[英]Limiting user input to a range in Python
在下面的代碼中,您將看到它要求“移位”值。 我的問題是我想將輸入限制為1到26。
For char in sentence:
if char in validLetters or char in space: #checks for
newString += char #useable characters
shift = input("Please enter your shift (1 - 26) : ")#choose a shift
resulta = []
for ch in newString:
x = ord(ch) #determines placement in ASCII code
x = x+shift #applies the shift from the Cipher
resulta.append(chr(x if 97 <= x <= 122 else 96+x%122) if ch != \
' ' else ch) # This line finds the character by its ASCII code
我怎么這么容易?
使用while
循環繼續詢問輸入,直到您收到您認為有效的內容:
shift = 0
while 1 > shift or 26 < shift:
try:
# Swap raw_input for input in Python 3.x
shift = int(raw_input("Please enter your shift (1 - 26) : "))
except ValueError:
# Remember, print is a function in 3.x
print "That wasn't an integer :("
你還希望在int()
調用周圍有一個try-except
塊,以防你得到一個ValueError
(如果他們輸入a
例子)。
請注意,如果使用Python 2.x,則需要使用raw_input()
而不是input()
。 后者將嘗試將輸入解釋為Python代碼 - 這可能非常糟糕。
另一個實現:
shift = 0
while not int(shift) in range(1,27):
shift = input("Please enter your shift (1 - 26) : ")#choose a shift
while True:
result = raw_input("Enter 1-26:")
if result.isdigit() and 1 <= int(result) <= 26:
break;
print "Error Invalid Input"
#result is now between 1 and 26 (inclusive)
嘗試這樣的事情
acceptable_values = list(range(1, 27))
if shift in acceptable_values:
#continue with program
else:
#return error and repeat input
可以放入while循環但是你應該限制用戶輸入,這樣它就不會變成無限
使用if條件:
if 1 <= int(shift) <= 26:
#code
else:
#wrong input
或者帶有if條件的while循環:
shift = input("Please enter your shift (1 - 26) : ")
while True:
if 1 <= int(shift) <= 26:
#code
#break or return at the end
shift = input("Try Again, Please enter your shift (1 - 26) : ")
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