python - 如何在Python中进行故障转移？

Question

我想问一下如何在 for 中进行故障转移，如下所示:

 again = 1
 start = 1
 try:
   def countthis():
          for i in range (start,200):
              again = i
              print i
 except:
    print "Failure occured, I will try again" 
    start = again
    countthis.run()

我希望如果 for 函数在 i 的 try 中失败，它会从最新的 i (而不是从 1)重新启动它

Answer 1

您可以使用这样的迭代器函数:

def countthis(start=0, end=100):
    for i in range(start, end):
        print i
        if i == 5:
            raise Exception('5 failed')
        yield i

然后在出错时使用下一个数字恢复计数器，跳过失败的:

ret = 0
end = 100
while ret < end - 1:
    try:
        for i in countthis(start=ret, end=end):
            ret = i
    except Exception, ex:
        print ex
        # when 5 reached, an exception will be raised, so here we restart at '6'
        ret = ret + 2

这最终将打印:

0
1
2
3
4
5
5 failed
6
7
......
99