問題描述

我在執(zhí)行以下 gui 時遇到問題.如果沒有消息框，它可以正常工作，但是當有消息框時，它會阻塞.知道為什么有消息時gui會阻塞.謝謝

I have a problem while executing the following gui. it works normally if there's no msgbox, but when there is a mesbox it blocks. any idea why the gui blocks when there is message. thank you

from PyQt5 import QtCore, QtGui, QtWidgets
import threading
import time
class Ui_MainWindow(object):
    def setupUi(self, MainWindow):
        MainWindow.setObjectName("MainWindow")
        MainWindow.resize(800, 600)
        self.centralwidget = QtWidgets.QWidget(MainWindow)
        self.centralwidget.setObjectName("centralwidget")
        self.verticalLayout = QtWidgets.QVBoxLayout(self.centralwidget)
        self.verticalLayout.setObjectName("verticalLayout")
        self.pushButton = QtWidgets.QPushButton(self.centralwidget)
        self.pushButton.setObjectName("pushButton")
        self.verticalLayout.addWidget(self.pushButton)
        MainWindow.setCentralWidget(self.centralwidget)
        self.menubar = QtWidgets.QMenuBar(MainWindow)
        self.menubar.setGeometry(QtCore.QRect(0, 0, 800, 26))
        self.menubar.setObjectName("menubar")
        MainWindow.setMenuBar(self.menubar)
        self.statusbar = QtWidgets.QStatusBar(MainWindow)
        self.statusbar.setObjectName("statusbar")
        MainWindow.setStatusBar(self.statusbar)
        self.retranslateUi(MainWindow)
        QtCore.QMetaObject.connectSlotsByName(MainWindow)
        self.pushButton.pressed.connect(self.threadingc)      
    def calculation(self):
        for i in range(10):
            time.sleep(1)
            print(i)
        msg = QtWidgets.QMessageBox()
        msg.setInformativeText('Finish')
        msg.exec_()   
    def threadingc(self):
        x=threading.Thread(target=self.calculation)
        x.start()
    def retranslateUi(self, MainWindow):
        _translate = QtCore.QCoreApplication.translate
        MainWindow.setWindowTitle(_translate("MainWindow", "MainWindow"))
        self.pushButton.setText(_translate("MainWindow", "PushButton"))
import sys
if __name__ == "__main__":
    app = QtWidgets.QApplication(sys.argv)
    MainWindow = QtWidgets.QMainWindow()
    ui = Ui_MainWindow()
    ui.setupUi(MainWindow)
    MainWindow.show()
    sys.exit(app.exec_())

推薦答案

任何訪問 UI 元素只允許在主 Qt 線程內.請注意，access 不僅意味著讀取/寫入小部件屬性，還意味著創(chuàng)建；任何從其他線程執(zhí)行此操作的嘗試在最好的情況下都會導致圖形問題或不一致的行為，而在最壞(和更常見)的情況下會導致崩潰.

Any access to UI elements is only allowed from within the main Qt thread. Note that access means not only reading/writing widget properties, but also creation; any attempt to do so from other threads results in graphical issues or incosistent behavior in the best case, and a crash in the worst (and more common) case.

這樣做的唯一正確方法是使用帶有(可能)自定義信號的 QThread:這允許 Qt 正確地對信號進行排隊，并在它可以實際處理它們時對它們做出反應.

The only correct way to do so is to use a QThread with (possibly) custom signals: this allows Qt to correctly queue signals and react to them when it can actually process them.

以下是一個非常簡單的情況，不需要創(chuàng)建 QThread 子類，但請考慮這只是為了教育目的.

The following is a very simple situation that doesn't require creating a QThread subclass, but consider that this is just for educational purposes.

class Ui_MainWindow(object):
    # ...

    def calculation(self):
        for i in range(10):
            time.sleep(1)
            print(i)

    def showMessage(self):
        msg = QtWidgets.QMessageBox()
        msg.setInformativeText('Finish')
        msg.exec_()   
        self.pushButton.setEnabled(True)

    def threadingc(self):
        self.pushButton.setEnabled(False)
        self.thread = QtCore.QThread()
        # override the `run` function with ours; this ensures that the function
        # will be executed in the new thread
        self.thread.run = self.calculation
        self.thread.finished.connect(self.showMessage)
        self.thread.start()

請考慮以下重要方面:

