我有一个循环,开头为for I,范围为(0,100)。正常情况下,它可以正常运行,但有时由于网络条件而出现故障。目前,我已经设置它,以便在失败时,它将继续在except子句中(继续到I的下一个数字)。

我是否可以将相同的数字重新分配给i,并再次运行失败的循环迭代?


当前回答

如果您正在寻找的是重新尝试x次失败的尝试,那么单个for else循环可能就是您想要的。考虑这个例子,尝试了3次:

attempts = 3

for attempt in range(1, attempts+1):
    try:
        if attempt < 4:
            raise TypeError(f"Error raised on attempt: {attempt}")
        else:
            print(f'Attempt {attempt} finally worked.')
    except (TypeError) as error:
        print(f'Attempt {attempt} hit the exception.')
        continue
    else:
        break
else:
    print(f'Exit after final attempt: {attempt}')

print(f'\nGo on to execute other code ...')

给出输出:

Attempt 1 hit the exception.
Attempt 2 hit the exception.
Attempt 3 hit the exception.
Exit after final attempt: 3

Go on to execute other code ...

再试一次它就成功了

attempts = 4

给出输出:

Attempt 1 hit the exception.
Attempt 2 hit the exception.
Attempt 3 hit the exception.
Attempt 4 finally worked.

Go on to execute other code ...

其他回答

使用这个装饰器,您可以轻松地控制错误

class catch:
    def __init__(self, max=1, callback=None):
        self.max = max 
        self.callback = callback 
    
    def set_max(self, max):
        self.max = max
    
    def handler(self, *args, **kwargs):
        self.index = 0
        while self.index < self.max: 
            self.index += 1
            try:
                self.func(self, *args, **kwargs)
        
            except Exception as error:
                if callable(self.callback):
                    self.callback(self, error, args, kwargs)
                
    def __call__(self, func):
        self.func = func
        return self.handler

import time
def callback(cls, error, args, kwargs):
    print('func args', args, 'func kwargs', kwargs)
    print('error', repr(error), 'trying', cls.index)
    if cls.index == 2:
        cls.set_max(4)
    
    else:
        time.sleep(1)
    
    
@catch(max=2, callback=callback)  
def test(cls, ok, **kwargs):
    raise ValueError('ok')

test(1, message='hello')

在Python装饰器库中也有类似的东西。

请记住,它不测试异常,而是测试返回值。它会重新尝试,直到被修饰的函数返回True。

稍微修改一下版本就可以了。

不使用那些丑陋的while循环的更“功能性”的方法:

def tryAgain(retries=0):
    if retries > 10: return
    try:
        # Do stuff
    except:
        retries+=1
        tryAgain(retries)

tryAgain()

如果您正在寻找的是重新尝试x次失败的尝试,那么单个for else循环可能就是您想要的。考虑这个例子,尝试了3次:

attempts = 3

for attempt in range(1, attempts+1):
    try:
        if attempt < 4:
            raise TypeError(f"Error raised on attempt: {attempt}")
        else:
            print(f'Attempt {attempt} finally worked.')
    except (TypeError) as error:
        print(f'Attempt {attempt} hit the exception.')
        continue
    else:
        break
else:
    print(f'Exit after final attempt: {attempt}')

print(f'\nGo on to execute other code ...')

给出输出:

Attempt 1 hit the exception.
Attempt 2 hit the exception.
Attempt 3 hit the exception.
Exit after final attempt: 3

Go on to execute other code ...

再试一次它就成功了

attempts = 4

给出输出:

Attempt 1 hit the exception.
Attempt 2 hit the exception.
Attempt 3 hit the exception.
Attempt 4 finally worked.

Go on to execute other code ...

我喜欢使用bool值,如下所示:

success = False
num_try = 0
while success is False:
    if num_try >= 10: # or any number
        # handle error how  you please
    try:
        # code
        success = True
    except Exception as e:
        # record or do something with exception if needed
        num_try += 1