[英]Breaking out of loop - Python
我試過谷歌搜索和搜索 SO,但我無法弄清楚為什么我在倒數第二行的中斷沒有離開 while 循環。 更好的是,我無法弄清楚為什么循環也沒有繼續。 我的目的是讓用戶有可能在最后一次選擇后前往主菜單(基本上是 menuchoice 的 while 循環(這是我在此處粘貼的內容之上的一個循環)。
有什么建議? 先感謝您。 感覺就像我錯過了一些必不可少的東西。
#this is going to show how many exercise weapons you need for next magic level
if menuchoice == "4":
#these functions returns the amount of magic wands/rods that is needed to be spent for next magic level
print("Select vocation")
print("Press P for Royal Paladin")
#ask user to input vocation:
while True:
vocationchoice = input()
if vocationchoice == "P" or vocationchoice == "p":
#ask user to input magic level for paladin
num1 = float (input("Enter your magic level: "))
#ask for own training dummy
print("Do you have your own exercise dummy? Type Y for yes and N for no.")
while True:
trainingdummy = input()
if trainingdummy == "y" or trainingdummy == "Y":
#list the different exercise weapons
print("Select exercise weapon:")
print("1. Training rod")
#loop, where we ask user to input what exercise weapon they want to calculate
while True:
while True:
weaponchoice = input()
if weaponchoice == "q":
sys.exit() #quit the program
if weaponchoice == "1" or weaponchoice == "2" or weaponchoice == "3" or weaponchoice == "f":
break #break out of the input loop
#User choice
if weaponchoice == "1":
print("The amount of training rods needed for next magic level is " + str((nextmaglvlpalwithdummy(num1))) + ".")
if trainingdummy == "n" or trainingdummy == "N":
#list the different exercise weapons
print("Select exercise weapon:")
print("1. Training rod")
#loop where ask user to input what exercise weapon they want to calculate
while True:
weaponchoice = input()
#User choice
if weaponchoice == "1":
print("The amount of training rods needed for next magic level is " + str((nextmaglvlpal(num1))) + ".")
elif weaponchoice == "f":
break
print("\nGo to main menu? Press F.")
為weaponchoice == "1"
添加一個break
以退出循環。
首先,您應該在input()
命令中打印內容,因為它會在 intent: input("Text to display")
被清除。
其次,如果你想退出到主菜單,你需要打破每一個嵌套的循環。 在這里,您只會打破最內部的循環。
由於在 Python 中沒有goto
指令或命名循環,您可以使用標志。 當用戶按下“F”時,標志設置為真,然后在每個外部嵌套循環開始時使用此標志來中斷它們。 它看起來像這樣:
while True: # This is your menu loop
menuFlag = False # Declare and set the flag to False here
menu = input("Choose the menu: ")
# ...
while True: # Choose character loop
if menuFlag: break # Do not forget to break all outer loops
character = input("Choose a character: ")
# ...
while True: # Any other loop (choose weapon, ...)
weapon = input("Choose weapon: ")
# Here you want to return to the menu if f is pressed
# Set the flag to True in this condition
if weapon == "f":
menuFlag = True
break
在您的游戲中,這類似於:
goToMainMenu = False
while True:
if goToMainMenu: break
vocationchoice = input("Select vocation.\nPress P for Royal Paladin: ")
if vocationchoice == "P" or vocationchoice == "p":
#ask user to input magic level for paladin
num1 = float (input("Enter your magic level: "))
#ask for own training dummy
while True:
if goToMainMenu: break
trainingdummy = input("Do you have your own exercise dummy?\nType Y for yes and N for no: ")
if trainingdummy == "y" or trainingdummy == "Y":
#loop, where we ask user to input what exercise weapon they want to calculate
while True:
while True:
weaponchoice = input("Select exercise weapon:\n1. Training rod: ")
if weaponchoice == "q":
sys.exit() #quit the program
if weaponchoice == "1" or weaponchoice == "2" or weaponchoice == "3" or weaponchoice == "f":
break #break out of the input loop
#User choice
if weaponchoice == "1":
print("The amount of training rods needed for next magic level is " + str((nextmaglvlpalwithdummy(num1))) + ".")
if trainingdummy == "n" or trainingdummy == "N":
#list the different exercise weapon
#loop where ask user to input what exercise weapon they want to calculate
while True:
weaponchoice = input("Select exercise weapon (press F for main menu):\n1. Training rod: ")
#User choice
if weaponchoice == "1":
print("The amount of training rods needed for next magic level is " + str((nextmaglvlpalwithdummy(num1))) + ".")
elif weaponchoice == "f" or weaponchoice == "F":
goToMainMenu = True
break
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.