Python正则表达式 - 匹配所有不在胡子括号中的东西

Question

I'm trying to create a regex in Python to match everything not in mustache brackets.

For example, this string:

This is an {{example}} for {{testing}}.

should produce this list of strings:

["This is an ", " for ", "."]

when used with re.findall .

For my mustache-matching regex, I am using this: {{(.*?)}} .

It seems like it should just be a simple matter of negating the above pattern, but I can't get it to work properly. I'm testing using: http://pythex.org

Thanks.

Answer 1

你需要拆分

re.split('{{.*?}}', s)

Answer 2

@Andrey solution is the most simple and clearly the way to go. You have another possible way with findall because when the pattern contains a capture group, it returns only the capture group content:

re.findall(r'(?:{{.*?}})*([^{]+)', s)

So, if you start the pattern with an optional non-capturing group for curly brackets parts followed by a capture group for other content, findall returns only the capture group content and all curly brackets parts are consumed.