PYGLET – Opening file using File Location

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 
 
Python3
| # importing pyglet module importpyglet importpyglet.window.key as key  # width of window width =500  # height of window height =500  # caption i.e title of the window title ="Geeksforzambiatek"  # creating a window window =pyglet.window.Window(width, height, title)   # text  text ="Welcome tozambiatek" # 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()# loadingzambiatek 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 defon_draw():           # clear the window     window.clear()           # draw the label    label.draw()         # draw the image on screen    sprite.draw()      # key press event     @window.event defon_key_press(symbol, modifier):       # key "C" get press     ifsymbol ==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) # creating a file location objectfile=pyglet.resource.FileLocation("E:/btech / certi/")# opening a filevalue =file.open("gfg.jpg")# showing value with the help of labellabel.text =str(value)   # start running the application pyglet.app.run()  | 
Output :
 
 
				 
					



