[英]Python - Input Validation
我正在尋找創建代碼,該代碼要求用戶在繼續之前輸入大於 2 的整數。 我正在使用 python 3.3。 這是我到目前為止所擁有的:
def is_integer(x):
try:
int(x)
return False
except ValueError:
print('Please enter an integer above 2')
return True
maximum_number_input = input("Maximum Number: ")
while is_integer(maximum_number_input):
maximum_number_input = input("Maximum Number: ")
print('You have successfully entered a valid number')
我不確定的是如何最好地設置整數必須大於 2 的條件。我剛剛開始學習 python,但想養成良好的習慣。
這應該可以完成這項工作:
def valid_user_input(x):
try:
return int(x) > 2
except ValueError:
return False
maximum_number_input = input("Maximum Number: ")
while valid_user_input(maximum_number_input):
maximum_number_input = input("Maximum Number: ")
print("You have successfully entered a valid number")
或者更短:
def valid_user_input():
try:
return int(input("Maximum Number: ")) > 2
except ValueError:
return False
while valid_user_input():
print('You have successfully entered a valid number')
def take_user_in():
try:
return int(raw_input("Enter a value greater than 2 -> ")) # Taking user input and converting to string
except ValueError as e: # Catching the exception, that possibly, a inconvertible string could be given
print "Please enter a number as" + str(e) + " as a number"
return None
if __name__ == '__main__': # Somethign akin to having a main function in Python
# Structure like a do-whole loop
# func()
# while()
# func()
var = take_user_in() # Taking user data
while not isinstance(var, int) or var < 2: # Making sure that data is an int and more than 2
var = take_user_in() # Taking user input again for invalid input
print "Thank you" # Success
我的看法:
from itertools import dropwhile
from numbers import Integral
from functools import partial
from ast import literal_eval
def value_if_type(obj, of_type=(Integral,)):
try:
value = literal_eval(obj)
if isinstance(value, of_type):
return value
except ValueError:
return None
inputs = map(partial(value_if_type), iter(lambda: input('Input int > 2'), object()))
gt2 = next(dropwhile(lambda L: L <= 2, inputs))
def check_value(some_value):
try:
y = int(some_value)
except ValueError:
return False
return y > 2
這驗證輸入是整數,但確實拒絕看起來像整數的值(如3.0
):
def is_valid(x):
return isinstance(x,int) and x > 2
x = 0
while not is_valid(x):
# In Python 2.x, use raw_input() instead of input()
x = input("Please enter an integer greater than 2: ")
try:
x = int(x)
except ValueError:
continue
希望這有幫助
import str
def validate(s):
return str.isdigit(s) and int(s) > 2
使用其他答案中顯示的內置int()
問題是它會將浮點數和布爾值轉換為整數,因此它並不是真正檢查您的參數是否為整數。
單獨使用內置的isinstance(value, int)
方法很誘人,但不幸的是,如果傳遞一個布爾值,它將返回 True。 所以如果你想要嚴格的類型檢查,這是我簡短而甜蜜的 Python 3.7 解決方案:
def is_integer(value):
if isinstance(value, bool):
return False
else:
return isinstance(value, int)
結果:
is_integer(True) --> False
is_integer(False) --> False
is_integer(0.0) --> False
is_integer(0) --> True
is_integer((12)) --> True
is_integer((12,)) --> False
is_integer([0]) --> False
等等...
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