• 我必須禁用按鈕，否則可以在前一個線程仍在執(zhí)行時創(chuàng)建一個新線程；這會產生問題，因為覆蓋 self.thread 會導致 python 在運行時嘗試垃圾收集(刪除)前一個線程，這是一個非常糟糕的 東西；
• 一個可能的解決方案是使用父線程創(chuàng)建線程，這通常使用簡單的 QThread(self) 來完成，但在您的情況下這是不可能的，因為 Qt 對象只能接受其他Qt 對象作為它們的父對象，而在您的情況下 self 將是一個 Ui_MainWindow 實例(這是一個基本的 python 對象)；
• 以上幾點是一個重要問題，因為您嘗試從 pyuic 生成的文件開始執(zhí)行程序，而這永遠不會這樣做:這些文件是打算保持原樣而無需任何手動修改，并且僅用作導入的模塊；閱讀有關 使用 Designer 的官方指南中有關此主題的更多信息；另請注意，試圖模仿這些文件的行為是沒有用的，因為這通常會導致對象結構非常混亂；
• 理論上，您可以添加對 qt 對象的引用(例如，通過在 setupUi() 函數中添加 self.mainWindow = MainWindow)并使用該引用 (thread = QThread(self.mainWindow))，或將線程添加到持久列表 (self.threads = []，再次在 setupUi())，但由于上述原因，我強烈建議您不要這樣做；

• I had to disable the pushbutton, otherwise it would be possible to create a new thread while the previous one still executing; this will create a problem, since overwriting self.thread will cause python to try to garbage collect (delete) the previous thread while running, which is a very bad thing;
• a possible solution to this is to create the thread with a parent, which is usually done with a simple QThread(self), but that's not possible in your case because Qt objects can accept only other Qt objects as their parent, while in your case self would be a Ui_MainWindow instance (which is a basic python object);
• the above point is an important issue, because you're trying to implement your program starting from a pyuic generated file, which should never be done: those files are intended to be left as they are without any manual modification, and used only as imported modules; read more about this topic on the official guidelines about using Designer; also note that trying to mimic the behavior of those files is useless, as normally leads to great confusion about object structure;
• you could theoretically add a reference to a qt object (for example, by adding self.mainWindow = MainWindow in the setupUi() function) and create the thread with that reference (thread = QThread(self.mainWindow)), or add the thread to a persistent list (self.threads = [], again in the setupUi()), but due to the above point I strongly discourage you to do so;

最后，更正確代碼實現將要求您再次生成 ui 文件，保持原樣并執(zhí)行類似以下示例的操作；請注意，我添加了一個非常基本的異常實現，它還展示了如何正確地與自定義信號交互.

Finally, a more correct implementation of your code would require you to generate again the ui file, leave it as it is and do something like the following example; note that I added a very basic exception implementation that also shows how to correctly interact with custom signals.

from PyQt5 import QtCore, QtGui, QtWidgets
from mainwindow import Ui_MainWindow
import time

class Calculation(QtCore.QThread):
    error = QtCore.pyqtSignal(object)
    def run(self):
        for i in range(10):
            time.sleep(1)
            print(i)
            try:
                10 / 0
            except Exception as e:
                self.error.emit(e)
                break


class MainWindow(QtWidgets.QMainWindow, Ui_MainWindow):
    def __init__(self):
        super().__init__()
        self.setupUi(self)
        self.pushButton.pressed.connect(self.threadingc)

    def showMessage(self):
        msg = QtWidgets.QMessageBox()
        msg.setInformativeText('Finish')
        msg.exec_()   

    def threadingc(self):
        # create the thread with the main window as a parent, this is possible 
        # since QMainWindow also inherits from QObject, and this also ensures
        # that python will not delete it if you want to start another thread
        thread = Calculation(self)
        thread.finished.connect(self.showMessage)
        thread.error.connect(self.showError)
        thread.start()


if __name__ == "__main__":
    import sys
    app = QtWidgets.QApplication(sys.argv)
    mainWindow = MainWindow()
    mainWindow.show()
    sys.exit(app.exec_())

在上述情況下，使用以下命令處理 ui 文件(顯然，假設 ui 名為mainwindow.ui"):

In the above case, the ui file was processed using the following command (assuming that the ui is named "mainwindow.ui", obviously):

pyuic mainwindow.ui -o mainwindow.py

這篇關于線程時出現 msgbox 錯誤，GUI 塊的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！