PyQt5 QCommandLinkButton – 进入下一个状态

在本文中，我们将看到如何进入可检查 QCommandLinkButton 的下一个状态。我们可以在setCheckable方法的帮助下使命令链接按钮可检查，这将使命令链接按钮的两种状态将被检查，即按下和未检查，即释放状态。 Next state 会根据当前状态改变状态。

为了做到这一点，我们使用带有命令链接按钮对象的nextCheckState方法

Syntax : button.nextCheckState()

Argument : It takes no argument

Return : It return None

下面是实现

# importing libraries
from PyQt5.QtWidgets import * 
from PyQt5 import QtCore, QtGui
from PyQt5.QtGui import * 
from PyQt5.QtCore import * 
import sys
  
  
class Window(QMainWindow):
  
    def __init__(self):
        super().__init__()
  
        # setting title
        self.setWindowTitle("Python ")
  
        # setting geometry
        self.setGeometry(100, 100, 500, 400)
  
        # calling method
        self.UiComponents()
  
        # showing all the widgets
        self.show()
  
    # method for components
    def UiComponents(self):
  
        # creating a command link button
        cl_button = QCommandLinkButton("Press", self)
  
        # setting geometry
        cl_button.setGeometry(250, 100, 200, 50)
  
        # making button checkable
        cl_button.setCheckable(True)
          
        # creating a push button
        push = QPushButton("Next State", self)
          
        # setting geometry to the push button
        push.setGeometry(50, 100, 120, 40)
          
        # adding action to the push button
        push.clicked.connect(lambda: next_method())
          
        # method called by push button
        def next_method():
              
            # going to next state
            cl_button.nextCheckState()
    
  
# create pyqt5 app
App = QApplication(sys.argv)
  
# create the instance of our Window
window = Window()
  
# start the app
sys.exit(App.exec())

输出 ：