如果语句仅打印其他

Question

def quantity():
    global amount 
    amount = input('How many tickets are you looking for? ')
    if amount == range(1,7):
        print('You have selected {} tickets '.format(amount))
    elif amount == (0):
        print('You have selected 0 tickets')
    else:
        print('Please choose an amount between 0-6')

I'm trying to create ticket cost calculator to jog my memory but I was stumped on the IF statement for the quantity section because it only prints the ELSE even when IF and ELIF are true.

Answer 1

You want

if amount in range(1,7):

rather than what you currently have,

if amount == range(1,7):

It's a bit more complicated in reality (because it returns a generator rather than a list), but you can conceptualize range(1,7) as a function that returns a list of numbers in that range. eg

range(1,7) ~~ [1, 2, 3, 4, 5, 6]

If your amount is an integer, you want to see if it's in that range, not if it is that range - after all, an integer cannot be a list at the same time.

---

[edit]: As one of the other answers pointed out, you might also want to cast the result of your input() to an int - as input() generally returns a string.

Answer 2

Try this code use in for ==

def quantity():
    global amount 
    amount = input('How many tickets are you looking for? ')
    if amount in range(1,7):
        print('You have selected {} tickets '.format(amount))
    elif amount == (0):
        print('You have selected 0 tickets')
    else:
        print('Please choose an amount between 0-6')

Answer 3

You have to convert the input() from a string into an integer, otherwise you will always run the else statement. You can do that with int() , but will likely have to do some error handling in case the user enters something that cannot be converted to an integer.

Answer 4

I have made some changes to your code. I believe this is what you are looking for.

def quantity():
    global amount
    amount = int(input('How many tickets are you looking for? '))
    if amount in range(1,7):
        print('You have selected {} tickets '.format(amount))
    elif amount ==0 :
        print('You have selected 0 tickets')
    else:
        print('Please choose an amount between 0-6')

Answer 5

Instead of using if amount == range(1,7): use if amount in range(1,7):

This is the modified code that you should use:

def quantity():
    global amount 
    amount = input('How many tickets are you looking for? ')
    if amount in range(1,7):
        print('You have selected {} tickets '.format(amount))
    elif amount == (0):
        print('You have selected 0 tickets')
    else:
        print('Please choose an amount between 0-6')

Answer 6

1.The range function in python3 returns an iteration type.

2.The input function in python3 returns a string and should be converted to an integer. Try this:

amount = int(input('How many tickets are you looking for? '))
if amount in range(1,7):
    print('You have selected {} tickets '.format(amount))
elif amount == (0):
    print('You have selected 0 tickets')
else:
    print('Please choose an amount between 0-6')

Answer 7

input返回一个字符串，所以我想你应该做amount = int(input('...')) 。