我正在尝试学习PyQt5 / Python 3.6.3。单击按钮时,我试图在状态栏中显示一条消息。问题是所述按钮位于QWidget内部,据我所知,statusBar()仅在QMainWindow中可用。这是我到目前为止拼凑的代码......
import sys
from PyQt5.QtCore import Qt
from PyQt5.QtWidgets import QStatusBar, QMainWindow, QApplication, QWidget,QHBoxLayout, QVBoxLayout, QPushButton, QSlider, QLCDNumber, QLabel
class MyMainWindow(QMainWindow):
def __init__(self, parent=None):
super().__init__()
self.main_widget = FormWidget(self)
self.setCentralWidget(self.main_widget)
self.init_UI()
def init_UI(self):
self.statusBar().showMessage('Ready')
self.setGeometry(200, 100, 300, 300)
self.setWindowTitle('Central Widget')
self.show()
class FormWidget(QWidget):
def __init__(self, parent):
super(FormWidget, self).__init__(parent)
self.init_UI()
def init_UI(self):
hbox = QHBoxLayout()
button_1 = QPushButton('Button 1', self)
button_1.clicked.connect(self.buttonClicked)
hbox.addWidget(button_1)
button_2 = QPushButton('Button 2', self)
button_2.clicked.connect(self.buttonClicked)
hbox.addWidget(button_2)
self.setLayout(hbox)
self.setGeometry(200, 100, 300, 300)
self.setWindowTitle('Slider and LCD')
self.show()
def buttonClicked(self):
sender = self.sender()
self.statusBar.showMessage(sender.text() + ' was clicked')
if __name__ == '__main__':
APP = QApplication(sys.argv)
ex = MyMainWindow()
sys.exit(APP.exec_())
当我运行它时,我收到以下错误:
Traceback (most recent call last):
File "central_widget_test.py", line 40, in buttonClicked
self.statusBar.showMessage(sender.text() + ' was clicked')
AttributeError: 'FormWidget' object has no attribute 'statusBar'
有人可以帮我解决这个问题吗?
答案 0 :(得分:1)
您应该初始化MyMainWindow
类的状态栏对象,以后可以更新。
您的FormWidget
可以通过引用MyMainWindow
对象来更新状态栏。
class MyMainWindow(QMainWindow):
. . .
def init_UI(self):
self.statusbar = self.statusBar()
self.statusbar.showMessage('Ready')
self.setGeometry(200, 100, 300, 300)
self.setWindowTitle('Central Widget')
self.show()
class FormWidget(QWidget):
def __init__(self, parent):
super(FormWidget, self).__init__(parent)
self.parent = parent
self.init_UI()
. . .
def buttonClicked(self):
sender = self.sender()
self.parent.statusbar.showMessage(sender.text() + ' was clicked')