Python:即使不满足条件，如果仍在运行

Question

import imaplib, re
import os

while(True):
    conn = imaplib.IMAP4_SSL("imap.gmail.com", 993)
    conn.login("xxx", "xxxx")
    unreadCount = re.search("UNSEEN (\d+)", conn.status("INBOX", "(UNSEEN)")[1][0]).group(1)
    print unreadCount

    if unreadCount > 10:
      os.system('ls')

即使 unreadCount 小于 10，它也会运行命令“ls”。为什么？

Answer 1

您可能希望将该值强制转换为整数，如下所示:

unreadCount = int (re.search (blah, blah, blah).group (1))

对 re.search 的调用返回一个字符串，如果您查看以下记录:

>>> x = "7"
>>> if x > 10:
...     print "yes"
... 
yes

>>> if int(x) > 10:
...     print "yes"
... 

>>> x = 7
>>> if x > 10:
...     print "yes"
... 
>>> 

你会明白为什么这不是一个好主意。

您看到这种(您可能称之为奇怪的)行为的原因可以从 manual 中找到。在 5.3 的底部:

CPython implementation detail: Objects of different types except numbers are ordered by their type names; objects of the same types that don’t support proper comparison are ordered by their address.

因为"7"的类型是str，10的类型是int，所以很简单比较类型名称("str" 在字母顺序中总是大于 "int")，导致一些有趣的事情，例如:

>>> "1" > 99999999999999999999999
True
>>> "1" == 1
False

该实现细节至少在 2.7.2 之前仍然有效。它可能在 Python 3000 流中发生了变化(该子句肯定已从相关文档部分中删除)，但 documentation there仍然说:

Most other objects of built-in types compare unequal unless they are the same object; the choice whether one object is considered smaller or larger than another one is made arbitrarily but consistently within one execution of a program.

所以这可能不是您应该依赖的东西。