Question

我是pygame的新手，当我在pygame.window中发布一个鼠标时，它打印出键码c = 99。我怎么能避免这个错误？

    if event.type == pygame.KEYDOWN:
        if chr(event.key) in symbols: #and event.key!=99 :
            text=text+chr(event.key)

编辑评论代码

import pygame as pg 
from pygame import draw, display

pg.init() 

depth = 16 
m = (640, 480) 

display.set_mode(m, 0) 

scr = display.get_surface() 
pg.display.set_caption("Example") 

cl = pg.time.Clock() 

sbs = ['a','b','c'] 

def printT(text): 
    font = pg.font.Font(None,30) 
    pr = font.render(t, 1,(255,0,0)) 
    scr.blit(pr, (0, 0)) 

t = '' 

do = True 

while do: 
    cl.tick(80) 

    for e in pg.event.get(): 
        if e.type == pg.QUIT: 
            do = False 
        if e.type == pg.KEYDOWN: 
            if chr(e.key) in sbs: 
                t += pg.key.name(e.key) 
                printT(t) 

    pg.display.flip() 

pg.quit()

我怎么能避免pygame keycode = 99错误？

0 个答案: