Linux 我想使用“awk”或 sed 打印文件中所有以“comm=”开头的行
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I want to use "awk" or sed to print all the lines that start with "comm=" in a file
提问by wael
I want to use "awk" or "sed" to print all the lines that start with comm=from the file filex, Note that each line contains "comm=somthing"
我想使用“awk”或“sed”来打印comm=从文件开始的所有行filex，注意每一行都包含“comm=somthing”
for example : comm=rm , comm=ll, comm=ls ....
How can i achieve that ?
我怎样才能做到这一点？
回答by jaypal singh
For lines that start with comm=
对于以开头的行 comm=
sed -n '/^comm=/p' filex
awk '/^comm=/' filex
If comm=is anywhere in the line then
如果comm=在该行中的任何位置，则
sed -n '/comm=/p' filex
awk '/comm=/' filex
回答by Cédric Julien
回答by Michael J. Barber
Here's an approach using grep:
这是使用 grep 的一种方法：
grep -o '\<comm=[[:alnum:]]*\>'
This treats a word as consisting of alphanumeric characters; extend the character class as needed.
这将一个单词视为由字母数字字符组成；根据需要扩展字符类。
回答by nbari
If grepis ok to use, you could give a try to:
如果grep可以使用，您可以尝试：
grep -E "^comm=" file
