Python等待用户输入的时间不够长
提问于 2022-05-29 05:34:31
我对python程序有问题。当我提示用户输入时，程序只等待2-3秒，如果我在这个时间框架内输入答案，它将加载其余的内容，但是如果我不这样做，它就会停止加载输入并变得不应答。我尝试过添加一个while循环，但是它不会改变任何东西。
import pygame as pg
pg.init()
pg.display.set_caption("Struktura Atoma")
(sirina, visina) = (250, 300)
prozor = pg.display.set_mode((sirina, visina))
kraj = True
r = 20 # poluprecnik
x = sirina // 2 # koordinata x centra kruznice
y = visina // 2 # koordinara y centra kruznice
izbor = input("Unesite 1 za atomsku strukturu atoma ugljenika, unesite 2 za atomsku strukturu magnezijuma: ")
if izbor == "1":
for i in range(3):
roze = (243, 58, 106)
siva = (220, 220, 220)
bojaAtom = (0, 255, 255)
font = pg.font.SysFont("Arial", 22) # podesavanje fonta
slC = font.render("C", True, pg.Color("black")) # formiranje slova
pg.draw.circle(prozor, bojaAtom, (x, y), r, 1) # crtanje kruznice
if r == 20:
pg.draw.circle(prozor, bojaAtom, (x, y), r, 0)
prozor.blit(slC, (x - 6, y - 12))
if r == 50:
pg.draw.circle(prozor, roze, (x - 45, y - 20), r - 40, 0)
pg.draw.circle(prozor, roze, (x + 45, y + 20), r - 40, 0)
if r == 80:
pg.draw.circle(prozor, siva, (x - 80, y), r - 70, 0)
pg.draw.circle(prozor, siva, (x + 80, y), r - 70, 0)
pg.draw.circle(prozor, siva, (x, y - 80), r - 70, 0)
pg.draw.circle(prozor, siva, (x, y + 80), r - 70, 0)
pg.display.update() # osvezavanje prozora
r = r + 30 # promena poluprecnika sledece kruznice
pg.time.wait(100) # cekaj 100ms
# provera da li je korisnik zatvorio prozor
if izbor == "2":
for i in range(4):
roze = (243, 58, 106)
siva = (220, 220, 220)
bojaAtom = (0, 255, 255)
font = pg.font.SysFont("Arial", 22) # podesavanje fonta
slMg = font.render("Mg", True, pg.Color("black")) # formiranje slova
pg.draw.circle(prozor, bojaAtom, (x, y), r, 1) # crtanje kruznice
if r == 20:
pg.draw.circle(prozor, bojaAtom, (x, y), r, 0)
prozor.blit(slMg, (x - 12, y - 13))
if r == 50:
pg.draw.circle(prozor, roze, (x - 45, y - 20), r - 40, 0)
pg.draw.circle(prozor, roze, (x + 45, y + 20), r - 40, 0)
if r == 80:
pg.draw.circle(prozor, siva, (x - 80, y), r - 70, 0)
pg.draw.circle(prozor, siva, (x + 80, y), r - 70, 0)
pg.draw.circle(prozor, siva, (x, y - 80), r - 70, 0)
pg.draw.circle(prozor, siva, (x, y + 80), r - 70, 0)
pg.draw.circle(prozor, siva, (x - 56, y - 56), r - 70, 0)
pg.draw.circle(prozor, siva, (x + 56, y - 56), r - 70, 0)
pg.draw.circle(prozor, siva, (x + 56, y + 56), r - 70, 0)
pg.draw.circle(prozor, siva, (x - 56, y + 56), r - 70, 0)
if r == 110:
pg.draw.circle(prozor, siva, (x - 105, y), r - 100, 0)
pg.draw.circle(prozor, siva, (x + 105, y), r - 100, 0)
pg.display.update() # osvezavanje prozora
r = r + 30 # promena poluprecnika sledece kruznice
pg.time.wait(100) # cekaj 100ms
while kraj:
for dogadjaj in pg.event.get():
if dogadjaj.type == pg.QUIT:
kraj = False
pg.quit()
回答 1
Stack Overflow用户
发布于 2022-05-29 06:25:49
您需要按如下方式更新事件循环中的屏幕，否则程序将在您看到发生的情况下挂起。
while kraj:
for dogadjaj in pg.event.get():
if dogadjaj.type == pg.QUIT:
kraj = False
pg.display.update()页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/72424108
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