Bash在指定字符之前获取n个字符

Question

This is the string:

Volume: 0:  120% 1:  128%

I want to get 3 characters before "%" character. So output must be like 120 128 for this example. How can i do that?

Answer 1

假设您实际上想要空格和％之间的所有内容（此示例与此相同），则可以执行以下操作：

awk '{print $NF}' RS=% ORS=' '

Answer 2

You can use this grep :

$ grep -Po '...(?=%)' file
120
128

Which is the same as:

grep -Po '.{3}(?=%)' file

or this if you want to make sure you get digits:

grep -Po '[0-9]{3}(?=%)' file

Explanation

The idea underlying is to print three characters before % is found. (?=%) indicates this and is called a "look-ahead".

With -Po we indicate grep that it can use Perl regular expressions ( -P ) and to just print the matches ( -o ).