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if else 流程判斷

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