如何仅从以x开头的行中剪切字符？

Question

我正在尝试使用cut -c-2-18- filename剪切第2和第18个角色之间的所有内容。这会在每一行上剪切指定的字符，但我只想从以>开头的行剪切它。

Answer 1

像

这样的东西

grep ^\> file | cut -c 2-18

应该这样做吗？

[92][11:46:21] vlazarenko@alluminium (~/tests) > grep ^\> file | cut -c 2-18
 asdf
 bcda
 ofpjg
[93][11:46:25] vlazarenko@alluminium (~/tests) > cat file
> asdf
> bcda
zxcv
lasdf
> ofpjg

Answer 2

您可以单独使用GNU grep执行此操作：

$ cat file
>grep from 2 to 18 character line 1
>grep from 2 to 18 character line 2
grep from 2 to 18 character line 3
grep from 2 to 18 character line 4

$ grep -Po '(?<=^>).{17}' file
grep from 2 to 18
grep from 2 to 18

这使用positive lookbehind仅匹配以>开头的行，.{17}匹配接下来的17个字符。

或awk：

$ awk '/^>/{print substr($0,2,18)}' file
grep from 2 to 18 
grep from 2 to 18 

Answer 3

awk '/^\>/{print substr($0,2,17)}' your_file