Question

由于某种原因，这段代码不起作用？我已经尝试了返回1并且中断了但是由于某种原因它给了我一个错误，我希望代码返回到开头，如果数字太长但没有理想的怎么做。

# Find the cube root of a perfect cube

x = int(input('Enter an integer: '))
if x > 5000:
     break:
     print('too long')
 ### this code is broken ^^^^^



ans = 0
while ans**3 < x:
    ans = ans + 1
if ans**3 != x:
    print(str(x) + ' is not a perfect cube')
else:
    print('Cube root of ' + str(x) + ' is ' + str(ans))


IndentationError: unexpected indent
>>> runfile('/home/dux/pyyyyy.py', wdir=r'/home/dux')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/dist-packages/spyderlib/widgets/externalshell/sitecustomize.py", line 540, in runfile
    execfile(filename, namespace)
  File "/home/dux/pyyyyy.py", line 7
    print('wrong'):
                  ^
SyntaxError: invalid syntax
>>> runfile('/home/dux/pyyyyy.py', wdir=r'/home/dux')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/dist-packages/spyderlib/widgets/externalshell/sitecustomize.py", line 540, in runfile
    execfile(filename, namespace)
  File "/home/dux/pyyyyy.py", line 7
    break:
         ^
SyntaxError: invalid syntax
>>> runfile('/home/dux/pyyyyy.py', wdir=r'/home/dux')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/dist-packages/spyderlib/widgets/externalshell/sitecustomize.py", line 540, in runfile
    execfile(filename, namespace)
  File "/home/dux/pyyyyy.py", line 8
    print('wrong')
    ^
IndentationError: unexpected indent
>>> runfile('/home/dux/pyyyyy.py', wdir=r'/home/dux')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/dist-packages/spyderlib/widgets/externalshell/sitecustomize.py", line 540, in runfile
    execfile(filename, namespace)
  File "/home/dux/pyyyyy.py", line 7
    break:
         ^
SyntaxError: invalid syntax
>>>

Answer 1

我认为你想要的是检查用户是否输入了有效的号码。尝试：

while True:
    x = int(input('Enter an integer: '))
    if x > 5000:
       print('too long')
    else:
       break

Answer 2

Break会停止一个循环。由于您的代码不在循环中，我无法理解您为什么要使用它。此外，休息后不需要冒号。只是让你知道我会给出一个例子。

count = 0
while True:
    print('Hello') #Prints Hello
    if count == 20:  #Checks if count is equal to 20
        break #If it is: break the loop
    count += 1 #Add 1 to count

当然，只需执行while count < 20:就可以轻松完成此操作，但我要说明一点。

编辑：此外，查看您收到的其他一些错误，不在print之后也需要冒号。

Python代码无法正常工作

2 个答案: