[英]Python: Error while making foolproof input method using for loop [duplicate]
digits = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '0']
def inp():
inp = input('Enter a value: ')
for i in inp:
if i not in digits:
print('Please enter a value from 2 to 100')
inp()
n = int(inp)
if not 2 <= n <= 100:
print('Please enter a value from 2 to 100')
inp()
else:
return n
number = inp()
我无法做到万无一失:用户可以输入一个空字符串""
或浮点数,它会给出一个错误。
例如:如果我输入10.5
然后输入10
那么它会显示错误,即 10.5 不能作为整数值。
如果有一个简单的函数,比如 VB6 中的val()
或 Python 的等效函数,我将不胜感激。
编辑:预期行为:我希望代码只采用整数值,2 到 100 作为输入,如果输入了不正确的值,那么它应该重复直到收到适当的值。
请帮忙。
我发现解决方案是这样的:
def inp_w():
while True:
try:
inp = round(float(input('Width of the maze (number of blocks): ')))
except:
print('Please enter a value from 2 to 100')
continue
else:
break
if not 2 <= inp <= 100:
print('Please enter a value from 2 to 100')
inp_w()
else:
return inp
a = inp_w()
Thanks to
"https://stackoverflow.com/users/16343464/mozway"
"https://stackoverflow.com/users/13000953/mahrkeenerh"
"https://stackoverflow.com/users/9635106/karl-wilhelm"
"https://stackoverflow.com/users/16601101/jeandemeusy"
你犯了三个错误:
这是一个工作代码:
valid_chars = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "."]
def input_val():
inp = input("Enter a value: ")
for i in inp:
if i not in valid_chars:
print("Please enter a value from 2 to 100")
return input_val()
n = float(inp)
if not 2 <= n <= 100:
print("Please enter a value from 2 to 100")
return input_val()
return n
number = input_val()
编辑:一个更好的答案,没有递归可能是这样的:
def is_float(string):
try:
float(string)
return True
except:
return False
def input_val(min_val: int, max_val: int):
while True:
inp = input("Enter a value: ")
if not is_float(inp):
print("Please enter a number")
else:
n = float(inp)
if not min_val <= n <= max_val:
print(f"Please enter a value from {min_val} to {max_val}")
else:
break
return n
number = input_val(2, 100)
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