python - 什么是识别 gmail 中的 "original message"前缀的好正则表达式？

Question

示例签名可能是:

On Tue, Mar 20, 2012 at 2:38 PM, Johnny Walker <johnny.talker@gmail.com> wrote:

然后是引用的回复。我确实有一种离散的感觉，这是特定于语言环境的，但这让我成为一个悲伤的程序员。

我问这个的原因是因为 roundup通过 gmail 回复问题时没有正确删除这些。我认为 origmsg_re是我需要与 keep_quoted_text = no 一起设置的 config.ini 变量解决这个问题。

现在是默认值 origmsg_re = ^[>|\s]*-----\s?Original Message\s?-----$

编辑:现在我正在使用 origmsg_re = ^On[^<]+<.+@.+>[ \n]wrote:[\n]它适用于一些断行太长的 gmail 客户端。

Answer 1

以下正则表达式将以非常安全的方式匹配 gmails 前缀。它确保有 3 个逗号和升文本 On ... wrote

On([^,]+,){3}.*?wrote:

如果正则表达式应该以不区分大小写的方式匹配，那么不要忘记添加修饰符。

if re.search("On([^,]+,){3}.*?wrote:", subject, re.IGNORECASE):
    # Successful match
else:
    # Match attempt failed

亲切的问候，巴克利

Match the characters “On” literally «On»
Match the regular expression below and capture its match into backreference number 1 «([^,]+,){3}»
   Exactly 3 times «{3}»
   Note: You repeated the capturing group itself.  The group will capture only the last iteration.  Put a capturing group around the repeated group to capture all iterations. «{3}»
   Match any character that is NOT a “,” «[^,]+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   Match the character “,” literally «,»
Match any single character that is not a line break character «.*?»
   Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the characters “wrote:” literally «wrote:»

Created with RegexBuddy