if else 流程判斷
阿新 • • 發佈:2017-09-22
any else語句 count 循環 冒號 bre span onf ldb
來個例子:
1 username=input("username:") 2 password=input("password:") 3 username1=‘whgvjp‘ 4 password1=‘0235‘ 5 if username==username1 and password==password1: 6 print(‘welcome user {name}logging in...‘.format(name=username)) 7 else: 8 print("The wrong username or password")
這裏的話,需要註意在if語句和else語句後面要有冒號,接著會強制縮進,默認縮進單位是四個空格,如果不縮進會出錯。Python這種風格可以使代碼更美觀,並且也能使程序員養成良好的寫代碼的習慣。
猜年齡的例子,這個例子可以實現只能猜三次,如果三次都猜不對,會打印一個“你已經嘗試太多次..”。
count=1 age_of_oldboy=56 while count<4: guess_age=int(input("guess age:")) if guess_age==age_of_oldboy: print("yes, you got it.") break elif guess_age>age_of_oldboy: print("think smaller...") else:print("Think biger!") count+=1
else:
print("you have tried too mang times..")
也可以用for循環來實現:
count=0 age_of_oldboy=56 for count in range (3): guess_age=int(input("guess age:")) if guess_age==age_of_oldboy: print("yes, you got it.") break elif guess_age>age_of_oldboy:print("think smaller...") else: print("Think biger!") count+=1 else: print(‘you have tried too many times..‘)
也可以count=1,然後for count in range(1,4):意義不大,都行。
下面這個代碼跟上面的基本相同,但是可以實現,如果你在每猜三次後,不想繼續猜了,那就輸入’n‘退出這個猜年齡遊戲;如果還想繼續玩下去,可以輸入除了‘n‘以外的任何一個鍵,繼續遊戲。
count=1 age_of_oldboy=56 for count in range (1,4): guess_age=int(input("guess age:")) if guess_age==age_of_oldboy: print("yes, you got it.") break elif guess_age>age_of_oldboy: print("think smaller...") else: print("Think biger!") count+=1 if count==4: countine_confirm=input(‘do you want to keep trying‘) if countine_confirm!=‘n‘: count=1
if else 流程判斷