Keyboard.is_pressed isn't working

[Enderg312]

if keyboard.is_pressed('enter'):
    enter()

def enter():
    if user.get() == admin and code.get() == password:
        sys.exit()
    else: 
        l3 = Label(root, text = "check login", font=("Courier", 22))
        l3.grid()

Every time I hit enter on the keyboard nothing happens, it doesn't execute the code.

[RockSolid1106]

Try defining the function enter() before the "if keyboard.is_pressed()" command?

 

def enter():
    if user.get() == admin and code.get() == password:
        sys.exit()
    else: 
        l3 = Label(root, text = "check login", font=("Courier", 22))
        l3.grid()
        
if keyboard.is_pressed('enter'):
    enter()

Like this?

[Enderg312]

All it does is disables the enter function.

[RockSolid1106]

Since I don't know your code, I made some alterations and the following code works for me:

import keyboard
def enter():
    print("Hey, you just pressed the Enter Key!")
while True:

    if keyboard.is_pressed('enter'):
        enter()
        exit()

And prints the text when I press the Enter key.

Maybe you've got some problem with other parts of the code?

[Enderg312]

from tkinter import *
import subprocess
import sys
import keyboard

root = Tk()
root.title("My Login")
root.geometry("650x650")
root.attributes('-fullscreen', True)

frame = Frame(root)

app = Frame(root)
app.grid

l = Label(root, text = "Login",font="Times 30", padx=5,  pady=5)
l.grid()

l1 = Label(root, text = "Username:",font="Times 30", padx=5,  pady=5)
l1.grid()

l2 = Label(root, text = "Password:", font="Times 30", padx=5,  pady=5)
l2.grid()

user = Entry(root)
user.grid( row= 1, column= 2)
user.configure(font = ("Courier", 44))
code = Entry(root)
code.grid( row= 2, column= 2)
code.configure(font = ("Courier", 44),show="*")

if keyboard.is_pressed('enter'):
    enter()

admin = "admin"  
password = "Password"  

def enter():
    if user.get() == admin and code.get() == password:
        sys.exit()
    else: 
        l3 = Label(root, text = "check login", font=("Courier", 22))
        l3.grid()

b1 = Button(root, text = "Enter", command = enter, font=("Courier", 44))
b1.grid()

root.mainloop()

Here's my code

[RockSolid1106]

from tkinter import *
import subprocess
import sys
import keyboard

root = Tk()
root.title("My Login")
root.geometry("650x650")
root.attributes('-fullscreen', True)

frame = Frame(root)

app = Frame(root)
app.grid

l = Label(root, text = "Login",font="Times 30", padx=5,  pady=5)
l.grid()

l1 = Label(root, text = "Username:",font="Times 30", padx=5,  pady=5)
l1.grid()

l2 = Label(root, text = "Password:", font="Times 30", padx=5,  pady=5)
l2.grid()

user = Entry(root)
user.grid( row= 1, column= 2)
user.configure(font = ("Courier", 44))
code = Entry(root)
code.grid( row= 2, column= 2)
code.configure(font = ("Courier", 44),show="*")



admin = "admin"  
password = "Password"  

def enter(event):
    if user.get() == admin and code.get() == password:
        sys.exit()
    else: 
        l3 = Label(root, text = "check login", font=("Courier", 22))
        l3.grid()

b1 = Button(root, text = "Enter", command = enter, font=("Courier", 44))
b1.grid()
root.bind('<Return>', enter)
root.mainloop()

Try this.

I have removed the keyboard.is_pressed() function, and instead, added root.bind(), which is used to detect key presses in Tkinter.

 

And just so you know, you CANNOT call a function before it is defined.