我一直在做游戏;其中一个主要组件是倒数计时器 - 但是这个计时器被延迟,我无法推断出原因。我希望它每秒减少一次,但它似乎每6秒钟减少一次。
以下是设置计时器的方法:
loops = 0
minute = 1
tens = 0
ones = 0
#Timer Calculation
screen.blit(cloudSky, (0,0))
if go == True:
loops = loops + 1
if (loops % 60)== 0:
if ones == 0 and tens == 0 and minute != 0:
tens = 6
ones = 0
minute = minute - 1
if ones == 0 and tens != 0:
ones = 9
tens = tens - 1
elif ones != 0:
ones = ones - 1
elif tens == 0 and ones == 0:
tens = 5
minute = minute - 1
elif ones == 0 and tens != 0:
tens = tens - 1
if minute <= 0 and tens == 0 and ones == 0:
go = False
我用下面的代码在屏幕上打印:
#Draw Clock Time
time = timeFont.render(str (minute)+ ":" + str (tens) + str (ones), True, WHITE)
screen.blit(time, (750,10))
非常感谢任何帮助!
答案 0 :(得分:0)
这可能是因为你依靠这个计时器计算正好60次/秒。如果你每次更新10次,它会每6秒实时增加1秒。您应该使用time
模块之类的东西来更精确地计时。
import time
# To remember current time
timer = time.time() # timer = 1497737106.913825
# To read time since the clock started
seconds_passed = time.time() - timer # seconds_passed = 11.117798089981079
# To get a tuple containing calculated time (minutes, seconds, etc.)
time_passed = time.gmtime(seconds_passed)
# time_passed = time.struct_time(tm_year=1970, tm_mon=1, tm_mday=1,
# tm_hour=0, tm_min=0, tm_sec=11, tm_wday=3, tm_yday=1, tm_isdst=0)
# As you can see year returned is 1970 because 0 means start of Unix Epoch time
# but you don't use it anyway
因为它的名字是元组,你可以使用索引:
time_passed[0] # 1970
或姓名:
time_passed.tm_year # 1970