匹配前后的 Grep 字符? [英] Grep characters before and after match?

问题描述

使用这个:

grep -A1 -B1 "test_pattern" file

将在文件中匹配模式前后产生一行.有没有办法不显示行而是显示指定数量的字符?

will produce one line before and after the matched pattern in the file. Is there a way to display not lines but a specified number of characters?

我的文件中的行非常大，所以我对打印整行不感兴趣，而只是在上下文中观察匹配.有关如何执行此操作的任何建议?

The lines in my file are pretty big so I am not interested in printing the entire line but rather only observe the match in context. Any suggestions on how to do this?

推荐答案

前 3 个字符后 4 个字符

3 characters before and 4 characters after

$> echo "some123_string_and_another" | grep -o -P '.{0,3}string.{0,4}'
23_string_and

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