Pygame.key.get_pressed未按下時被識別

Question

我是編碼新手（大約3周），並且我一直在嘗試使用pygame制作基於回合的RPG戰斗系統。 我遇到了涉及pygame的key.get_pressed的問題。 我正在制作一個下拉菜單，其中如果您按ENTER鍵，則會出現一個較小的黑框，而當您再次按ENTER鍵時，該框應消失。 不幸的是，黑框出現了，但幾乎立即消失了。 我將打印功能用作調試程序以查找錯誤，並且似乎程序注冊了第二個ENTER鍵（應關閉該框），而沒有被按下。 任何幫助將不勝感激。 我也在使用Windows 10和Python 3.6

import pygame
pygame.init()

win = pygame.display.set_mode((820,567))
pygame.display.set_caption('Basic battle')

drop_menu_attack = pygame.image.load('drop menu attack.png')
health_pp = pygame.image.load('hp and pp.png')
menu_img = pygame.image.load('menu.png')
attack_img = pygame.image.load('attack.png')
defend_img = pygame.image.load('defend.png')
bag_img = pygame.image.load('bag.png')
background_img = pygame.image.load('rpg background.gif')
Snake_enemy_img = pygame.image.load('pixil-frame-0.png')

clock = pygame.time.Clock()

class player(object):
    def __init__ (self, x, y, width, height):
        self.x = x
        self.y = y
        self.width = width
        self.height = height
        self.cursor_state = 1
        self.attack_x = 100
        self.defend_x = 325
        self.bag_x = 575
        self.top_menu_y = 70

class drop_menu(object):
    def __init__ (self, x, y):
        self.x = x
        self.y = y
        self.drop_menu_state = False

def cursor_position():
    if c_p.cursor_state == 1:
       print ('it should be on attack')
       c_p.x = c_p.attack_x
       c_p.y = c_p.top_menu_y
    elif c_p.cursor_state == 2:
        print ('it hould be on defend')
        c_p.x = c_p.defend_x
        c_p.y = c_p.top_menu_y
    elif c_p.cursor_state == 3:
        print ('it should be on bag')
        c_p.x = c_p.bag_x
        c_p.y = c_p.top_menu_y
    elif c_p.cursor_state > 3:
        c_p.cursor_state = 3
        print ('it should be on bag')
        c_p.x = c_p.bag_x
        c_p.y = c_p.top_menu_y

    elif c_p.cursor_state < 1:
        c_p.cursor_state = 1
        print ('it should be on attack')
        c_p.x = c_p.attack_x
        c_p.y = c_p.top_menu_y

def select():
    if c_p.cursor_state == 1:
        d_m.drop_menu_state = True
        print (d_m.y)
        while d_m.drop_menu_state:
            pygame.time.delay(150)
            win.blit (drop_menu_attack, (d_m.x, d_m.y))
            pygame.display.update()
            print ('dooooog')
            keys = pygame.key.get_pressed()
            if d_m.drop_menu_state == True:
                if keys[pygame.K_RETURN]:
                    d_m.drop_menu_state = False
                    print('SSSSSS')
            pygame.event.pump()

def redraw_gamewindow():
    win.blit (background_img, (0,0))
    win.blit (Snake_enemy_img, (300, 174))
    win.blit (menu_img, (25, 425 ))
    win.blit (menu_img, (25, 10))
    win.blit (health_pp, (70, 450))
    win.blit (attack_img, (c_p.attack_x + 30, c_p.top_menu_y - 15))
    win.blit (defend_img, (c_p.defend_x + 30, c_p.top_menu_y - 15))
    win.blit (bag_img, (c_p.bag_x + 30, c_p.top_menu_y - 15))
    pygame.draw.rect(win, (255, 255, 255), (c_p.x, c_p.y, 7 , 7))
    pygame.display.update()

d_m = drop_menu(25, 125)
print (d_m.x)
c_p = player(100, 70, 7,7)
run = True

while run:
    clock.tick(27)

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            run = False

    keys = pygame.key.get_pressed()

        if keys[pygame.K_RIGHT]:
        c_p.cursor_state = c_p.cursor_state + 1
        cursor_position()
        pygame.time.delay(150)

    if keys[pygame.K_LEFT]:
        c_p.cursor_state = c_p.cursor_state -1
        cursor_position()
        pygame.time.delay(150)
    if keys[pygame.K_RETURN]:
        select()
        pygame.time.delay(500)

    redraw_gamewindow()

Answer 1

所以發生的事情是您按住ENTER鍵。 在while循環中，當您按Enter鍵時，它將調用select()函數，然后該函數將檢測仍在按的ENTER鍵。 讓我畫一個視覺效果：

按下ENTER鍵-> select（）被調用， drop_menu_state變為drop_menu_state > select（）檢測到ENTER仍被按下-> drop_menu_state變為false

您可能會問，它如何檢測？ 好了，這些幀以每秒想要的27幀的速度運行。 因此，在按ENTER鍵時，它將每隔1/27秒更改一次放置菜單狀態。 如果您可以按住ENTER鍵的時間不超過1/27秒，它將正常工作。

有3個解決方案。 將菜單的關閉綁定到另一個鍵，添加短計時器或使用事件循環。 我的編程風格是添加一個計時器。

import time

pre_time = time.time()
def select():
    ... # code in select()
    if d_m.drop_menu_state and pre_time + 5 > time.time() and keys[pygame.K_RETURN]:
        pre_time = time.time()
        d_m.drop_menu_state = False

我簡化了您的代碼（否== True） pre_time + 5 > time.time()所說的是，自上次將pre_time定義為當前時間以來已經有5秒了。 我希望這有幫助！