PYGLET – Getting Location of Resource

In this article we will see how we can get resource location 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 load a file i.e resource 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. Getting the location of a resource is useful for opening files referenced from a resource. For example, an HTML file loaded as a resource might reference some images. These images should be located relative to the HTML file, not looked up individually in the loader’s path.
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 do this we use location method with the pyglet.resource
Syntax : resource.location(resource_name)
Argument : It takes string i.e resource name as argument
Return : It returns Location object

Below is the implementation

Python3

# 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 properties
# 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: 
        
        # printing 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) 

# getting resource location
value = pyglet.resource.location("gfg.png")

# setting text  of label
label.text = str(value)

# start running the application 
pyglet.app.run()

Output :

PYGLET – Getting Location of Resource

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