Question

我有一个带有文字的文件，其中一些重要的项目标有开始和结束括号，例如：

Once upon a {time}, there lived a rabbit.
The {rabbit lived} in a small house.
One day, the {rabbit visited }the {mountains}.
In the mountains, he {found} a tree with 10{,000} branches.

我需要将{x}表单中的所有项目替换为{x::x}，例如：

Once upon a {time::time}, there lived a rabbit.
The {rabbit lived::rabbit lived} in a small house.
One day, the {rabbit visited ::rabbit visited }the {mountains::mountains}.
In the mountains, he {found::found} a tree with 10{,000::,000} branches.

每个开头{在同一行都有匹配的}。
大括号永远不会分开。
大括号永远不会嵌套。
{和}之间可能会出现任何类型的符号。

我尝试了sed的几种方法，但没有任何效果，例如：

sed 's/{(.*)}/{&::&}/g' file.txt

如何更换大括号中的所有项目，例如{some word}模式为{some word::some word}？

Answer 1

这是修复

sed 's/{\([^}]*\)}/{\1::\1}/g' file

Once upon a {time::time}, there lived a rabbit.
The {rabbit lived::rabbit lived} in a small house.
One day, the {rabbit visited ::rabbit visited }the {mountains::mountains}.
In the mountains, he {found::found} a tree with 10{,000::,000} branches.

解释

[^}]*匹配非}字符
$...$将捕获在parens中指定的字符，\ 1将用于引用第一个匹配，这是正则表达式的一部分。

Answer 2

如果您可以使用perl：

，则会更容易

$ perl -ple 's/{(.*?)}/{$1::$1}/g' file
Once upon a {time::time}, there lived a rabbit.
The {rabbit lived::rabbit lived} in a small house.
One day, the {rabbit visited ::rabbit visited }the {mountains::mountains}.
In the mountains, he {found::found} a tree with 10{,000::,000} branches.

它匹配大括号{...} 非贪婪中的所有内容，然后将其替换为所需的字符串{$1::$1}。

Answer 3

你应该使用

sed 's/\([^{]*{\)\([^}]*\)\(}.*\)/\1\2::\2\3/'

未经测试

Answer 4

awk变体：

$ awk 'BEGIN{ORS=""} NR%2==0{$0="{"$0"::"$0"}"} 1' RS='[{}]' file.txt

Once upon a {time::time}, there lived a rabbit.
The {rabbit lived::rabbit lived} in a small house.
One day, the {rabbit visited ::rabbit visited }the {mountains::mountains}.
In the mountains, he {found::found} a tree with 10{,000::,000} branches.

如何在BASH中替换匹配`{+（anything）+}`与`{+（anything）+ :: +（anything）+}`的模式？

4 个答案:

解释