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?
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
我最近用我的python解决了这个问题,我很高兴与stackoverflow的访问者分享,如果需要请给予反馈。
print("\nmonthly salary per day and year converter".title())
print('==' * 25)
def income_counter(day, salary, month):
global result2, result, is_ready, result3
result = salary / month
result2 = result * day
result3 = salary * 12
is_ready = True
return result, result2, result3, is_ready
i = 0
for i in range(5):
try:
month = int(input("\ntotal days of the current month: "))
salary = int(input("total salary per month: "))
day = int(input("Total Days to calculate> "))
income_counter(day=day, salary=salary, month=month)
if is_ready:
print(f'Your Salary per one day is: {round(result)}')
print(f'your income in {day} days will be: {round(result2)}')
print(f'your total income in one year will be: {round(result3)}')
break
else:
continue
except ZeroDivisionError:
is_ready = False
i += 1
print("a month does'nt have 0 days, please try again")
print(f'total chances left: {5 - i}')
except ValueError:
is_ready = False
i += 1
print("Invalid value, please type a number")
print(f'total chances left: {5 - i}')
其他回答
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()
在for循环中执行while True,将try代码放入其中,只有当代码成功时才退出while循环。
for i in range(0,100):
while True:
try:
# do stuff
except SomeSpecificException:
continue
break
尝试次数= 3 而尝试: 试一试: ... ... <状态好了> 打破 除了: 尝试- = 1 Else: #已执行的唯一break未被引发 < >失败状态
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
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')