我们如何在pyqt中单击按钮打开特定于选项卡的选项卡?

setCurrentIndex() 方法 QTabWidget ,对于此方法,您必须指示索引。每次触发信号与 QAction 菜单的

class MainWindow(QMainWindow):
    def __init__(self, parent=None):
        QMainWindow.__init__(self, parent)

        widget = QTabWidget(self)
        for i in range(10):
            widget.addTab(QListWidget(), "tab{}".format(i+1))

        self.setCentralWidget(widget)

        menubar = self.menuBar()
        action = menubar.addAction("Select tab5")
        action.triggered.connect(lambda: widget.setCurrentIndex(4))


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

加上:

    self.Add_GroupD.triggered.connect(lambda checked, index1=4, index2=1 : self.someslot(index1, index2))
def someslot(self, index1, index2)
    self.tabWidget_4.setCurrentIndex(index1)
    self.tabs.setCurrentIndex(index2)