you wish.")


class GuiHandler:
    def __init__(self,gui,main_window,):
        # Declarations
        self.gui = gui
        self.main_window = main_window

        self.gui.setupUi(self.main_window)
        self.gui.pb_start_simu.clicked.connect(self.pb_start_simu_clicked)
        self.gui.pb_stop_simu.clicked.connect(self.pb_stop_simu_clicked)

        # Create instance of the second class
        app = Solver()

        # Connect signal to emit signal from second class to ihm later
        app.data_changed.connect(self.myfunction)

        # Initialize some parameters of my second class
        app.init_parameters("",json_file="",output_file="")

        # Create thread object
        self.t = threading.Thread(target=app.transient_solve,args=("task",))

    def pb_start_simu_clicked(self):
        print("Start thread")
        self.t.start()

    def pb_stop_simu_clicked(self):
        print("Stop thread")
        self.t.do_run = False
        self.t.join()

    def myfunction(self,string="this_is_a_test",integer=0):
        print(str(string))
        print(str(integer))


if __name__ == "__main__":
    app = QApplication(sys.argv)
    handler = GuiHandler(Ui_SaftMainWindow(),QtWidgets.QWidget())
    handler.main_window.show()
    sys.exit(app.exec_())

这个简单的例子有效。但是，当我通过调用更大的函数来替换睡眠时（调用编译为 Python 的 Fortran 文件的 time_step 函数，GUI 函数被阻止）

知道它是否与我创建和启动线程的方式有关，或者关于我必须检查/查找的内容才能使我的代码起作用吗？

我暂时无法为您提供另一个代码示例。

解决方法

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multithreading nonblocking python