如何在需要时激活信号答案

【问题标题】：How to activate signals when required如何在需要时激活信号
【发布时间】：2021-06-11 15:22:35
【问题描述】：

在这个最小的可重现示例中，我有一个组合框和一个按钮。我正在尝试根据从组合框中选择的当前文本激活按钮，但是当我尝试在 elif else 条件下首先验证它时，我无法激活按钮，如何根据当前文本激活正确的功能。

from PyQt5.QtGui import *
from PyQt5.QtCore import *
from PyQt5.QtWidgets import *

import sys

class MainWindow(QWidget):
    def __init__(self):
        super(QWidget, self).__init__()

        main_layout = QVBoxLayout(self)

        self.buttons = []

        # Works:
        self.combo = QComboBox()
        main_layout.addWidget(self.combo)
        self.combo.addItems(['PHC'])
        self.combo.addItems(['CHC'])
        self.combo.addItems(['HSC'])
        self.combo.addItems(['DH'])
        self.combo.addItems(['LSH'])
        # # Connecting comboBox to VerifyFType function
        self.combo.currentIndexChanged[str].connect(self.VerifyFType)

        self.button_2 = QPushButton('Validate', self)
        main_layout.addWidget(self.button_2)
        self.buttons.append(self.button_2)

    def VerifyFType(self):
        print("Entered VerifyFType")

        FType = self.combo.currentText()
        print(FType)

        if(FType == "PHC"):
            self.button_2.clicked.connect(lambda: self.PHC_Validate)
        elif(FType == "CHC"):
            self.button_2.clicked.connect(lambda: self.CHC_Validate)
        elif(FType == "DH"):
            self.button_2.clicked.connect(lambda: self.DH_Validate)
        elif(FType == "HSC"):
            self.button_2.clicked.connect(lambda: self.HSC_Validate)
        elif(FType == "LSH"):
            self.button_2.clicked.connect(lambda: self.LSH_Validate)
        else:
            "Nothing Matched , No such FType"

    
    def PHC_Validate(self):
        print('Entered PHC_Validate')

    def CHC_Validate(self):
        print('Entered CHC_Validate')

    def DH_Validate(self):
        print('Entered DH_Validate')

    def HSC_Validate(self):
        print('Entered HSC_Validate')

    def LSH_Validate(self):
        print('Entered LSH_Validate')


app = QApplication(sys.argv)
w = MainWindow()
w.show()
sys.exit(app.exec_())

【问题讨论】：

标签： python python-3.x pyqt5

【解决方案1】：

您的逻辑是错误的，因为您似乎认为将信号连接到另一个函数会断开信号与前一个函数的连接。

解决方法是在按下按钮时使用 QComboBox 的 currentText 调用相应的函数。

class MainWindow(QWidget):
    def __init__(self):
        super(MainWindow, self).__init__()

        self.combo = QComboBox()
        self.combo.addItems(["PHC", "CHC", "HSC", "DH", "LSH"])

        self.button_2 = QPushButton("Validate", self)
        self.button_2.clicked.connect(self.handle_clicked)

        main_layout = QVBoxLayout(self)
        main_layout.addWidget(self.combo)
        main_layout.addWidget(self.button_2)

    def handle_clicked(self):
        FType = self.combo.currentText()
        if FType == "PHC":
            self.PHC_Validate()
        elif FType == "CHC":
            self.CHC_Validate()
        elif FType == "DH":
            self.DH_Validate()
        elif FType == "HSC":
            self.HSC_Validate()
        elif FType == "LSH":
            self.LSH_Validate()
        else:
            "Nothing Matched , No such FType"

    def PHC_Validate(self):
        print("Entered PHC_Validate")

    def CHC_Validate(self):
        print("Entered CHC_Validate")

    def DH_Validate(self):
        print("Entered DH_Validate")

    def HSC_Validate(self):
        print("Entered HSC_Validate")

    def LSH_Validate(self):
        print("Entered LSH_Validate")

【讨论】：