bash 如何删除方括号和里面的任何文字？
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How to remove square brackets and any text inside?
提问by Village
I have a document containing some text inside square brackets, e.g.:
我有一个包含方括号内的一些文本的文档，例如：
The fish [ate] the bird.
[This is some] text.
Here is a number [1001] and another [1201].
I need to delete all of the information contained inside the square brack and the brakets, e.g.:
我需要删除方括号和刹车中包含的所有信息，例如：
The fish the bird.
text.
Here is a number and another .
- I tried
sed -r 's/\[[+]\]//g' file.txt, but this did not work.
- 我试过了
sed -r 's/\[[+]\]//g' file.txt，但这没有用。
How can I delete anything in the pattern [<anything>]?
如何删除模式中的任何内容[<anything>]？
回答by Kent
try this sed line:
试试这个 sed 行：
sed 's/\[[^]]*\]//g'
example:
例子：
kent$ echo "The fish [ate] the bird.
[This is some] text.
Here is a number [1001] and another [1201]."|sed 's/\[[^]]*\]//g'
The fish the bird.
text.
Here is a number and another .
explanation:
解释：
the regex is actually straightforward:
正则表达式实际上很简单：
\[ #match [
[^]]* #match any non "]" chars
\] #match ]
so it is
原来如此
match string, starting with [then all chars but ]and ending with ]
匹配字符串，以[所有字符开头，但]以]
回答by Amitanshu Gupta
sed 's/([^])*)/replacement text/g' in case of Parenthesis ()
sed 's/([^])*)/在括号 () 的情况下替换文本/g'
