英文:
Output strings via grep excluding matches within them
问题
有没有一种方法，可以输出使用grep找到的字符串，但排除匹配项？
假设我有一个带有字符串的文件：
>A.123 TextTextTextText....
>B.123 OtherTextTextText....
我想要得到以下结果：
>TextTextTextText...
>OtherTextTextText...
我使用以下命令来搜索这样的字符串：
grep -P '(?<=>)[A-Z0-9\.]*\s' File.txt
它能正确找到匹配项，但我无法弄清如何将所有行输出到单独的文件中，同时排除其中的匹配项。
我知道有一个-o标志，可以用于仅获取匹配项。是否可能"反转"它？
英文:
Is there a way, to output strings found with grep, excluding matches?
Let's say I have a file with strings:
>A.123 TextTextTextText....
>B.123 OtherTextTextText....
I would like to get the following:
>TextTextTextText...
>OtherTextTextText...
I use the following command to search for such strings:
grep -P '(?<=>)[A-Z0-9\.]*\s' File.txt
It finds matches correctly, but I can't figure out how to output all lines in a separate file, excluding matches within them.
I know that there is a -o flag that can be used to get only matches. Is it possible to "invert" it?
答案1
得分: 2
使用 awk：
$ awk '/^>/ {print ">" $2}' file
>TextTextTextText...
>OtherTextTextText...
英文:
Using awk:
$ awk '/^>/ {print ">" $2}' file
>TextTextTextText...
>OtherTextTextText...
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