Question

在我的Python项目中，下面有很多代码如下：

# the codes I with to simplify
for _ in range(3):
    try:
        # do something
        break
    except:
        print "Exception. Retrying..."
        continue
 else:
    print "ERROR!"

我知道我可以使用contextmanager来简化“try ... except ...”部分：

# define the exception handler function first
@contextmanager
def ignore_exception():
    try:
        yield
    except:
        print "Exception. Retrying..."
        continue

# then use "with" to simplify the codes
with ignore_exception:
    # do something

但是，我不能以同样的方式添加“for”循环，即

# define the exception handler function with for loop first
@contextmanager
def ignore_exception():
    try:
        for _ in range(3):
            yield
            break
        else:
            print "ERROR!"
    except:
        print "Exception. Retrying..."
        continue

# then use "with" to simplify the codes
with ignore_exception:
    # do something

将会引发异常：

/usr/lib/python2.7/contextlib.pyc in __exit__(self, type, value, traceback)
     26                 return
     27             else:
---> 28                 raise RuntimeError("generator didn't stop")
     29         else:
     30             if value is None:

RuntimeError: generator didn't stop

有没有办法解决这样的错误，并将我的代码简化为：

# ideal codes
with xxxxxx: # don't have to use "with" statement
    # do something

Answer 1

只需编写一个装饰器，它可以处理循环和异常忽略部分，就像这样

def trys(maximum_tries=3):
    def decorator(func):
        def inner(*args, **kwargs):
            for _ in range(1, maximum_tries + 1):
                try:
                    return func(*args, **kwargs)
                except ArithmeticError, e:
                    print("Error : [{}], Retrying Attempt {}...".format(e, _))
                else:
                    break
        return inner
    return decorator

现在，您可以像这样调用它

outer_a = 0


@trys(maximum_tries=4)
def adder(a):
    global outer_a
    outer_a += 1
    if outer_a < 4:
        raise ArithmeticError("Cannot add with {}".format(outer_a))
    else:
        return a + outer_a

print(adder(0))

它一直在尝试将当前值添加到outer_a，并且只有当它大于或等于4时才能添加它。因此，它会一直重试，直到最大尝试次数用完为止。它捕获抛出的异常，将其打印到stdout并再次尝试。所以，输出将是

Error : [Cannot add with 1], Retrying Attempt 1...
Error : [Cannot add with 2], Retrying Attempt 2...
Error : [Cannot add with 3], Retrying Attempt 3...
4

注意：有一个名为retrying的开源库，可以更好，更灵活地执行相同的任务。如果可能，您可能需要检查并使用它，而不是滚动自己的装饰器。不要重复自己： - ）

是否可以重试特定的代码而不管异常？

1 个答案: