Answers for "tkinter.filedialog"

3

python tkinter filedialog

from tkinter import filedialog

						# Where it open to.					# What the window is called.	# What file types the user can choose between. first one is the defualt. (("what ever", "*.format"), ("what ever 2", "*.format2"))
filedialog.askopenfilename(initialdir=os.path.normpath("C://"), title="Example", filetypes =(("PNG", "*.png"),("JPG", "*.jpg"),("All Files","*.*")))
Posted by: Guest on January-02-2021
3

python tkinter filedialog folder

from tkinter import filedialog

								# Where it open to.					# What the window is called.
folder = filedialog.askdirectory(initialdir=os.path.normpath("C://"), title="Example")
Posted by: Guest on January-02-2021
1

How to open dialog box to select files in python

# open_file = filedialog.askopenfilenames(filetypes=[("File type", "Image's .extesnions spearated by space")])
# In this example I'll be opening dialog box to select only images with (.jpg, .jpeg, .png, .jfif) extensions.
>>> from tkinter import Tk, filedialog
>>> 
>>> root = Tk() # pointing root to Tk() to use it as Tk() in program.
>>> root.withdraw() # Hides small tkinter window.
''
>>> root.attributes('-topmost', True) # Opened windows will be active. above all windows despite of selection.
''
>>># Open dialog box to select images with certain extensions.
>>> open_file = filedialog.askopenfilenames(filetypes=[("Image Files", ".png .jfif, .jpg, .jpeg")]) # returns a tuple with opened file's complete path
>>> print(open_file)
('C:/Users/User/OneDrive/Documents/Images/Image (1).png', 'C:/Users/User/OneDrive/Documents/Images/Image (2).jpeg', 'C:/Users/User/OneDrive/Documents/Images/Image (3).jpg', 'C:/Users/User/OneDrive/Documents/Images/Image (4).png', 'C:/Users/User/OneDrive/Documents/Images/Image (5).jfif')
Posted by: Guest on January-05-2021
0

tkinter filedialog filename

def open_file():
    file = askopenfile(mode='r', filetypes=[
                       ('Text files', '*.txt'), ('CSV Files', '*.csv')])
    if file is not None:
        print(file.name.split("/")[-1]) # this will print the file name

btn = Button(root, text='Open', command=lambda: open_file())
btn.pack(side=TOP, pady=10)
Posted by: Guest on May-01-2021

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