pyqt5如何获取当前文件并切换到下一个，一个按钮可以对应多个事件。

import sys
from PyQt5.QtWidgets import QApplication, QMainWindow, QPushButton, QFileDialog, QLabel, QVBoxLayout, QWidget

class MainWindow(QMainWindow):
    def __init__(self):
        super().__init__()

        self.current_file_index = 0
        self.file_list = ['file1.txt', 'file2.txt', 'file3.txt']  # 假设这是你的文件列表
        self.current_file_label = QLabel(f'Current file: {self.file_list[self.current_file_index]}')

        self.button = QPushButton('Select File and Switch')
        self.button.clicked.connect(self.select_file_and_switch)
        self.button.clicked.connect(self.another_action)

        layout = QVBoxLayout()
        layout.addWidget(self.current_file_label)
        layout.addWidget(self.button)

        container = QWidget()
        container.setLayout(layout)
        self.setCentralWidget(container)

    def select_file_and_switch(self):
        options = QFileDialog.Options()
        file_path, _ = QFileDialog.getOpenFileName(self, "QFileDialog.getOpenFileName()", "",
                                                  "All Files (*);;Python Files (*.py)", options=options)
        if file_path:
            self.current_file_label.setText(f'Selected file: {file_path}')
            # 这里可以添加切换文件的逻辑

    def another_action(self):
        # 这里是按钮点击的另一个动作
        print("Another action triggered")

if __name__ == '__main__':
    app = QApplication(sys.argv)
    window = MainWindow()
    window.show()
    sys.exit(app.exec_())