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PyQt5 QSpinBox – Finding child using ID

  • Last Updated : 19 May, 2020

In this article we will see how we can find the child of spin box using the child ID, we can get the child id using winId method with the object of child. Spin box is the parent widget it has child like line edit widget and those up and down button.

In order to do this we use find method

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Syntax : spin_box.find(id)



Argument : It takes child id as argument

Return : It returns the child object

Below is the implementation




# 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, 600, 400)
  
        # calling method
        self.UiComponents()
  
        # showing all the widgets
        self.show()
  
        # method for widgets
    def UiComponents(self):
        # creating spin box
        self.spin = QSpinBox(self)
  
        # setting geometry to spin box
        self.spin.setGeometry(100, 100, 250, 40)
  
        # setting range to the spin box
        self.spin.setRange(0, 999999)
  
        # setting prefix to spin
        self.spin.setPrefix("Prefix ")
  
        # setting suffix to spin
        self.spin.setSuffix(" Suffix")
  
        # ensuring spin box is polished
        i_d = self.spin.lineEdit().winId()
  
        # getting child using ID
        child = self.spin.find(i_d)
  
        # creating a label
        label = QLabel(self)
  
        # making it multi line
        label.setWordWrap(True)
  
        # setting its geometry
        label.setGeometry(100, 200, 200, 60)
  
        # setting text to the label
        label.setText(str(child))
  
  
  
# create pyqt5 app
App = QApplication(sys.argv)
  
# create the instance of our Window
window = Window()
  
# start the app
sys.exit(App.exec())

Output :




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