如何解析前导空格的行

I am very new to Grep and regular expressions. From below text output, how can I get list groups. I just want "company, it and "fun guys"". This list of groups varies.

我是Grep和正则表达式的新手。从文本输出下方，我如何获得列表组。我只想要“公司，它和”有趣的家伙“”。这个组列表各不相同。

So basically, I am looking for all the lines after "Groups:" and before "Licenses:". Also do not want anything between the the brackets (for example, i want to remove <company@domain.com>)

基本上，我正在寻找“Groups：”之后和“Licenses：”之前的所有行。也不要在括号之间有任何东西（例如，我想删除） @domain.com>

1 个解决方案

#1

This command will do it,

这个命令会这样做，

sed -n '/Groups:/,/Licenses/p' sample |sed '1d'|sed '$d' |cut -f1 -d'<'

Here is explanation,

这是解释，

sed -n '/Groups:/,/Licenses/p' sample

sed -n'/ Groups：/，/ Licenses / p'样本

Extracting text from the file sample in between two patterns as mentioned in regex above.

如上面的正则表达式中所提到的，在两个模式之间从文件样本中提取文本。

sed '1d' Deletes the line containing "Groups:" keyword.

sed'1d'删除包含“Groups：”关键字的行。

sed '$d' Deletes the line containing "Licenses" keyword.

sed'$ d'删除包含“Licenses”关键字的行。

cut -f1 -d'<' Extracting the OPs mentioned patterns finally.

cut -f1 -d'<'最后提取OP提到的模式。

Output:

输出：

$ sed -n '/Groups:/,/Licenses/p' sample |sed '1d'|sed '$d' |cut -f1 -d'<'
        company 
        it 
        fun guys

#1