bash 使用sed删除括号之间的字符串
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Using sed to delete a string between parentheses
提问by Teddy T
I'm having trouble figuring how to delete a string in between parentheses only when it is in parentheses. For example, I want to delete the string "(Laughter)", but only when it's in parenthesis and not just make it a case sensitive deletion since a sentence within that string starts with Laughter.
我无法确定如何仅在括号中删除括号之间的字符串。例如，我想删除字符串“(Laughter)”，但仅当它在括号中时，而不是仅仅将其设为区分大小写的删除，因为该字符串中的句子以 Laughter 开头。
回答by MightyPork
I'm not sure I understand you correctly, but the following will remove text between parentheses:
我不确定我是否理解正确，但以下内容将删除括号之间的文本：
sed "s/[(][^)]*[)]/()/g" myfile
(or as Llama pointed out in comments, just:)
（或者正如 Llama 在评论中指出的那样，只是:)
sed "s/([^)]*)/()/g" myfile
It matches a literal open paren [(]followed by any number of non-)characters [^)]*followed by a literal close paren [)].
它匹配文字打开括号，[(]后跟任意数量的非)字符，[^)]*然后是文字关闭括号[)]。
Example:
例子：
$ echo "Blah blah (potato) moo (cow is a pretty bird)(hello)" | sed "s/[(][^)]*[)]/()/g"
Blah blah () moo ()()
Use //instead of /()/there if you don't want the empty parens in the output.
如果您不想在输出中使用空括号，请使用//而不是/()/那里。
回答by buydadip
Here's an awksolution because, why not?
这是一个awk解决方案，因为，为什么不呢？
awk '{gsub("[(][^)]*[)]","")}1' file
OR
或者
awk -F '[(][^)]*[)]' '{for(i=1; i<=NF; i+=1)printf("%s",$i);printf("\n")}' file
Remember that in order to escape parentheses use must enclose them in []brackets.
请记住，为了转义括号，必须使用括号将它们括起来[]。
By using [^)]*, as @MightyPork did in his answer, which indicates any number of characters that are not parentheses, will reduce the greed of the match. In contrast, using .*instead, would result in a greedy match.
通过使用[^)]*，正如@MightyPork 在他的回答中所做的那样，表示任意数量的非括号字符，将减少匹配的贪婪。相反，使用.*相反，会导致贪婪匹配。
