Regex 正则表达式:查找任何用方括号括起来的字符串
我正在努力找到一个正则表达式,它可以找到任何用方括号括起来的字符串。 我有一份从维基百科上刮下来的大文件,里面有很多这样的段落:Regex 正则表达式:查找任何用方括号括起来的字符串,regex,Regex,我正在努力找到一个正则表达式,它可以找到任何用方括号括起来的字符串。 我有一份从维基百科上刮下来的大文件,里面有很多这样的段落: "Theology translates into English from the Greek theologia (θεολογία) which derived from Τheos (Θεός), meaning "God," and -logia (-λογία),**[12]** meaning "utterances, sayings, or oracl
"Theology translates into English from the Greek theologia (θεολογία) which derived from Τheos (Θεός), meaning "God," and -logia (-λογία),**[12]** meaning "utterances, sayings, or oracles" (a word related to logos **[λόγος][Citation needed]**".
预期的结果将是:
"Theology translates into English from the Greek theologia (θεολογία) which derived from Τheos (Θεός), meaning "God," and -logia (-λογία), meaning "utterances, sayings, or oracles" (a word related to logos".
如有任何建议,将不胜感激。
谢谢 这些**也在文档中吗?你只提到方括号
如果是的话,正则表达式就是
(\*\*\[.*?\]\*\*)
如果它真的只是方括号,那么这与您所追求的匹配:
(\[.*?\])
您没有提到语言,但在Python中,这将由
import re
my_re = r'(\*\*\[.*?\]\*\*)'
my_string = '"Theology translates into English from the Greek theologia (θεολογία) which derived from Τheos (Θεός), meaning "God," and -logia (-λογία),**[12]** meaning "utterances, sayings, or oracles" (a word related to logos **[λόγος][Citation needed]**".'
my_corrected_string = re.sub(my_re, '', my_string)
print(my_corrected_string)
最好的建议是让你的问题更加精确,向我们展示你迄今为止所做的尝试,并阅读一些编程语言和正则表达式方面的知识。谢谢!这正是我需要的。抱歉,没有提供足够的信息。我在JS中使用了这个,它工作得非常好。