Adding a delay in "virus"

Question

We're doing a "virus" now and I was told to make a delay in my code. Help me, please. Here's the code:

import pygame         
from tkinter import *
from tkinter import messagebox 
import time               
import random
pygame.init()
screen = pygame.display.set_mode((800, 300))
pygame.display.set_caption("ЗАРАЖЕНИЕ АЧЕ)")

Tk().wm_withdraw()
font = pygame.font.SysFont("Lucida Console", 35)
label2 = font.render("DELETING ALL PASSWORDS. . . . .", 1, (12, 140, 0, 1))

a = 0

while True:
    screen.fill((0, 0, 0))
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            pygame.quit()
            time.sleep(0.5)
            screen = pygame.display.set_mode((800, 300))
            pygame.display.set_caption("ЗАРАЖЕНИЕ АЧЕ)")
            messagebox.showerror("ERROR!", " Вы попытались закрыть окно! Начинаю удаление файлов")
    warning = f"Процент удаления: {a}"
    label3 = font.render(warning, 1, (12, 140, 0, 1))
    a = a + 1
    time.sleep(0.1)
    if a >= 100:
        break

    screen.blit(label2, (50, 50))
    screen.blit(label3, (50, 100))
    pygame.display.update()

My question is, I don't know which command to use to delay and where to put it.

Thank you in advance!

Please give textual information and code as text, not as link to picture of text. https://meta.stackoverflow.com/a/285557/7733418 That is part of providing a helpful [mre]. — Yunnosch, May 30 '20 at 10:09
@Sadap warning = f"Процент удаления: {a}" label3 = font.render(warning, 1, (12, 140, 0, 1)) a = a + 1 time.sleep(0.1) if a >= 100: break — Andriy, May 30 '20 at 10:34
@Andrew then place `time.sleep(2)` before that line. `2` is the number of seconds to wait — bb1950328, May 30 '20 at 10:37

score 3 · Accepted Answer · answered May 30 '20 at 10:38

With pygame, you don't use time.sleep, you need to replace your time.sleep code with pygame.time.wait, which is written in milliseconds, so your updated code would look like so:

import pygame         
from tkinter import *
from tkinter import messagebox 
import time               
import random
pygame.init()
screen = pygame.display.set_mode((800, 300))
pygame.display.set_caption("ЗАРАЖЕНИЕ АЧЕ)")

Tk().wm_withdraw()
font = pygame.font.SysFont("Lucida Console", 35)
label2 = font.render("DELETING ALL PASSWORDS. . . . .", 1, (12, 140, 0, 1))

a = 0

while True:
    screen.fill((0, 0, 0))
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            pygame.display.quit()
            pygame.time.wait(500)
            screen = pygame.display.set_mode((800, 300))
            pygame.display.set_caption("ЗАРАЖЕНИЕ АЧЕ)")
            messagebox.showerror("ERROR!", " Вы попытались закрыть окно! Начинаю удаление файлов")
    warning = f"Процент удаления: {a}"
    label3 = font.render(warning, 1, (12, 140, 0, 1))
    a = a + 1
    pygame.time.wait(100)
    if a >= 100:
        break

    screen.blit(label2, (50, 50))
    screen.blit(label3, (50, 100))
    pygame.display.update()

Also with your code I replaced pygame.quit() with pygame.display.quit() so the code will continue after the so called "error".

Adding a delay in "virus"

1 Answers1