将 PyQt5 按钮连接到函数时如何使用 lambda答案

【问题标题】：How to use lambda when connecting PyQt5 buttons to functions将 PyQt5 按钮连接到函数时如何使用 lambda
【发布时间】：2021-10-01 18:57:18
【问题描述】：

在我的代码中，当单击一个按钮 (self.longRunningBtn) 时，它会连接到一个不带参数的函数，并且工作正常。当该函数被修改为采用额外参数a, b, c 时，我收到错误TypeError: argument 1 has unexpected type 'NoneType'。然后通过在调用中添加 lambda 来解决此问题，但现在该函数似乎根本没有运行，即使它位于不同的线程上，整个 GUI 也会冻结。我希望能够单击带有参数a, b, c 的self.longRunningBtn 而不会看到该错误并且不会冻结GUI。

没有参数的原始代码，完美运行：

from PyQt5.QtCore import QObject, QThread, pyqtSignal
import sys
from time import sleep

from PyQt5.QtCore import Qt
from PyQt5.QtWidgets import (
    QApplication,
    QLabel,
    QMainWindow,
    QPushButton,
    QVBoxLayout,
    QWidget,
)

class Worker(QObject):
    finished = pyqtSignal()
    progress = pyqtSignal(str)

    def run(self):
        for i in range(5):
            sleep(1)
            self.progress.emit(f"{i+1}")
        self.finished.emit()


class Window(QMainWindow):
    def __init__(self, parent=None):
        super().__init__(parent)
        self.clicksCount = 0
        self.setupUi()

    def setupUi(self):
        self.setWindowTitle("Freezing GUI")
        self.resize(300, 150)
        self.centralWidget = QWidget()
        self.setCentralWidget(self.centralWidget)
        # Create and connect widgets
        self.clicksLabel = QLabel("Counting: 0 clicks", self)
        self.clicksLabel.setAlignment(Qt.AlignHCenter | Qt.AlignVCenter)
        self.stepLabel = QLabel("Long-Running Step: 0")
        self.stepLabel.setAlignment(Qt.AlignHCenter | Qt.AlignVCenter)
        self.countBtn = QPushButton("Click me!", self)
        self.countBtn.clicked.connect(self.countClicks)
        self.longRunningBtn = QPushButton("Long-Running Task!", self)
        self.longRunningBtn.clicked.connect(self.runLongTask)
        # Set the layout
        layout = QVBoxLayout()
        layout.addWidget(self.clicksLabel)
        layout.addWidget(self.countBtn)
        layout.addStretch()
        layout.addWidget(self.stepLabel)
        layout.addWidget(self.longRunningBtn)
        self.centralWidget.setLayout(layout)

    def countClicks(self):
        self.clicksCount += 1
        self.clicksLabel.setText(f"Counting: {self.clicksCount} clicks")

    def reportProgress(self, n):
        self.stepLabel.setText(f"Long-Running Step: {n}")

    def runLongTask(self):
        # Step 2: Create a QThread object
        self.thread = QThread()
        # Step 3: Create a worker object
        self.worker = Worker()
        self.worker.moveToThread(self.thread)
        self.thread.started.connect(self.worker.run)
        self.worker.finished.connect(self.thread.quit)
        self.worker.finished.connect(self.worker.deleteLater)
        self.thread.finished.connect(self.thread.deleteLater)
        self.worker.progress.connect(self.reportProgress)
        # Step 6: Start the thread
        self.thread.start()

        # Final resets
        self.longRunningBtn.setEnabled(False)
        self.thread.finished.connect(
            lambda: self.longRunningBtn.setEnabled(True)
        )
        self.thread.finished.connect(
            lambda: self.stepLabel.setText("Long-Running Step: 0")
        )

app = QApplication(sys.argv)
win = Window()
win.show()
sys.exit(app.exec())

添加参数的代码：

    def run(self, a, b, c):
        for i in range(5):
            sleep(1)
            self.progress.emit(f"{i+1} {a} {b} {c}")
        self.finished.emit()

        self.thread.started.connect(self.worker.run("alpha", "beta", "charlie"))

返回TypeError: argument 1 has unexpected type 'NoneType'

修复类型错误：

        self.thread.started.connect(lambda: self.worker.run("alpha", "beta", "charlie"))

但现在 GUI 冻结了，我看不到该函数正在执行的任何操作，我希望它的功能与原始代码中的完全相同，但标签末尾也附加了 a, b, c。

【问题讨论】：

标签： python python-3.x pyqt pyqt5

【解决方案1】：

问题在于最终 lambda 等效于直接在主线程中调用 run() 方法，而不是在接收者线程中发出信号来执行。

一种可能的解决方案是使用 functools.partial：

from functools import partial

self.thread.started.connect(partial(self.worker.run, "alpha", "beta", "charlie"))

【讨论】：