In this article, we will see how we can access default resource path 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. Locations in the search path are searched in order and are always case-sensitive. After changing the path you must call reindex.
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 path attribute with the pyglet.resource
Syntax : resource.path
Argument : It takes no argument
Return : It returns string
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 = "Geeksforneveropen" # creating a window window = pyglet.window.Window(width, height, title) # text text = "Welcome to neveropen" # creating label with following properties# font = cooper# position = 250, 150# anchor position = centerlabel = 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 neveropen imageimage = pyglet.image.load('gfg.png')# creating sprite object# it is instance of an image displayed on-screensprite = 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) # accessing default resource pathvalue = pyglet.resource.path# setting text of labellabel.text = "Default path : " + str(value)# start running the application pyglet.app.run() |
Output :
