CODEWITHSUNDEEP
pythonpyqtpyqt5qt-designer
damada
damada

Reputation: 141

Load widget with signal-slot defintions from UI file in main window

I am having difficulties to load a widget generated by Qt Designer into my MainWindow. It fails when I have also defined signals and slots in the UI file.

# -*- coding: utf-8 -*-
import sys

from PyQt5 import QtWidgets
from PyQt5 import uic as pyuic

class MainWindow(QtWidgets.QMainWindow):
    def __init__(self, *args):
        super().__init__(*args)
        self.uifile = 'serialTandemGUI.ui'
        self.form_widget = pyuic.loadUi(self.uifile)
        _widget = QtWidgets.QWidget()
        _layout = QtWidgets.QVBoxLayout(_widget)
        _layout.addWidget(self.form_widget)
        self.setCentralWidget(_widget)

if __name__ == '__main__':
    app = QtWidgets.QApplication.instance()
    if app is None:
        app = QtWidgets.QApplication(sys.argv)
    app.aboutToQuit.connect(app.deleteLater)

    w = MainWindow()
    w.show()

    app.exec_()

The error I get is: AttributeError: 'QWidget' object has no attribute 'init'

Is it possible at all to inherit a widget this way? Tia!

Edit: Here's a simple UI file with a clicked() signal to demonstrate. It works as long as no signal definitions have been made (i.e. really just UI, but that is only half of the work, or even less). 

<?xml version="1.0" encoding="UTF-8"?>
<ui version="4.0">
 <class>Form</class>
 <widget class="QWidget" name="Form">
  <property name="geometry">
   <rect>
    <x>0</x>
    <y>0</y>
    <width>400</width>
    <height>300</height>
   </rect>
  </property>
  <property name="windowTitle">
   <string>Form</string>
  </property>
  <layout class="QGridLayout" name="gridLayout">
   <item row="0" column="0">
    <widget class="QGroupBox" name="groupBox">
     <property name="title">
      <string>GroupBox</string>
     </property>
     <layout class="QGridLayout" name="gridLayout_2">
      <item row="0" column="0">
       <widget class="QPushButton" name="pushButton">
        <property name="text">
         <string>PushButton</string>
        </property>
       </widget>
      </item>
     </layout>
    </widget>
   </item>
  </layout>
 </widget>
 <resources/>
 <connections>
  <connection>
   <sender>pushButton</sender>
   <signal>clicked()</signal>
   <receiver>Form</receiver>
   <slot>init()</slot>
   <hints>
    <hint type="sourcelabel">
     <x>203</x>
     <y>162</y>
    </hint>
    <hint type="destinationlabel">
     <x>396</x>
     <y>201</y>
    </hint>
   </hints>
  </connection>
 </connections>
 <slots>
  <slot>init()</slot>
 </slots>
</ui>

Upvotes: 1

Views: 1299

Answers (1)

eyllanesc
eyllanesc

Reputation: 244282

Explanation:

When you have created the .ui you have only pointed out that there is a connection between the clicked signal and the init method of the widget, but when you load it using loadUi() and do not pass it as a second parameter a QWidget will use the base class of the design, in your case QWidget, which clearly has no "init" method throwing you that error.

Solution:

You must create a class that inherits from QWidget and has the init method, and then use loadUi() to fill it out.

# -*- coding: utf-8 -*-
import os
import sys

from PyQt5 import QtCore, QtWidgets, uic as pyuic


class Widget(QtWidgets.QWidget):
    def __init__(self, parent=None):
        super().__init__(parent)
        current_dir = os.path.dirname(os.path.realpath(__file__))
        uifile = os.path.join(current_dir, "serialTandemGUI.ui")
        pyuic.loadUi(uifile, self)

    @QtCore.pyqtSlot()
    def init(self):
        print("init")


class MainWindow(QtWidgets.QMainWindow):
    def __init__(self, parent=None):
        super().__init__(parent)
        self.form_widget = Widget()
        _widget = QtWidgets.QWidget()
        _layout = QtWidgets.QVBoxLayout(_widget)
        _layout.addWidget(self.form_widget)
        self.setCentralWidget(_widget)


if __name__ == "__main__":
    app = QtWidgets.QApplication.instance() or QtWidgets.QApplication(sys.argv)
    app.aboutToQuit.connect(app.deleteLater)

    w = MainWindow()
    w.show()

    sys.exit(app.exec_())

Upvotes: 2

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