Python 正则表达式匹配:## ## [英] Python Regular Expression Matching: ## ##

问题描述

我正在逐行搜索文件中是否出现##random_string##.除了多个 #...

I'm searching a file line by line for the occurrence of ##random_string##. It works except for the case of multiple #...

pattern='##(.*?)##'
prog=re.compile(pattern)

string='lala ###hey## there'
result=prog.search(string)

print re.sub(result.group(1), 'FOUND', string)

期望的输出:

"lala #FOUND there"

相反，我得到以下内容，因为它抓住了整个###hey##:

Instead I get the following because its grabbing the whole ###hey##:

"lala FOUND there"

那么我如何忽略开头或结尾的任意数量的#，而只捕获##string##".

So how would I ignore any number of # at the beginning or end, and only capture "##string##".

推荐答案

要在任一端匹配至少两个哈希值:

To match at least two hashes at either end:

pattern='##+(.*?)##+'

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