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PYGLET – Opening file using File Location

  • Difficulty Level : Hard
  • Last Updated : 24 Sep, 2020

In this article we will see how we can open a file using file location object in PYGLET module in python. Pyglet is easy to use but powerful library for developing visually rich GUI applications like games, multimedia etc. A window is a “heavyweight” object occupying operating system resources. Windows may appear as floating regions or can be set to fill an entire screen (fullscreen). In order to create a file location object we use resource module of pyglet. This module allows applications to specify a search path for resources. Relative paths are taken to be relative to the application’s __main__ module. ZIP files can appear on the path; they will be searched inside. File can be opened which is on the same folder of the file location is pointing using the file name to be opened.

We can create a window object with the help of command given below

# creating a window
window = pyglet.window.Window(width, height, title)

In order to create window we use open method with the file location object

Syntax : file.open(name)

Argument : It takes file name i.e string as argument



Return : It returns class ‘_io.BufferedReader

Below is the implementation




# importing pyglet module 
import pyglet 
import pyglet.window.key as key
    
# width of window 
width = 500
    
# height of window 
height = 500
    
# caption i.e title of the window 
title = "Geeksforgeeks"
    
# creating a window 
window = pyglet.window.Window(width, height, title) 
    
# text  
text = "Welcome to GeeksforGeeks"
   
# creating label with following proeprties
# font = cooper
# position = 250, 150
# anchor position = center
label = pyglet.text.Label(text, 
                          font_name ='Cooper'
                          font_size = 16
                          x = 250,  
                          y = 150
                          anchor_x ='center',  
                          anchor_y ='center')
  
  
# creating a batch 
batch = pyglet.graphics.Batch()
  
# loading geeksforgeeks image
image = pyglet.image.load('gfg.png')
  
  
# creating sprite object
# it is instance of an image displayed on-screen
sprite = pyglet.sprite.Sprite(image, x = 200, y = 230)
    
# on draw event 
@window.event 
def on_draw(): 
        
    # clear the window 
    window.clear() 
        
    # draw the label
    label.draw() 
      
    # draw the image on screen
    sprite.draw()
        
# key press event     
@window.event 
def on_key_press(symbol, modifier): 
    
    # key "C" get press 
    if symbol == key.C: 
          
        # printng the message
        print("Key : C is pressed")
          
# image for icon 
img = image = pyglet.resource.image("gfg.png"
  
# setting image as icon 
window.set_icon(img) 
  
# creating a file location object
file = pyglet.resource.FileLocation("E:/btech / certi/")
  
# opening a file
value = file.open("gfg.jpg")
  
# showing value with the help of label
label.text = str(value)
     
# start running the application 
pyglet.app.run() 

Output :

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